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18.102 Introduction to Functional Analysis Spring 2009

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4 LECTURE NOTES FOR 18.102, SPRING 2009

Lecture 1. Tuesday, 3 Feb.

Linear spaces, metric spaces, normed spaces. Banach spaces. Examples – Euclidean spaces, continuous functions on a closed interval – C0([0, 1]) with supremum norm. The (Riemannian) L1 norm, outline that this is not complete on C0([0, 1]). Brief description of l2 – This is Hilbert space, but not explained.

What is it all for? Main aims:- To be able to carry out ‘standard’ constructions in (linear) functional analysis:

Abstract Hilbert space – one in each dimension Concrete Hilbert space – Many, such as L2([0, 1]) Example of a theorem:- The Dirichlet problem. Let V : [0, 1] −→ R be a real-valued function. We are interested in ‘oscillating modes’ on the interval; something like this arises in quantum mechanics for instance. Namely we want to know about functions u(x) – twice continuously differentiable on [0, 1] which satisfy the differential equation

d2u(1.1) −

dx2 (x) + V (x)u(x) = λu(x)

where λ is an ‘unknown’ constant – that is we want to know which λ’s can occur. Well, of course all λ’s can occur with u ≡ 0 but this is the ‘trivial solution’ which will always be there for such an equation. What other solutions are there? Well, there is an infinite sequence of λ’s for which there is a nontrivial solution of (1.1) λj ∈ R – they are all real no non-real complex λ’s can occur. For each of these there is at least one (and maybe more) ‘independent’ solution uj . We can say a lot more about everything here but one main aim of this course is to get at least to this point.

Now, in fact (1.1) is just the eigenvalue equation. What we are dealing with here is an ‘infinite matrix’. This is not obvious, and in fact is not a very good way of looking at things (there was such a matrix approach to quantum mechanics in the early days but it was replaced by the sort of ‘operator’ theory on Hilbert space that we will use here.) Still, we are in some sense dealing with infinite dimensional matrices. One of the crucial differences between infinite and finite dimensional settings is that topology is encountered. This is enshrined here in the notion of a normed linear space

Linear space:- Should I break out the axioms? It is a space V in which we can add elements and multiply by scalars with rules very similar to the basic examples of Rn or Cn . Note that for us the ‘scalars’ are either the real numbers of the complex numbers – usually the latter. Let’s be neutral and denote by K either R or C but of course consistently. Then our set V – the set of vectors with which we will deal, comes with two ‘laws’. These are maps

(1.2) + : V × V −→ V, : K × V −→ V. ·

which we denote not by +(v, w) and (s, v) but by v + w and sv. Then we ·impose the axioms of a vector space – look them up! These are commutative group axioms for + axioms for the action of K and the distributive law.

Our examples: The ‘trivial’ case of a finite dimensional vector space. The lp spaces. These are spaces of sequences – the first problem set is all about them. Thus l2 – which is a Hilbert space – consists of all the sequences

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a : N −→ C also denoted {aj }∞ where aj = a(j), such that j=1

∞

(1.3) |aj |2 < ∞. j=1

Seriously non-trivial examples such as C([0, 1]) the space of continuous functions on [0, 1] (say with complex values).

All these vector spaces carry norms, and that is what we want to talk about for most of the semester.

Definition 1. A norm on a vector space V is a function

(1.4) � · � : V −→ [0, ∞)

which satisfies the three properties: (1) �v� = 0 iff v = 0. (2) �tv� = |t|�v� for all t ∈ K and v ∈ V. (3) Triangle inequality. �v + w� ≤ �v� + �w� for all v, w ∈ V.

Then show that d(v, w) = �v − w� is a metric on V. It follows that all the notions from 18.100 come into play – open sets, balls, closed

sets, convergence of sequences, compact sets, connected sets, complete metric spaces. We will use them all!

Definition 2. A normed space which is complete with respect to the induced metric is called a Banach space.

I then discussed the supremum norm on C([0, 1]);

(1.5) = sup u(x) .�u�∞ x∈[0,1]

| |

I said, but did not prove, that it this gives a Banach space. I also disussed the L1 norm on C([0, 1]) : � 1

(1.6) �u�L1 = |u(x)|dx 0

and indicated why it is not complete. This is basically why we need to study the Lebesgue integral.

One might ask ‘is the space of Riemann integrable functions on [0, 1] complete with respect to the L1 norm’ – indeed someone did. The answer is that �u�L1 is not even a norm on the space of Riemann integable functions. Namely it is only a ‘seminorm’ – the second two conditions are fine but there are non-zero functions with integral zero. So, the question is not quite precise, which is why I did not talk about it. One the other hand one can fiddle with the space (take a quotient) so that one gets a norm and then it is not complete. So the morally correct answer is NO, it is not a Banach space and this is true for a better (or worse) reason than it simply not being a normed space in the first place! We will get to this.

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Problem set 1, Due 11AM Tuesday 10 Feb.

Full marks will be given to anyone who makes a good faith attempt to answer each question. The first four problems concern the ‘little L p’ spaces lp. Note that you have the choice of doing everything for p = 2 or for all 1 ≤ p < ∞.

Problem 1.1 Write out a proof (you can steal it from one of many places but at least write it out in your own hand) either for p = 2 or for each p with 1 ≤ p < ∞that

∞

lp = {a : N −→ C; |aj |p < ∞, aj = a(j)}j=1

is a normed space with the norm ⎛ ⎞ 1

∞ p

p�a�p = ⎝ |aj | ⎠ . j=1

This means writing out the proof that this is a linear space and that the three conditions required of a norm hold.

Problem 1.2 The ‘tricky’ part in Problem 1.1 is the triangle inequality. Suppose you knew – meaning I tell you – that for each N ⎛ ⎞ 1

N p

p⎝ |aj | ⎠ is a norm on CN

j=1

would that help? Problem 1.3 Prove directly that each lp as defined in Problem 1.1 – or just l2 – is complete,

i.e. it is a Banach space. At the risk of offending some, let me say that this means showing that each Cauchy sequence converges. The problem here is to find the limit of a given Cauchy sequence. Show that for each N the sequence in CN obtained by truncating each of the elements at point N is Cauchy with respect to the norm in Problem 1.2 on CN . Show that this is the same as being Cauchy in CN in the usual sense (if you are doing p = 2 it is already the usual sense) and hence, this cut-off sequence converges. Use this to find a putative limit of the Cauchy sequence and then check that it works.

Problem 1.4 Consider the ‘unit sphere’ in lp – where if you want you can set p = 2. This is

the set of vectors of length 1 :

S = {a ∈ lp; �a�p = 1}. (1) Show that S is closed. (2) Recall the sequential (so not the open covering definition) characterization

of compactness of a set in a metric space (e.g. by checking in Rudin). (3) Show that S is not compact by considering the sequence in lp with kth

element the sequence which is all zeros except for a 1 in the kth slot. Note that the main problem is not to get yourself confused about sequences of sequences!

Problem 1.5 Show that the norm on any normed space is continuous.





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Lecture 2. Thursday, 5 Feb.

Plan:- Linear maps between normed spaces are continuous iff they are bounded. The best bound gives a norm. If the second space is Banach the space of linear operators is Banach. Corollary – the dual space of a normed space is a Banach space. Examples: Integral operators on C0([0, 1]) with respect to supremum or L1

norms. Differentiation as an operator from C1([0, 1]) to C0([0, 1]) In practice:- I did not get to the part about differentiation. Reading: (1) Wilde:- Chapter 2 to 2.7 (2) Chen:- First part of Chapter 6 and of Chapter 7. (3) Ward:- Chapter 3, first 2 sections.





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Lecture 3. Tuesday, 10 Feb.

Recalled the proof from last time that the bounded operators from a normed space into a Banach space form a Banach space – mainly to suggest that it is not so hard to remember how such a proof goes. Then proved that a normed space is Banach iff every ‘absolutely summable’ series is convergent. Absolute summability means that the sum of the norms is finite. Then did most of the proof that every normed space can be completed to a Banach space using this notion of absolutely summable sequences. The last part – and a guide to how to attempt the part of the proof that is the first question on the next homework. The proof of the result about completeness from the early part of the leture is in

(1) Wilde:- Proposition 1.6 (2) Chen:- I didn’t find it. (3) Ward:- Lemma 2.1 (easy way only)

Here is a slightly abbreviated version of what I did in lecture today on the completion of a normed space. The very last part I asked you to finish as the first part of the second problem set, not due until February 24 due to the vagaries of the MIT calendar (but up later today). This problem may seem rather heavy sledding but if you can work through it all you will understand, before we get to it, the main sorts of arguments needed to prove most of the integrability results we will encounter later.

Let V be a normed space with norm � · �V . A completion of V is a Banach space B with the following properties:

(1) There is an injective (1-1)linear map I : V −→ B (2) The norms satisfy

(3.1) �I(v)�B = �v�V ∀ v ∈ V.

(3) The range I(V ) ⊂ B is dense in B.

Notice that if V is itself a Banach space then we can take B = V with I the identity map.

So, the main result is:

Theorem 1. Each normed space has a completion.

‘Proof’ (the last bit is left to you). First we introduce the rather large space ∞

(3.2) V� = {uk}∞k=1; uk ∈ V and �uk� < ∞ k=1

the elements of which I called the absolutely summable series in V. Now, I showed in the earlier result that each element of V� is a Cauchy sequence

N

– meaning the corresponding sequence of partial sums vN = uk is Cauchy if k=1

{uk} is absolutely summable. Now V� is a linear space, where we add sequences, and multiply by constants, by doing the operations on each component:

(3.3) t1{uk} + t2{u�k} = {t1uk + t2u�k}.

This always gives an absolutely summable series by the triangle inequality:

(3.4) �t1uk + t2u�k� ≤ |t1| �uk� + |t2| �u�k�.

k k k

�

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� �

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LECTURE NOTES FOR 18.102, SPRING 2009 9

Within �V

(3.5)

consider the linear subspace �

S = {uk}; �

k

�uk� < ∞,

� �

k

uk = 0

of those which converge to 0. As always for a linear subspace of a linear space we can form the quotient

(3.6) B = V /S

the elements of which are the ‘cosets’ of the form {uk} + S ⊂ V� V . where {uk} ∈ �We proceed to check the following properties of this B.

(1) A norm on B is defined by n

(3.7) �b�B = lim uk�, {uk} ∈ b. n→∞

� k=1

(2) The original space V is imbedded in B by

(3.8) V � v �−→ I(v) = {uk} + S, u1 = v, uk = 0 ∀ k > 1

and the norm satisfies (3.1). (3) I(V ) ⊂ B is dense. (4) B is a Banach space with the norm (3.7).

So, first that (3.7) is a norm. The limit on the right does exist since the limit of the norm of a Cauchy sequence always exists – namely the sequence of norms is itself Cauchy but now in R. Moreover, adding an element of S to {uk} does not change the norm of the sequence of partial sums, since the addtional term tends to zero in norm. Thus �b�B is well-defined for each element b ∈ B and �b�B = 0 means exactly that the sequence {uk} used to define it tends to 0 in norm, hence is in S hence b = 0. The other two properties of norm are reasonably clear, since if b, b� ∈ B are represented by {uk}, {u�k} in V� then tb and b + b� are reprented by {tuk} and {uk + u�k} and

n n

lim tuk� = t lim n→∞

� k=1

| | n→∞

� k=1

uk�,

n

lim (uk + u� )� = A = ⇒kn→∞

� k=1

n

(3.9) for � > 0 ∃ N s.t. ∀ n ≥ N, A − � ≤ � (uk + u�k)� = ⇒ k=1

n n

A − � ≤ � uk� + � uk� )� ∀ n ≥ N = ⇒

k=1 k=1

A − � ≤ �b�B + �b��B ∀ � > 0 =⇒

�b + b��B ≤ �b�B + �b��B .

Now the norm of the element I(v) = v, 0, 0, · · · , is the limit of the norms of the sequence of partial sums and hence is �v�V so �I(v)�B = �v�V and I(v) = 0 therefore implies v = 0 and hence I is also injective.

So, we need to check that B is complete, and also that I(V ) is dense. Here is an extended discussion of the difficulty – of course maybe you can see it directly

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yourself (or have a better scheme). Note that I want you to write out your own version of it carefully for the next problem set.

Okay, what does it mean for B to be a Banach space, well as we saw in class today it means that every absolutely summable series in B is convergent. Such a series {bn} is given by bn = {uk

(n)} + S where {uk (n)} ∈ V� and the summability

condition is that N

(3.10) = lim u(n)∞ >

n

�bn�B

nN→∞

� k=1

k �V .

So, we want to show that bn = b converges, and to do so we need to find the n

limit b. It is supposed to be given by an absolutely summable series. The ‘problem’ is that this series should look like

�� u

(kn) in some sense – because it is supposed

n k to represent the sum of the bn’s. Now, it would be very nice if we had the estimate

(3.11) �u(kn)�V < ∞

n k

since this should allow us to break up the double sum in some nice way so as to get an absolutely summable series out of the whole thing. The trouble is that (3.11) need not hold. We know that each of the sums over k – for given n – converges, but not the sum of the sums. All we know here is that the sum of the ‘limits of the norms’ in (3.10) converges.

So, that is the problem! One way to see the solution is to note that we do not have to choose the original {u(n)} to ‘represent’ bn – we can add to it any element k

of S. One idea is to rearrange the u(kn) – I am thinking here of fixed n – so that it

‘converges even faster.’ Given � > 0 we can choose N1 so that for all N ≥ N1,

(3.12) |� u(kN)�V − �bn�B | ≤ �, �u(

kn)�V ≤ �.

k≤N k≥N

Then in fact we can choose successive Nj < Nj−1 (remember that little n is fixed here) so that

(3.13) |� u(kN)�V − �bn�B | ≤ 2−j �, �u(

kn)�V ≤ 2−j �.

k≤Nj k≥Nj

N1 � Now, ‘resum the series’ defining instead v1

(n) = �

u(kn), vj

(n) = Nj

u(kn) and do

k=1 k=Nj−1

this setting � = 2−n for the nth series. Check that now

(3.14) �vk (n)�V < ∞.

n k

Of course, you should also check that bn = {vk (n)} + S so that these new summable

series work just as well as the old ones. After this fiddling you can now try to find a limit for the sequence as

(3.15) b = {wk} + S, wk = v(p) ∈ V. l

l+p=k

So, you need to check that this {wk} is absolutely summable in V and that bn b→ as n →∞.

I may add some more to discussion of completeness if needed.

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Finally then there is the question of showing that I(V ) is dense in B. You can do this using the same idea as above – in fact it might be better to do it first. Given an element b ∈ B we need to find elements in V, vk such that �I(vk) − b�B → 0 as

Nj

k →∞. Take an absolutely summable series uk representing b and take vj = uk k=1

where the Nj ’s are constructed as above and check that I(vj ) b by computing � � →

(3.16) �I(vj ) − b�B = lim up�V ≤ �up�V . p→∞

� p>Nj p>Nj

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Problem set 2, Due 11AM Tuesday 24 Feb.

I was originally going to make this problem set longer, since there is a missing Tuesday. However, I would prefer you to concentrate on getting all four of these questions really right!

Problem 2.1 Finish the proof of the completeness of the space B constructed in lecture on February 10. The description of that construction can be found in the notes to Lecture 3 as well as an indication of one way to proceed.

Problem 2.2 Let’s consider an example of an absolutely summable sequence of step functions. For the interval [0, 1) (remember there is a strong preference for left-closed but right-open intervals for the moment) consider a variant of the construction of the standard Cantor subset based on 3 proceeding in steps. Thus, remove the ‘central interval [1/3, 2/3). This leave C1 = [0, 1/3) ∪ [2/3, 1). Then remove the central interval from each of the remaining two intervals to get C2 = [0, 1/9) ∪ [2/9, 1/3) ∪ [2/3, 7/9) ∪ [8/9, 1). Carry on in this way to define successive sets Ck ⊂ Ck−1, each consisting of a finite union of semi-open intervals. Now, consider the series of step functions fk where fk(x) = 1 on Ck and 0 otherwise.

(1) Check that this is an absolutely summable series. (2) For which x ∈ [0, 1) does |fk(x)| converge?

k (3) Describe a function on [0, 1) which is shown to be Lebesgue integrable

(as defined in Lecture 4) by the existence of this series and compute its Lebesgue integral.

(4) Is this function Riemann integrable (this is easy, not hard, if you check the definition of Riemann integrability)?

(5) Finally consider the function g which is equal to one on the union of all the intervals which are removed in the construction and zero elsewhere. Show that g is Lebesgue integrable and compute its integral.

Problem 2.3 The covering lemma for R2 . By a rectangle we will mean a set of the form [a1, b1) × [a2, b2) in R2 . The area of a rectangle is (b1 − a1) × (b2 − a2).

(1) We may subdivide a rectangle by subdividing either of the intervals – replacing [a1, b1) by [a1, c1) ∪ [c1, b1). Show that the sum of the areas of rectangles made by any repeated subdivision is always the same as that of the original.

(2) Suppose that a finite collection of disjoint rectangles has union a rectangle (always in this same half-open sense). Show, and I really mean prove, that the sum of the areas is the area of the whole rectange. Hint:- proceed by subdivision.

(3) Now show that for any countable collection of disjoint rectangles contained in a given rectange the sum of the areas is less than or equal to that of the containing rectangle.

(4) Show that if a finite collection of rectangles has union containing a given rectange then the sum of the areas of the rectangles is at least as large of that of the rectangle contained in the union.

(5) Prove the extension of the preceeding result to a countable collection of rectangles with union containing a given rectangle.

Problem 2.4

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(1) Show that any continuous function on [0, 1] is the uniform limit on [0, 1) of a sequence of step functions. Hint:- Reduce to the real case, divide the interval into 2n equal pieces and define the step functions to take infimim of the continuous function on the corresponding interval. Then use uniform convergence.

(2) By using the ‘telescoping trick’ show that any continuous function on [0, 1) can be written as the sum

(3.17) fj (x) ∀ x ∈ [0, 1) i

where the fj are step functions and |fj (x)| < ∞ for all x ∈ [0, 1). j

(3) Conclude that any continuous function on [0, 1], extended to be 0 outside this interval, is a Lebesgue integrable function on R.

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Solutions to Problem set 1

Full marks will be given to anyone who makes a good faith attempt to answer each question. The first four problems concern the ‘little L p’ spaces lp. Note that you have the choice of doing everything for p = 2 or for all 1 ≤ p < ∞.

Everyone who handed in a script received full marks. Problem 1.1 Write out a proof (you can steal it from one of many places but at

least write it out in your own hand) either for p = 2 or for each p with 1 ≤ p < ∞that

∞

lp = {a : N −→ C; |aj |p < ∞, aj = a(j)}j=1

is a normed space with the norm ⎛ ⎞ 1

∞ p

p�a�p = ⎝ |aj | ⎠ . j=1

This means writing out the proof that this is a linear space and that the three conditions required of a norm hold.

Solution:- We know that the functions from any set with values in a linear space form a linear space – under addition of values (don’t feel bad if you wrote this out, it is a good thing to do once). So, to see that lp is a linear space it suffices to see that it is closed under addition and scalar multiplication. For scalar multiples this is clear:

(3.18) |tai| = |t||ai| so �ta�p = |t|�a�p

which is part of what is needed for the proof that � · �p is a norm anyway. The fact that a, b ∈ lp imples a + b ∈ lp follows once we show the triangle inequality or we can be a little cruder and observe that

|ai + bi|p ≤ (2 max(|a|i, |bi|))p = 2p max(|a|pi , |bi|p) ≤ 2p(|ai| + |bi|)

(3.19) �a + b�pp = � |ai + bi|p ≤ 2p(�a�p + �b�p),

j

where we use the fact that tp is an increasing function of t ≥ 0. Now, to see that lp is a normed space we need to check that �a�p is indeed a

norm. It is non-negative and �a�p = 0 implies ai = 0 for all i which is to say a = 0. So, only the triangle inequality remains. For p = 1 this is a direct consequence of the usual triangle inequality:

(3.20) �a + b�1 = |ai + bi| ≤ (|ai| + |bi|) = �a�1 + �b�1. i i

For 1 < p < ∞ it is known as Minkowski’s inequality. This in turn is deduced from Holder’s inequality – which follows from Young’s inequality! The latter says if 1/p + 1/q = 1, so q = p/(p − 1), then

αp βq

(3.21) αβ ≤ + ∀ α, β ≥ 0. p q

To check it, observe that as a function of α = x,

xp βq

(3.22) f(x) = − xβ + p q

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if non-negative at x = 0 and clearly positive when x >> 0, since xp grows faster than xβ. Moreover, it is differentiable and the derivative only vanishes at xp−1 = β, where it must have a global minimum in x > 0. At this point f(x) = 0 so Young’s inequality follows. Now, applying this with α = |ai|/�a�p and β = |bi|/�b�q

(assuming both are non-zero) and summing over i gives Holder’s inequality � � � q

| i

aibi|/�a�p�b�q ≤ i

|ai||bi|/�a�p�b�q ≤ i �|a

a

�i|pp

p

p + �|b

b

�i|qqq

= 1

(3.23) � = ⇒ | aibi| ≤ �a�p�b�q.

i

Of course, if either �a�p = 0 or �b�q = 0 this inequality holds anyway. Now, from this Minkowski’s inequality follows. Namely from the ordinary trian

gle inequality and then Minkowski’s inequality (with q power in the first factor)

(3.24) |ai + bi|p = |ai + bi|(p−1)|ai + bi|i i

≤ |ai + bi|(p−1)|ai| + |ai + bi|(p−1)|bi|i i � �1/q

≤ |ai + bi|p (�a�p + �b�q) i

gives after division by the first factor on the right

(3.25) �a + b�p ≤ �a�p + �b�p.

Thus, lp is indeed a normed space. I did not necessarily expect you to go through the proof of Young-Holder-

Minkowksi, but I think you should do so at some point since I will not do it in class.

Problem 1.2 The ‘tricky’ part in Problem 1.1 is the triangle inequality. Suppose you knew – meaning I tell you – that for each N ⎛ ⎞ 1

N p

p⎝ |aj | ⎠ is a norm on CN

j=1

would that help? Solution:- Yes indeed it helps. If we know that for each N ⎛ ⎞ 1 ⎛ ⎞ 1 ⎛ ⎞ 1

N p N p N p

p p(3.26) ⎝ |aj + bj | ⎠ ≤ ⎝ |aj |p⎠ + ⎝ |bj | ⎠ j=1 j=1 j=1

then for elements of lp the norms always bounds the right side from above, meaning ⎛ ⎞ 1

N p

p(3.27) ⎝ |aj + bj | ⎠ ≤ �a�p + �b�p. j=1

Since the left side is increasing with N it must converge and be bounded by the right, which is independent of N. That is, the triangle inequality follows. Really

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this just means it is enough to go through the discussion in the first problem for finite, but arbitrary, N.

Problem 1.3 Prove directly that each lp as defined in Problem 1.1 – or just l2 – is complete,

i.e. it is a Banach space. At the risk of offending some, let me say that this means showing that each Cauchy sequence converges. The problem here is to find the limit of a given Cauchy sequence. Show that for each N the sequence in CN obtained by truncating each of the elements at point N is Cauchy with respect to the norm in Problem 1.2 on CN . Show that this is the same as being Cauchy in CN in the usual sense (if you are doing p = 2 it is already the usual sense) and hence, this cut-off sequence converges. Use this to find a putative limit of the Cauchy sequence and then check that it works.

Solution:- So, suppose we are given a Cauchy sequence a(n) in lp. Thus, each element is a sequence {a(n)}∞ in lp. From the continuity of the norm in Problem j j=1

1.5 below, �a(n)� must be Cauchy in R and so converges. In particular the sequence is norm bounded, there exists A such that �a(n)�p ≤ A for all n. The Cauchy condition itself is that given � > 0 there exists M such that for all m,n > M, � 1

p

(n) (m) p(3.28) �a(n) − a(m)�p = < �/2.| − a |ai i i

Now for each i, |ai (n) − ai

(m)| ≤ �a(n) − a(m)�p so each of the sequences ai (n) must

be Cauchy in C. Since C is complete

(3.29) lim ai (n) = ai exists for each i = 1, 2, . . . .

n→∞

So, our putative limit is a, the sequence {ai}∞ The boundedness of the norms i=1. shows that

N

(3.30) |ai (n)|p ≤ Ap

i=1

and we can pass to the limit here as n → ∞ since there are only finitely many terms. Thus

N

(3.31) |ai|p ≤ Ap ∀ N =⇒ �a�p ≤ A. i=1

Thus, a ∈ lp as we hoped. Similarly, we can pass to the limit as m → ∞ in the finite inequality which follows from the Cauchy conditions

N(n) (m) 1

(3.32) ( p) < �/2| − a |a pi i

i=1

�

to see that for each N N

(n) 1(3.33) (

i=1

and hence

p) ≤ �/2|a − ai| pi

(3.34) �a(n) − a� < � ∀ n > M.

Thus indeed, a(n) a in lp as we were trying to show. →

�


Notice that the trick is to ‘back off’ to finite sums to avoid any issues of interchanging limits.

Problem 1.4 Consider the ‘unit sphere’ in lp – where if you want you can set p = 2. This is

the set of vectors of length 1 :

S = {a ∈ lp; �a�p = 1}. (1) Show that S is closed. (2) Recall the sequential (so not the open covering definition) characterization

of compactness of a set in a metric space (e.g. by checking in Rudin). (3) Show that S is not compact by considering the sequence in lp with kth

element the sequence which is all zeros except for a 1 in the kth slot. Note that the main problem is not to get yourself confused about sequences of sequences!

Solution:- By the next problem, the norm is continuous as a function, so

(3.35) S = {a; �a� = 1}is the inverse image of the closed subset {1}, hence closed.

Now, the standard result on metric spaces is that a subset is compact if and only if every sequence with values in the subset has a convergent subsequence with limit in the subset (if you drop the last condition then the closure is compact).

In this case we consider the sequence (of sequences)

(n) 0 i = n(3.36) ai =

�.

1 i = n 1

This has the property that �a(n) − a(m)�p = 2 p whenever n =� m. Thus, it cannot have any Cauchy subsequence, and hence cannot have a convergent subsequence, so S is not compact.

This is important. In fact it is a major difference between finite-dimensional and infinite-dimensional normed spaces. In the latter case the unit sphere cannot be compact whereas in the former it is.

Problem 1.5 Show that the norm on any normed space is continuous. Solution:- Right, so I should have put this problem earlier! The triangle inequality shows that for any u, v in a normed space

(3.37) �u� ≤ �u − v� + �v�, �v� ≤ �u − v� + �u� which implies that

(3.38) |�u� − �v�| ≤ �u − v�.This shows that � · � is continuous, indeed it is Lipschitz continuous.





http://ocw.mit.edu

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Lecture 4. Thursday, 12 Feb

I talked about step functions, then the covering lemmas which are the basis of the definition of Lebesgue measure – which we will do after the integral – then properties of monotone sequences of step functions.

To be definite, but brief, by an interval we will mean [a, b) – an interval closed on the left and open on the right – at least for a little while. This is just so the length of the interval, b − a, only vanishes when the interval is empty (not true for closed intervals of course) and also so that we can decompose an interval, in this sense, into two disjoint intervals by choosing any interior point:

(4.1) [a, b) = [a, t) ∪ [t, b), a < t < b.

Now, by a step function

(4.2) f : R −→ C

(although often we will restrict to functions with real values) we mean a function which vanishes outside a finite union of disjoint ‘intervals’ and is constant on each of them. Thus f(R) is finite – the function only takes finitely many values – and

(4.3) f−1(c) is a finite union of disjoint intervals, c = 0� .

It is also often convenient to write a step function as a sum N

(4.4) f = ciχ[ai,bi)

i=1

of multiples of the characteristic functions of our intervals. Note that such a ‘presentation’ is not unique but can be made so by demanding that the intervals be disjoint and ‘maximal’ – so f is does not take the same value on two intervals with a common endpoint.

Now, a constant multiple of a step function is a step function and so is the sum of two step functions – clearly the range is finite. Really this reduces to checking that the difference [a, b) \ [a�, b�) and the union of two intervals is always a union of intervals. The absolute value of a step function is also a step function.

A similar argument shows that the integral, defined by

(4.5) f = ci(bi − ai) R i

from (4.4) is independent of the ‘presentation’ used to define it. It is of course equal to the Riemann integral which is one way of seeing that it is well-defined (but of course the result is much more elementary).

Proposition 1. The step functions on R in the sense defined above form a normed space with the L1 norm

(4.6) �f�L1 = |f |. R

So, we will complete this space instead of the continuous functions – it is both more standard and a little easier. The fact that we can directly construct a ‘concrete’ completion is due to Mikusinski. Already at this stage we can define a Lebesgue integrable function, however we need to do some work to flesh out the definition.

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Definition 3. A function g : R −→ C is Lebesgue integrable if there exists an absolutely summable sequence of step functions fn, i.e. satisfying � (4.7)

n

|fn| < ∞

such that � � (4.8) f(x) =

n

fn(x) ∀ x ∈ R such that n

|fn(x)| < ∞.

So, the definition is a little convoluted – there must exist a sequence of step functions the sum the integrals of the absolute values of which converges and such that the sum itself converges to the function but only at the points of absolute converge of the (pointwise) series. Tricky, but one can get used to it – and ultimately simplify it. The main attraction of this definition is that it is self-contained and in principle ‘everything’ can be deduced from it.

Now, the first thing we need is the covering lemma – basically some properties of countable collections of our ‘intervals’ such as will arise when we look at a sequence, or series, of step functions. I leave it to you to devise careful proofs of the following two facts.

Lemma 1. If Ci = [ai, bi), i = 1, . . . , N, is a finite collection of intervals then

N

(4.9) Ci ⊂ [a, b) ∀ i and Ci ∩ Cj = ∅ ∀ i =� j = ⇒ (bi − ai) ≤ (b − a). i=1

On the other hand N N

(4.10) [a, b) ⊂ Ci = ⇒ (bi − ai) ≥ (b − a). i=1 i=1

You can prove this by inserting division points etc. Now, what we want is the same thing for a countable collection of intervals.

Proposition 2. If Ci = [ai, bi), i ∈ N, is a countable collection of intervals then ∞

(4.11) Ci ⊂ [a, b) ∀ i and Ci ∩ Cj = ∅ ∀ i =� j = ⇒ (bi − ai) ≤ (b − a) i=1

or N � ∞

(4.12) [a, b) ⊂ Ci = ⇒ (bi − ai) ≥ (b − a). i=1 i=1

Proof. You might think these are completely obvious, and the first is – the hypothesis (4.9) holds for any finite subcollection and hence the finite sum is always less than the fixed number b − a and hence so is the infinite sum – which therefore converges.

On the other hand (4.12) is not quite so obvious since it depends on Heine-Borel. To be able to apply (4.10) choose δ > 0. Now extend each interval by replacing the lower limit by ai − 2−iδ and consider the open intervals which therefore have the property (4.13) [a, b − δ] ⊂ [a, b) ⊂ (ai − 2−iδ, bi).

i

� �

�

� �


Now, by Heine-Borel – the compactness of closed bounded intervals – a finite subcollection of these open intervals covers [a, b − δ) so (4.10) does apply to the semi-open intervals and shows that for some finite N (hence including the finite subcollection)

N ∞

(4.14) (bi − ai) − 2−iδ ≥ b − a − δ = ⇒ (bi − ai) ≥ b − a − 2δ i=1 i=1

where the sum here might be infinite, but then it is all the more true. The fact that this is true for all δ > 0 now proves (4.12). �

Welcome to measure theory. Okay, now the basis result on which most of the properties of integrable functions

hinge is the following monotonicity lemma.

Lemma 2. Let fn be a sequence of step functions which decreases monotonically to 0

(4.15) fn(x) ↓ 0 ∀ x ∈ R,

then

(4.16) lim fn = 0.

Note that ‘decreases’ here means that fn(x) is, for each x, a non-increasing sequence which has limit 0. Of course this means that all the fn are non-negative. Moreover the first one vanishes outside some interval [a, b) hence so do all of them. The crucial thing about this lemma is that we get the vanishing of the limit of the sequence of integrals without having to assume uniformity of the limit.

Proof. Going back to the definition of the integral of step functions, clearly fn(x) ≥fn+1(x) for all x implies that fn is a decreasing (meaning non-increasing) sequence. So there are only two possibilities, it converges to 0, as we claim, or it converges to some positive value. This means that there is some δ > such that fn > δ for all n, so we just need to show that this is not so.Given an � > 0 consider the sets

(4.17) Sj = {x ∈ [a, b); fj (x) ≤ �}.

(Here [a, b) is an interval outside which f1 vanishes and hence all the fn vanish outside it). Each of the Sj is a finite union of intervals. Moreover (4.18) Sj = [a, b)

j

since fn(x) 0 for each x. In fact the Sj increase with j and if we set B1 = S1→and

(4.19) Bj = Sj+1 \ Sj

then the Bj are all disjoint, each consists of finitely many intervals and (4.20) Bj = [a, b).

Thus, both halves of Proposition 2 apply to the intervals forming the Bj . If we let l(Bj ) be the length of Bj – the sum of the lengths of the finitely many intervals of

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which it is composed – then

(4.21) l(Bj ) = b − a. j

Now, let A be such that f1(x) ≤ A – and hence it is an upper bound for all the fn’s. We can choose N so large that

(4.22) l(Bj ) < �. j≥N

Dividing the integral for fk, k ≥ N, into the part over SN and the rest we see that

(4.23) fk ≤ (b − a)� + �A.

The first estimate comes for the fact the fact that fk ≤ � on SN , and the second that the total lengths of the remaining intervals is no more than � (and the function is no bigger than A.)

Thus the integral is eventually small! �

We can make this result look stronger as follows.

Proposition 3. Let gn be a sequence of real-valued step functions which is non-decreasing and such that

(4.24) lim gn(x) ∈ [0, ∞] ∀ x ∈ R n→∞

then

(4.25) lim gn ∈ [0, ∞]. n→∞

I have written out [0, ∞] on the right here to make sure that it is clear that we only demand that the non-decreasing sequence gn(x) ‘becomes 0 or positive’ in the limit – including the possibility that it ‘converges’ to +∞ and then the same is true of the sequence of integrals.

Proof. Consider the functions

(4.26) fn(x) = max(0, −gn(x)) ∀ x ∈ R. This is a sequence of non-negative step functions which decreases to 0 at each point and

(4.27) gn ≥ − fn.

Thus, as claimed, the result follows from the lemma. �

Next time we will use this to show that the definition of a Lebesgue integrable function above makes some sense. In particular we will show that the integral is well-defined!





http://ocw.mit.edu

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Lecture 5. Thursday, 19 Feb

I may not get quite this far, since I do not want to rush unduly!Let me denote by L1(R) the space of Lebesgue integrable functions on the line

– as defined last time.

Proposition 4. L1(R) is a linear space.

Proof. The space of all functions on R is linear, so we just need to check that L1(R) is closed under multiplication by constants and addition. The former is easy enough – multiplication by 0 gives the zero function which is integrable. If g ∈ L1(R) then by definition there is an absolutely summable series of step function with elements fn such that

(5.1) |fn| < ∞, f(x) = fn(x) ∀ x s.t. |fn(x)| < ∞. n n n

Then if c = 0� , cfn ‘works’ for cg. The sum of two functions g and g� ∈ L1(R) is a little trickier. The ‘obvious’

thing to do is to take the sum of the series of step functions. This will lead to trouble! Instead, suppose that fn and fn

� are series of step functions showing that g, g� ∈ L1(R). Then consider

(5.2) hn(x) = fk n = 2k − 1

fk� n = 2k.

This is absolutely summable since

(5.3) |hn| = |fk| + |fk� | < ∞.

n k k

More significantly, the series |hn(x)| converges if and only if both the series n

|fk(x)| and the series |fk� (x) converge. Then, because absolutely convergent

k k series can be rearranged, it follows that

(5.4) |hn(x)| < ∞ = ⇒ hn(x) = fk(x) + fk� (x) = g(x) + g�(x).

n n k k

Thus, g + g� ∈ L1(R). �

So, the message here is to be a bit careful about the selection of the ‘approximating’ absolutely summable series. Here is another example.

Definition 4. A set E ⊂ R is of measure zero if there exists an absolutely summable series of step functions fn such that

(5.5) |fn(x)| = ∞ ∀ x ∈ E. n

Proposition 5. If g ∈ L1(R) and g� : R −→ C is such that g� = g on R \ E where E is of measure zero, then g� ∈ L1(R).

Proof. What do we have to play with here – an absolutely summable series fn

of step functions which approximates g and another one, fn� which is such that

�

� � � � � �

� �

�

�

� � �

�

� �

� � � �


n |fn� (x)| = ∞ for all x ∈ E. So, to approximate g� consider the interlaced series

with terms ⎧ ⎪fk(x) n = 3k − 2⎨ (5.6) hn(x) = ⎪f � (x) n = 3k − 1k⎩ −fn

� (x) n = 3k.

Thus we add, but then subtract, fk� . This series is absolutely summable with

(5.7) |hn| = |fk| + 2 |fk� |.

n k k

When does the pointwise series converge absolutely? We must have

(5.8) |fk(x)| + 2 |fk� (x)| < ∞.

k k

The finiteness of the second term implies that x /∈ E and the finiteness of the first means that

(5.9) hn(x) = g(x) = g�(x) when (5.8) holds n

N

since the finite sum is always fk(x), or this plus f � (x) – which tends to zero N k=1

with N by the absolute convergence of the series in (5.8). So indeed g� ∈ L1(R). �

This certainly makes one conclude that sets of measure zero are small, except that we have not yet shown that L1(R) really makes any sense. That it does starts to become clear when we check:

Proposition 6. For any element f ∈ L1(R) the integral

(5.10) f = fn

n

is well-defined independent of which approximating absolutely summable series of step functions satisfying (5.1) is used to define it.

Proof. We can suppose that fn and fn� are two absolutely summable series as in

(5.1). Now that we have a little experience, it is probably natural to look at

(5.11) hn(x) = fk(x) n = 2k − 1

−fk� (x) n = 2k.

This is absolutely summable and the pointwise series is absolutely convergent only when both series are absolutely convergent. The individual terms then tend to zero and so we see that

(5.12) |hn(x)| < ∞ = ⇒ hn(x) = 0. n n

Moreover, from the absolutely convergence of the sequence of integrals –

(5.13) | hn| ≤ |hn| < ∞ n n

� � � � � �

� �

�

� � � �

�

�

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it follows that we can rearrange the series to see that

(5.14) hn = fk − fk� .

n k k

Now, what we want is that these two sums are equal, so we want to see that the left side of this equality vanishes. This follows directly from the next result which is just a little more general than needed here so is separated off. �

Proposition 7. If an absolutely summable series of step functions satisfies (5.1) with f ≡ 0 then

(5.15) fn = 0. n

Proof. So, the only thing we have at our disposal is the monotonicity result from last time. The trick is to use it! The ‘trick’ is to choose and N ∈ N and consider the new series of step functions with terms

N

(5.16) g1(x) = fj (x), gk(x) = |fN+k−1(x)|, k > 1. j=1

Now, this is absolutely summable, since convergence is a property of the ‘tail’ and in any case

(5.17) |gk| ≤ |fn|. k n

Moreover, since all the terms after the first are non-negative, the partial sums of the series

p

(5.18) Gp(x) = gp(x) is non-decreasing. k=1

It is again a sequence of step functions. Note that there are two possibilities, depending on x. If the original series

n |fn(x)| diverges, i.e. converges to +∞, then

the same is true of Gp – since this is also a property of the tails. On the other hand, if

n |fn(x)| is finite, then for large p,

N �� p−1 p+�N−1

(5.19) Gp(x) = fk(x) + |fN +j (x)| ≥ fk(x). k=1 j=1 k=1

The right side converges to zero, so the limit of this series (which is finite) is nonnegative. So, the monotonicity proposition from last time applies to Gp and shows that

(5.20) lim Gp ≥ 0 p→∞

where divergence to +∞ is a possibility. This however means that, for the N we originally chose,

N � �∞

(5.21) fk + |fN +k| ≥ 0. j=1 k=1

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This then is true for every N. On the other hand, the series of integrals is finite, so given δ > 0 there exists M such that if N > M, � N � (5.22) |fk| < δ = ⇒ fk ≥ −δ ∀ N > M.

k≥M j=1

This then implies that

(5.23) fk ≥ 0. k

This is half of what we want, but the other half follows by applying the same reasoning to −fk.

This the Proposition is proved. �

So, this is a pretty fine example of measure-integration reasoning.

Corollary 1. The integral is a well-defined map

(5.24) : L1(R) −→ C

defined by setting

(5.25) f = fn

n

for any approximating sequence as in (5.1).

In particular this integral is not trivial. Namely, if f is actually a step function then the sequence f1 = f, fj = 0 for all j > 1 is absoutely summable and approximates f in the sense of (5.1) so

(5.26) f is consistent with the integral on step functions.

So, we must be onto something here! It seems I was pretty carrie away today – no doubt because there were not enough

questions to slow me own. Hence I even went as far as to prove:

Proposition 8. A countable union of sets of measure zero has measure zero.

Proof. By definition, a set E is of measure zero if there exists an absolutely summable series of step functions, fn, so

n|fn| < ∞ such that

(5.27) |fn(x)| = +∞ on E. n

So, the data here gives us a countable collection Ej , j = 1, . . . , of sets and for each

of them we have an absolutely summable series of step functions fn (j) such that

(5.28) |fn (j)| < ∞, Ej ⊂ {x ∈ R; |fn

(j)(x)| = +∞}. n n

The idea is to look for one series of step functions which is absolutely summable and which diverges absolutely on each Ej . The trick is to first ‘improve’ the fn

(j) .

Namely, the divergence in (5.28) is a property of the ‘tail’ – it persists if we toss

� �

� � � � �

� �

�


out any finite number of terms. The absolute convergence means that for each j we can choose an Nj such that

(5.29) |f (j)| < 2−j ∀ j. n n≥Nj

Now, simply throw away the all the terms in fn (j) before n = Nj . If we relabel this

new sequence as fn (j) again, then we still have (5.28) but in addition we have not

only absolute summability but also

(5.30) |fn (j)| ≤ 2−j ∀ j = ⇒ |fn

(j)| < ∞. n j n

Thus, the double sum (of integrals of absolutely values) is absolutely convergent. Now, let hk be the fn

(j) ordered in some reasonable way – say by working along each row j + n = p in turn – in fact any enumeration of the double sequence will work. This is an absolutely summable series, because of (5.30). Moreover the pointwise series

(5.31) |hk(x)| = +∞ if |f (j)(x)| = +∞ for any jnk n

since the second sum is contained in the first. Thus |hk(x)| diverges at each k

point of each Ej , so the union has measure zero. �





http://ocw.mit.edu

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Lecture 6. Tuesday, Feb 24

By now the structure of the proofs should be getting somewhat routine – but I will go on to the point that I hope it all becomes clear!

So, recall the definitions of a Lebesgue integrable function on the line (forming the linear space L1(R)) and of a set of measure zero E ⊂ R.

The first thing we want to show is that the putative norm on L1 does make sense.

Proposition 9. If f ∈ L1(R) then |f | ∈ L1(R) and if fn is an absolutely summable series of step functions converging to f almost everywhere then � � N

(6.1) f = lim fk .| | N→∞

| k=1

|

So in some sense the definition of the Lebesgue integral ‘involves no cancellations’. There are extensions of the integral, we may even see one, which do exploit cancellations.

Proof. By definition if f ∈ L1(R) then it is the limit, on the set of absolute convergence, of a summable series of step functions, {fn}. We need to make such a series for |f |. The idea in this case is the ‘obvious’ one. We know that

n

(6.2) fj (x) → f(x) if |fj (x)| < ∞. j=1 j

So, set k �� k−1

(6.3) g1(x) = |f1(x)|, gk(x) = | fj (x)| − | fj (x)| ∀ x ∈ R. j=1 j=1

Then, for sure, N N

(6.4) gk(x) = | fj (x)| → |f(x)| if |fj (x)| < ∞. k=1 j=1 j

So, what we need to check, for a start, is that {gj } is an absolutely summable series of step functions.

The triangle inequality in the form ||v| − |w|| ≤ |v − w| shows that, for k > 1,

k k−1

(6.5) |gk(x)| = || fj (x)| − | fj (x)|| ≤ |fk(x)|. j=1 j=1

Thus

(6.6) |gk| ≤ |fk| < ∞ k k

so the gk’s do indeed form an absolutely summable series. From its construction we know that

N N

(6.7) gk(x) = | fj (x)| → |f(x)| if |fn(x)| < ∞. k=1 j=1 n

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So, this is what we want except that the set on which |gk(x)| < ∞ may be larger k

than the set for which we have convergence here. Now, in the notes there is a result to handle this, but we can simply make the series converge less rapidly by adding a ‘pointless’ subseries. Namely replace gk by ⎧ ⎪gk(x) if n = 3k − 2⎨ (6.8) hn(x) = ⎪fk(x) if n = 3k − 1 ⎩ −fk(x) if n = 3k.

This series converges absolutely if and only if both the |gk(x)| and |fk(x)| series converge – the convergence of the latter implies the convergence of the former so

(6.9) |hn(x)| < ∞⇐⇒ |fk(x)|. n k

On the other hand when this holds,

(6.10) hn(x) = |f(x)|n

since each partial sum is either a sum for gk, or this with fn(x) added. Since fn(x) → 0, (6.10) holds whenever the series converges absolutely, so indeed |f | ∈

1(R).LThen (6.1) follows. �

Now, the next thing we want to know is when ‘norm’ vanishes. That is, when does |f | = 0? One way is fairly easy, I think I actually skated over this a bit in the lecture, so let me write it out carefully here. The result we are after is:

Proposition 10. For an integrable function f ∈ L1(R), the vanishing of |f |implies that f is a null function in the sense that

(6.11) f(x) = 0 ∀ x ∈ R \ E where E is of measure zero.

Conversely, if (6.11) holds then f ∈ L1(R) and |f | = 0.

Proof. The main part of this is the first part, that the vanishing of |f | implies that f is null. This I will prove using the next Proposition. The converse is the easier direction.

Namely, if f is null in the sense of (6.11) then, by the definition of a set of measure zero, there exists an absolutely summable series of step functions, fn, such that

(6.12) E ⊂ {x ∈ R; |fn(x)| = ∞}. n

Note that it is possible that the absolute series here diverges on a larger set than E. Still, if we consider the alternating series

(6.13) gn(x) = fk(x) if n = 2k − 1

−fk(x) if n = 2k

then

(6.14) gn(x) = 0 whenever n

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since the latter condition is equivalent to n |fn(x)| < ∞. So in fact

(6.15) |gn(x)| < ∞ = ⇒ f(x) = gn(x) = 0 n n

because of (6.12). Thus the null function f ∈ L1(R), and so is |f | and from (6.15)

(6.16) |f | = gk = lim fk = 0 k

k→∞

where the last statement follows from the absolute summability. �

For the converse argument we will use the following result, which is also closely related to the completeness of L1(R).

Proposition 11. If fn ∈ L1(R) is an absolutely summable series, in the sense that

n |fn| < ∞ then

(6.17) E = {x ∈ R; |fn(x)| = ∞} has measure zero n

and if f : R −→ C and

(6.18) f(x) = fn(x) ∀ x ∈ R \ E n

then f ∈ L1(R) and

(6.19) f = fn. n

So this basically says we can replace ‘step function’ by ‘integrable function’ in the definition and get the same result. Of course this makes no sense without the original definition.

Proof. The proof is very like the proof of completeness of the ‘completion’ of a normed space that was in Problems 2, here it is a little more concrete.

Thus, by assumption each fn ∈ L1(R), so there exists an absolutely summable series of step functions fn,j , so |fn,j | < ∞ and

j

(6.20) |fn,j (x)| < ∞ = ⇒ fn(x) = fn,j (x). j j

We can expect f(x) to be given by the sum of the fn,j (x) over both n and j, but in general, this double series is not absolutely summable. However we can make it so. Simply choose Nn for each n so that

(6.21) |fn,j | < 2−n . j>Nn

This is possible by the assumed absolute summability – so the tail of the series is small. Having done this, we replace the series fn,j by

(6.22) f � = fn,j (x), n,j +k−1(x) ∀ j ≥ 2.n,1 f � (x) = fn,Nn

j≤Nn

� � �

� �

�

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� � � �

�

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This still converges to fn on the same set as in (6.20). So in fact we can simply replace fn,j by f � and we have in addition the estimate n,j

(6.23) f � fn + 2−n+1 ∀ n.| n,j | ≤ | |j

This follows from the triangle inequality since, using (6.21), � N � � � (6.24) f � f � f � f � f �| n,1 + n,j | ≥ | n,1| − | n,j | ≥ | n,1| − 2−n

k=1 j≥2

and the left side converges to |fn| by (6.1) as N → ∞. Using (6.21) again gives (6.23).

So, now dropping the primes from the notation and using the new series as fn,j

we can set

(6.25) gk(x) = fn,j . n+j=k

This gives a new series of step functions which is absolutely summable since N � � �

(6.26) |gk| ≤ |fn,j | ≤ ( |fn| + 2−n+1) < ∞. k=1 n,j n

Now, using rearrangement of absolutely convergent series we see that

(6.27) |fn,j (x)| < ∞ = ⇒ f(x) = |gk(x)| = fn,j (x). n,j k n j

From a result last week, we know that the set on the left here is of the form R \ E where E is of measure zero; E is the union of the sets En on which |fn,j (x)| = ∞.

j

So, take another absolutel summable series of step functions hk which diverges on E (at least) and insert hk and −hk between successive gk’s as before. This new series still coverges to f by (6.27) and shows that f ∈ L1(R) as well as (6.17). The final result (6.19) also follows by rearranging the double series for the integral (which is also absolutely convergent). �

Now for the moment we only need the weakest part, (6.17), of this. That is for any absolutely summable series of integrable functions the absolute pointwise series converges off a set of measure zero – can only diverge on a set of measure zero. It is rather shocking but this allows us to prove the rest of (10)! Namely, suppose f ∈ L1(R) and |f | = 0. Then Proposition 11 applies to the series with each term being |f |. This is absolutely summable since all the integrals are zero. So it must converge pointwise except on a set of measure zero. Clearly it diverges whenever f(x) = 0, so

(6.28) |f | = o =⇒ {x; f(x) = 0� } has measure zero

which is what we wanted to show. Finally this allows us to define the standard Lebesgue space

(6.29) L1(R) = L1(R)/N , N = {null functions}and to check that |f | is indeed a norm on this space.

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Problem set 3, Due 11AM Tuesday 3 Mar.

This problem set is also intended to be a guide to what will be on the in-class test on March 5. In particular I will ask you to prove some of the properties of the Lebesgue integral, as below, plus one more abstract proof. Recall that equality a.e (almost everywhere) means equality on the complement of a set of measure zero.

Problem 3.1 If f and g ∈ L1(R) are Lebesgue integrable functions on the line show that

(1) If f(x) ≥ 0 a.e. then f ≥ 0. (2) If f(x) ≤ g(x) a.e. then f ≤ g. (3) If f is complex valued then its real part, Re f, is Lebesgue integrable and | Re f | ≤ |f |.

(4) For a general complex-valued Lebesgue integrable function

(6.30) | f | ≤ |f |.

Hint: You can look up a proof of this easily enough, but the usual trick is to choose θ ∈ [0, 2π) so that eiθ f = (eiθf) ≥ 0. Then apply the preceeding estimate to g = eiθf.

(5) Show that the integral is a continuous linear functional

(6.31) : L1(R) −→ C.

Problem 3.2 If I ⊂ R is an interval, including possibly (−∞, a) or (a, ∞), we define Lebesgue integrability of a function f : I −→ C to mean the Lebesgue integrability of

(6.32) f(x) = f(x) x ∈ I

f : R −→ C, 0 x ∈ R \ I.

The integral of f on I is then defined to be

(6.33) f = f. ˜I

(1) Show that the space of such integrable functions on I is linear, denote it L1(I).

(2) Show that is f is integrable on I then so is |f |. (3) Show that if f is integrable on I and

I |f | = 0 then f = 0 a.e. in the sense that f(x) = 0 for all x ∈ I \ E where E ⊂ I is of measure zero as a subset of R.

(4) Show that the set of null functions as in the preceeding question is a linear space, denote it N (I).

(5) Show that I |f | defines a norm on L1(I) = L1(I)/N (I).

(6) Show that if f ∈ L1(R) then �

(6.34) g : I −→ C, g(x) = f(x) 0

x ∈ I

x ∈ R \ I

is integrable on I.

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(7) Show that the preceeding construction gives a surjective and continuous linear map ‘restriction to I’

(6.35) L1(R) −→ L1(I).

(Notice that these are the quotient spaces of integrable functions modulo equality a.e.)

Problem 3.3 Really continuing the previous one.(1) Show that if I = [a, b) and f ∈ L1(I) then the restriction of f to Ix = [x, b)

is an element of L1(Ix) for all a ≤ x < b. (2) Show that the function

(6.36) F (x) = f : [a, b) −→ C Ix

is continuous. (3) Prove that the function x−1 cos(1/x) is not Lebesgue integrable on the

interval (0, 1]. Hint: Think about it a bit and use what you have shown above.

Problem 3.4 [Harder but still doable] Suppose f ∈ L1(R). (1) Show that for each t ∈ R the translates

(6.37) ft(x) = f(x − t) : R −→ C

are elements of L1(R). (2) Show that

(6.38)tlim

0 |ft − f | = 0.

→

This is called ‘Continuity in the mean for integrable functions’. Hint: I will add one!

(3) Conclude that for each f ∈ L1(R) the map (it is a ‘curve’)

(6.39) R � t �−→ [ft] ∈ L1(R)

is continuous. Problem 3.5 In the last problem set you showed that a continuous function on a

compact interval, extended to be zero outside, is Lebesgue integrable. Using this, and the fact that step functions are dense in L1(R) show that the linear space of continuous functions on R each of which vanishes outside a compact set (which depends on the function) form a dense subset of L1(R).

Problem 3.6

(1) If g : R −→ C is bounded and continuous and f ∈ L1(R) show that gf ∈ L1(R) and that � �

(6.40) gf ≤ sup g f .| | R | | · | |

(2) Suppose now that G ∈ C([0, 1] × [0, 1]) is a continuous function (I use C(K) to denote the continuous functions on a compact metric space). Recall from the preceeding discussion that we have defined L1([0, 1]). Now, using the first part show that if f ∈ L1([0, 1]) then

(6.41) F (x) = G(x, )f( ) ∈ C· ·[0,1]


(where is the variable in which the integral is taken) is well-defined for · each x ∈ [0, 1].

(3) Show that for each f ∈ L1([0, 1]), F is a continuous function on [0, 1]. (4) Show that

(6.42) L1([0, 1]) � f �−→ F ∈ C([0, 1])

is a bounded (i.e. continuous) linear map into the Banach space of continuous functions, with supremum norm, on [0, 1].

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I was originally going to make this problem set longer, since there is a missing Tuesday. However, I would prefer you to concentrate on getting all four of these questions really right!

Problem 2.1 Finish the proof of the completeness of the space B constructed in lecture on February 10. The description of that construction can be found in the notes to Lecture 3 as well as an indication of one way to proceed.

Solution. The proof could be shorter than this, I have tried to be fairly complete. To recap. We start of with a normed space V. From this normed space we

construct the new linear space V with points the absolutely summable series in V. Then we consider the subspace S ⊂ V of those absolutely summable series which converge to 0 in V. We are interested in the quotient space

˜(6.43) B = V /S.

What we know already is that this is a normed space where the norm of b = {vn}+S – where {vn} is an absolutely summable series in V is

N

(6.44) �b�B = lim vn�V . N →∞

� n=1

This is independent of which series is used to represent b – i.e. is the same if an element of S is added to the series.

Now, what is an absolutely summable series in B? It is a sequence {bn}, thought of a series, with the property that

(6.45) �bn�B < ∞. n

We have to show that it converges in B. The first task is to guess what the limit should be. The idea is that it should be a series which adds up to ‘the sum of the bn’s’. Each bn is represented by an absolutely summable series vk

(n) in V. So, we can just look for the usual diagonal sum of the double series and set

(6.46) wj = vk (n) .

n+k=j

The problem is that this will not in generall be absolutely summable as a series in V. What we want is the estimate

(6.47) �wj � = � vk (n)� < ∞.

j j j=n+k

The only way we can really estimate this is to use the triangle inequality and conclude that

(6.48)∞

�wj � ≤ �vk (n)�V .

j=1 k,n

Each of the sums over k on the right is finite, but we do not know that the sum over k is then finite. This is where the first suggestion comes in:

We can choose the absolutely summable series vk (n) representing bn such that

(6.49) �vk (n)� ≤ �bn�B + 2−n .

k

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Suppose an initial choice of absolutely summable series representing bn is uk, so N

�bn� = limN→∞ � uk� and �uk�V < ∞. Choosing M large it follows that k=1 k

(6.50) �uk�V ≤ 2−n−1 . k>M

M

With this choice of M set v(n) = uk and v(n) = uM+k−1 for all k ≥ 2. This does 1 k k=1

still represent bn since the difference of the sums,

N N N+M−1

(6.51) vk (n) − uk = − uk

k=1 k=1 k=N

for all N. The sum on the right tends to 0 in V (since it is a fixed number of terms). Moreover, because of (6.50), (6.52)

M N N(n)�vk �V = � uj �V + �uk� ≤ � uj � + 2 �uk� ≤ � uj � + 2−n

k j=1 k>M j=1 k>M j=1

for all N. Passing to the limit as N →∞ gives (6.49). Once we have chosen these ‘nice’ representatives of each of the bn’s if we define

the wj ’s by (6.46) then (6.47) means that

(6.53) �wj �V ≤ �bn�B + 2−n < ∞ j n n

because the series bn is absolutely summable. Thus {wj } defines an element of Vand hence b ∈ B.

Finally then we want to show that bn = b in B. This just means that we need n

to show

N

(6.54) lim bn�B = 0. N→∞

�b − n=1

N

The norm here is itself a limit – b − bn is represented by the summable series n=1

with nth term

N

(6.55) wk − vk (n)

n=1

and the norm is then p N

(6.56) lim v(n)

p→∞ �

k=1

(wk − n=1

k )�V .

Then we need to understand what happens as N → ∞! Now, wk is the diagonal sum of the vj

(n)’s so sum over k gives the difference of the sum of the vj (n) over the

first p anti-diagonals minus the sum over a square with height N (in n) and width

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p. So, using the triangle inequality the norm of the difference can be estimated by the sum of the norms of all the ‘missing terms’ and then some so

p N

(6.57) � (wk − vk (n))�V ≤ �vl

(m)�V

k=1 n=1 l+m≥L

where L = min(p,N). This sum is finite and letting p →∞ is replaced by the sum over l + m ≥ N. Then letting N → ∞ it tends to zero by the absolute (double) summability. Thus

N

(6.58) lim bn�B = 0 N →∞

�b − n=1

which is the statelent we wanted, that bn = b. � n

Problem 2.2 Let’s consider an example of an absolutely summable sequence of step functions. For the interval [0, 1) (remember there is a strong preference for left-closed but right-open intervals for the moment) consider a variant of the construction of the standard Cantor subset based on 3 proceeding in steps. Thus, remove the ‘central interval [1/3, 2/3). This leave C1 = [0, 1/3) ∪ [2/3, 1). Then remove the central interval from each of the remaining two intervals to get C2 = [0, 1/9) ∪ [2/9, 1/3) ∪ [2/3, 7/9) ∪ [8/9, 1). Carry on in this way to define successive sets Ck ⊂ Ck−1, each consisting of a finite union of semi-open intervals. Now, consider the series of step functions fk where fk(x) = 1 on Ck and 0 otherwise.

(1) Check that this is an absolutely summable series. (2) For which x ∈ [0, 1) does |fk(x)| converge?

k (3) Describe a function on [0, 1) which is shown to be Lebesgue integrable

(as defined in Lecture 4) by the existence of this series and compute its Lebesgue integral.

(4) Is this function Riemann integrable (this is easy, not hard, if you check the definition of Riemann integrability)?

(5) Finally consider the function g which is equal to one on the union of all the subintervals of [0, 1) which are removed in the construction and zero elsewhere. Show that g is Lebesgue integrable and compute its integral.

Solution. (1) The total length of the intervals is being reduced by a factor of 2k

1/3 each time. Thus l(Ck) = 3k . Thus the integral of f, which is non

negative, is actually

2k � �∞ 2k

(6.59) fk = 3k = ⇒ |fk| = 3k

= 2 k k=1

Thus the series is absolutely summable. (2) Since the Ck are decreasing, Ck ⊃ Ck+1, only if

(6.60) x ∈ E = Ck

k

does the series |fk(x)| diverge (to +∞) otherwise it converges. k

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(3) The function defined as the sum of the series where it converges and zero otherwise ⎧�⎨ fk(x) x ∈ R \ E

(6.61) f(x) = k ⎩0 x ∈ E

is integrable by definition. Its integral is by definition

(6.62) f = fk = 2 k

from the discussion above. (4) The function f is not Riemann integrable since it is not bounded – and this

is part of the definition. In particular for x ∈ Ck \ Ck+1, which is not an empty set, f(x) = k.

(5) The set F, which is the union of the intervals removed is [0, 1) \ E. Taking step functions equal to 1 on each of the intervals removed gives an absolutely summable series, since they are non-negative and the kth one has integral 1/3×(2/3)k−1 for k = 1, . . . . This series converges to g on F so g is Lebesgue integrable and hence

(6.63) g = 1.

Problem 2.3 The covering lemma for R2 . By a rectangle we will mean a set of the form [a1, b1) × [a2, b2) in R2 . The area of a rectangle is (b1 − a1) × (b2 − a2).

(1) We may subdivide a rectangle by subdividing either of the intervals – replacing [a1, b1) by [a1, c1) ∪ [c1, b1). Show that the sum of the areas of rectangles made by any repeated subdivision is always the same as that of the original.

(2) Suppose that a finite collection of disjoint rectangles has union a rectangle (always in this same half-open sense). Show, and I really mean prove, that the sum of the areas is the area of the whole rectange. Hint:- proceed by subdivision.

(3) Now show that for any countable collection of disjoint rectangles contained in a given rectange the sum of the areas is less than or equal to that of the containing rectangle.

(4) Show that if a finite collection of rectangles has union containing a given rectange then the sum of the areas of the rectangles is at least as large of that of the rectangle contained in the union.

(5) Prove the extension of the preceeding result to a countable collection of rectangles with union containing a given rectangle.

Solution. (1) For subdivision of one rectangle this is clear enough. Namely we either divide the first side into two or the second side in two at an intermediate point c. After subdivision the area of the two rectanges is either

(c − a1)(b2 − a2) + (b1 − c)(b2 − a2) = (b1 − c1)(b2 − a2) or (6.64)

(b1 − a1)(c − a2) + (b1 − a1)(b2 − c) = (b1 − c1)(b2 − a2).


this shows by induction that the sum of the areas of any the rectangles made by repeated subdivision is always the same as the original.

(2) If a finite collection of disjoint rectangles has union a rectangle, say [a1, b2)×[a−2, b2) then the same is true after any subdivision of any of the rectangles, by the previous result. Moreover after such subdivision the sum of the areas is always the same. Look at all the points C1 ⊂ [a1, b1) which occur as an endpoint of the first interval of one of the rectangles. Similarly let C2 be the corresponding set of end-points of the second intervals of the rectangles. Now divide each of the rectangles repeatedly using the finite number of points in C1 and the finite number of points in C2. The total area remains the same and now the rectangles covering [a1, b1) × [A2, b2) are precisely the Ai × Bj where the Ai are a set of disjoint intervals covering [a1, b1) and the Bj are a similar set covering [a2, b2). Applying the one-dimensional result from class we see that the sum of the areas of the rectangles with first interval Ai is the product

(6.65) length of Ai × (b2 − a2).

Then we can sum over i and use the same result again to prove what we want.

(3) For any finite collection of disjoint rectangles contained in [a1, b1) × [a2, b2) we can use the same division process to show that we can add more disjoint rectangles to cover the whole big rectangle. Thus, from the preceeding result the sum of the areas must be less than or equal to (b1 − a1)(b2 − a2). For a countable collection of disjoint rectangles the sum of the areas is therefore bounded above by this constant.

(4) Let the rectangles beDi, i = 1, . . . , N the union of which contains the rectangle D. Subdivide D1 using all the endpoints of the intervals of D. Each of the resulting rectangles is either contained in D or is disjoint from it. Replace D1 by the (one in fact) subrectangle contained in D. Proceeding by induction we can suppose that the first N −k of the rectangles are disjoint and all contained in D and together all the rectangles cover D. Now look at the next one, DN−k+1. Subdivide it using all the endpoints of the intervals for the earlier rectangles D1, . . . , Dk and D. After subdivision of DN −k+1

each resulting rectangle is either contained in one of the Dj , j ≤ N − k or is not contained in D. All these can be discarded and the result is to decrease k by 1 (maybe increasing N but that is okay). So, by induction we can decompose and throw away rectangles until what is left are disjoint and individually contained in D but still cover. The sum of the areas of the remaining rectangles is precisely the area of D by the previous result, so the sum of the areas must originally have been at least this large.

(5) Now, for a countable collection of rectangles covering D = [a1, b1) × [a2, b2) we proceed as in the one-dimensional case. First, we can assume that there is a fixed upper bound C on the lengths of the sides. Make the kth rectangle a little larger by extending both the upper limits by 2−kδ where δ > 0. The area increases, but by no more than 2C2−k . After extension the interiors of the countable collection cover the compact set [a1, b1 − δ] × [a2, b1 − δ]. By compactness, a finite number of these open rectangles cover, and hence there semi-closed version, with the same endpoints, covers [a1, b1 − δ) ×

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[a2, b1 − δ). Applying the preceeding finite result we see that

(6.66) Sum of areas + 2Cδ ≥ Area D − 2Cδ.

Since this is true for all δ > 0 the result follows.

I encourage you to go through the discussion of integrals of step functions – now based on rectangles instead of intervals – and see that everything we have done can be extended to the case of two dimensions. In fact if you want you can go ahead and see that everything works in Rn!

Problem 2.4

(1) Show that any continuous function on [0, 1] is the uniform limit on [0, 1) of a sequence of step functions. Hint:- Reduce to the real case, divide the interval into 2n equal pieces and define the step functions to take infimim of the continuous function on the corresponding interval. Then use uniform convergence.

(2) By using the ‘telescoping trick’ show that any continuous function on [0, 1) can be written as the sum

(6.67) fj (x) ∀ x ∈ [0, 1) i

where the fj are step functions and |fj (x)| < ∞ for all x ∈ [0, 1). j

(3) Conclude that any continuous function on [0, 1], extended to be 0 outside this interval, is a Lebesgue integrable function on R.

Solution. (1) Since the real and imaginary parts of a continuous function are continuous, it suffices to consider a real continous function f and then add afterwards. By the uniform continuity of a continuous function on a compact set, in this case [0, 1], given n there exists N such that |x − y| ≤ 2−N = ⇒ |f(x)−f(y)| ≤ 2−n . So, if we divide into 2N equal intervals, where N depends on n and we insist that it be non-decreasing as a function of n and take the step function fn on each interval which is equal to min f = inf f on the closure of the interval then

(6.68) |f(x) − Fn(x)| ≤ 2−n ∀ x ∈ [0, 1)

since this even works at the endpoints. Thus Fn f uniformly on [0, 1).→(2) Now just define f1 = F1 and fk = Fk − Fk−1 for all k > 1. It follows that

these are step functions and that n

(6.69) = fn. k=1

Moreover, each interval for Fn+1 is a subinterval for Fn. Since f can varying by no more than 2−n on each of the intervals for Fn it follows that

(6.70) |fn(x)| = |Fn+1(x) − Fn(x)| ≤ 2−n ∀ n > 1.

Thus |fn| ≤ 2−n and so the series is absolutely summable. Moreover, it actually converges everywhere on [0, 1) and uniformly to f by (6.68).

(3) Hence f is Lebesgue integrable.

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(4) For some reason I did not ask you to check that � � 1

(6.71) f = f(x)dx 0

where on the right is the Riemann integral. However this follows from the fact that

(6.72) f = lim Fn n→∞

and the integral of the step function is between the Riemann upper and lower sums for the corresponding partition of [0, 1].





http://ocw.mit.edu

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Lecture 7. Thursday, Feb 26

So, what was it with my little melt-down? I went too cheap on the monotonicity theorem and so was under-powered for Fatou’s Lemma. In my defense, I was trying to modify things on-the-fly to conform to how we are doing things here. I should also point out that at least one person in the audience made a comment which amounted to pointing out my error.

So, here is something closer to what I should have said – it is not far from what I did say of course.

Proposition 12. [Montonicity again] If fj ∈ L1(R) is a monotone sequence, either fj (x) ≥ fj+1(x) for all x ∈ R and all j or fj (x) ≤ fj+1(x) for all x ∈ R and all j, and fj is bounded then

(7.1) {x ∈ R; lim fj (x) is finite} = R \ E j→∞

where E has measure zero and

f = lim fj (x) a.e. is an element of L1(R)

(7.2)j→∞ �

with lim f − fj = 0. j→∞

| |

Moral of the story – drop the assumption of positivity and replace it with the bound on the integral. In the approach through measure theory this is not necessary because one has the concept of a measureable, non-negative, function for which the integral ‘exists but is infinite’ – we do not have this.

Proof. Since we can change the sign of the fi (now) it suffices to assume that the fi

are monotonically increasing. The sequence of integrals is therefore also montonic increasing and, being bounded, converges. Thus we can pass to a subsequence gi = fni with the property that

(7.3) |gj − gj−1| = gj − gj−1 < 2−j ∀ j > 1.

This means that the series h1 = g1, hj = gj − gj−1, j > 1, is absolutely summable. So we know for the result last time that it converges a.e., that the limit, f, is integrable and that � � j � � (7.4) f = lim hk = lim gj = lim fj .

j→∞ k=1

j→∞ n→∞

In fact, everywhere that the series hj (x), which is to say the sequence gk(x), j

converges so does fn(x), since the former is a subsequence of the latter which is monotonic. So we have (7.1) and the first part of (7.2). The second part, corresponding to convergence for the equivalence classes in L1(R) follows from monotonicity, since

(7.5) |f − fj | = f − fj → 0 as j →∞.

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Now, to Fatou’s Lemma. This really just takes the monotonicity result and applies it to a general sequence of integrable functions with bounded integral. You should recall – as I did – that the max and min of two integrable functions is integrable and that

(7.6) min(f, g) ≤ min( f, g).

Lemma 3. [Fatou]. Let fj L1(R) be a sequence of non-negative (so real-valued integrable) functions such that fj is bounded above in R, then

f(x) = lim inf fn(x) exists a.e., f ∈ L1(R) and n→∞

(7.7) � � lim inf fn ≤ lim inf fn.

Proof. You should remind yourself of the properties of liminf as necessary! Fix k and consider

(7.8) Fk,n = min fp1(R)

k≤p≤k+n (x) ∈ L

as discussed briefly above. Moreover, this is a decreasing sequence, as n increases, because the minimum is over an increasing set an all elements are non-negative. Thus the integrals are bounded below by 0 so the monotonicity result above applies and shows that

(7.9) gk(x) = inf fp(x) ∈ L1(R), gk ≤ fn ∀ n ≥ k. p≥k

Note that for a decreasing sequence of non-negative numbers the limit exists everywhere and is indeed the infimum. Thus in fact,

(7.10) gk ≤ lim inf fn.

Now, let k vary. Then, the infimum in (7.9) is over a set which decreases as k increases. Thus the gk(x) are increasing. The integral is always bounded by one of the fn� and hence is bounded above independent of k since we assumed a bound on the fn’s. So, now we can apply the monotonicity result again to see that

f(x) = lim gk(x) exists a.e and f ∈ L1(R) has (7.11)

k→∞ � � f ≤ lim inf fn.

Since f(x) = lim inf fn(x), by definition of the latter, we have proved the Lemma.

Now, we apply Fatou’s Lemma to prove what we are really after:

Theorem 2. [Lebesgue’s dominated convergence]. Suppose fj ∈ L1(R) is a sequence of integrable functions such that

(7.12) ∃ h ∈ L1(R) with |fj (x)| ≤ h(x) a.e. and

f(x) = lim fj (x) exists a.e. � n→∞� Then f ∈ L1(R) and f = limn→∞ fn (including the assertion that this limit exists).

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� � �

�

�


Proof. First, we can assume that the fj are real since the hypotheses hold for its real and imaginary parts and together give the desired result. Moroever, we can change all the fj ’s to make them zero on the set on which the estimate in (7.12) does not hold. Then this bound on the fj ’s becomes

(7.13) −h(x) ≤ fj (x) ≤ h(x) ∀ x ∈ R.

In particular this means that gj = g − fj is a non-negative sequence of integrable functions and the sequence of integrals is also bounded, since (7.12) also implies that |fj | ≤ h, so gj ≤ 2 h. Thus Fatou’s Lemma applies to the gj . Since we have assumed that the sequence gj (x) converges a.e. to f we know that

h − f(x) = lim inf gj (x) a.e. and (7.14) � � � � �

h − f ≤ lim inf (h − fj ) = h − lim sup fj .

Notice the change on the right from liminf to limsup because of the sign. Now we can apply the same argument to gj

� (x) = h(x) + fj (x) since this is also non-negative and has integrals bounded above. This converges a.e. to h(x) + f(x) so this time we conclude that

(7.15) h + f ≤ lim inf (h + fj ) = h + lim inf fj .

In both inequalities (7.14) and (7.15) we can cancel and h and combining them we find

(7.16) lim sup fj ≤ f ≤ lim inf fj .

In particular the limsup on the left is smaller than, or equal to, the liminf on the right, for the same real sequence. This however implies that the are equal and that the sequence fj converges (look up properties of liminf and limsup if necessary ...). Thus indeed

(7.17) f = lim fn. n→∞

Generally in applications it is Lebesgue’s dominated convergence which is used to prove that some function is integrable.

Finally I want to make sure that we agree that L1(R) is a Banach space. Note once again that I have used the somewhat non-standard notation

(7.18) L1(R) = {f : R −→ C; f is integrable.}

This is a curly ‘L’. We know that f ∈ L1(R) implies that |f | ∈ L1(R) (if you are wondering the converse might not be true if f oscillates badly enough). Now, we know exactly when the integral of the abolute value vanishes. Namely

(7.19) N = {f ∈ L1(R); |f | = 0}

= {f : R −→ C; f(x) = 0 ∀ x ∈ R \ E, E of measure zero}.

Namely, this is the linear space of null functions. We then defined

(7.20) L1(R) = L1(R)/N .

�

� �

� �

� �

� � � � �

�

� �


This has a non-curly ‘L’ – the notation is by no means standard but the definition (7.20) certainly is. Thus the elements of L1(R) are equivalence classes of functions

(7.21) [f ] = f + N , f ∈ L1(R).

That is, we ‘identify’ to element of L1(R) if (and only if) there difference is null, which is to say they are equal off a set of measure zero. Note that the set is not fixed, but can depend on the functions. Anyway, for an element of L1(R) the integral of the absolute value is well-defined:

(7.22) �[f ]�L1 = |f |

since the right side is independent of which representative we choose.

Theorem 3. The function � · �L1 in (7.22) is a norm on L1(R) with respect to which it is a Banach space.

The integral of the absolute value, f is a semi-norm on L1(R) – it satisfies all the properties of a norm except that

| ||f | = 0 does not imply f = 0, only f ∈ N .

We are ‘killing’ this problem by taking the quotient.

Proof. I will not go through the proofs of the norm properties but you should. So, the only issue remaining is the completeness of L1(R) with respect to �·}L1 .

The completeness is a direct consequence of the Theorem in the last lecture on absolutely summable series of Lebesgue functions, so remind yourself of what this says. Also recall how we showed that if f is integrable, so is |f |. Namely, if fj is an absolutely summable series (originally of step functions, now of Lebesgue integrable functions) then we defined

(7.23) g1 = |f1|, gj = | fk| − | fk|k≤j k≤j−1

and observed that

(7.24) |gj | ≤ |fj | ∀ j. � Thus, gj is also absolutely summable and everywhere fj (x) converges,

j

(7.25) (x) = | fj (x)| → |f(x)| as N →∞. j≤N j≤N

This shows that |f | ∈ L1(R), but more than that since

(7.26) f = lim fj (x) fj .| | N→∞

| | ≤ j

| |j≤N

Roughly speaking this is why we have been using absolutely summable series from the beginning.

So, going back to fj and absolutely summable series in L1(R), in the sense that |fj |, we can apply the discussion above to the truncated series starting at point

j

N. Namely, the fj for j ≥ N give an absolutely convergent series which sums a.e. to

(7.27) f(x) − fj (x) = fj (x). j<N j>N

� � � �

� �

� � �

� � �

�


Now, applying (7.26) we see that

(7.28) |f(x) − fj (x)| ≤ |fj |. j<N j>N

However, the absolute convergence means that the tail on the right is small with N, that is,

(7.29) lim fj = 0. N→∞

|f − j<N

|

So, finally it is only necessary to think about L1(R) instead of L1(R). An absolutely summable sequence in Fj in L1(R) is a series of equivalence classes fj + Nwhere fj ∈ L1(R). The absolutely summability condition is

(7.30) �Fj �L1 = |fj | < ∞ j j

is what we need to start the discussion above. Namely, we have shown that the sum a.e. f of the series fj is an element of L1(R) and (7.29) holds. But this just means that the equivalenct class F = f + N satisfies

(7.31) lim = lim = 0. N →∞

�F − j<N

Fj �L1 N →∞

|f − j<N

fj |

N

Thus, Fj = F in L1(R) which is therefore complete. � j=1

Note that despite the fact that it is technically incorrect, everyone says ‘L1(R) is the space of Lebesgue integrable functions’ even though it is really the space of equivalence classes of these functions modulo equality almost everywhere. Not much harm can come from doing this.





http://ocw.mit.edu

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Lecture 8. Tuesday, Mar 3: Cauchy’s inequality and Lebesgue measure

I first discussed the definition of preHilbert and Hilbert spaces and proved Cauchy’s inequality and the parallelogram law. This can be found in all the lecture notes and many other places so I will not repeat it here. Another nice source is the book of G.F. Simmons, “Introduction to topology and modern analysis”. I like it – but I think it is out of print.

In case anyone is interested in how to define Lebesgue measure from where we are now – and I may have time to do this later – we can just use the integral as I outlined on Tuesday. First, we define locally integrable functions. Thus f : R −→ C is locally integrable if

(8.1) F[−N,N ] = f(x) x ∈ [−N,N ] 1(R) ∀ N. 0 x if |x| > N

∈ L

For example any continuous function on R is locally integrable.

Lemma 4. The locally integrable functions form a linear space.

Proof. Follows from the linearity of L1(R). �

Definition 5. A set A ⊂ R is measurable if its characteristic function χA is locally integrable. A measurable set A has finite measure if χA ∈ L1(R) and then

(8.2) µ(A) = χN

is the Lebesgue measure of A. If A is measurable but not of finite measure then µ(A) = ∞ by definition.

We know immediately that any interval (a, b) is measurable (whether open, semi-open or closed) and has finite measure if and only if it is bounded – then the measure is b − a. Some things to check:

Proposition 13. The complement of a measurable set is measureable and any countable union of measurable sets is measurable.

Proof. The first part follows from the fact that the constant function 1 is locally integrable and hence χR\A = 1 − χA is locally integrable if and only if χA is locally integrable.

Notice the relationship between characteristic functions and the sets they define:

(8.3) χA∪B = max(χA, χB ), χA∩B = min(χA, χB ). � If we have a sequence of sets An then Bn = Ak is clearly an increasing

sequence of sets and k≤n

(8.4) χBn χB , B = An→ n

is an increasing sequence which converges pointwise (at each point it jumps to 1 somewhere and then stays or else stays at 0.) Now, if we multiply by χ[−N,N ] then

(8.5) fn = χ[−N,N ]χBn → χB∩[−N,N ]

is an increasing sequence of integrable functions – assuming that is that the Ak’s are measurable – with integral bounded above, by 2N. Thus by our monotonicity � theorem the limit is integrable so χB is locally integrable and hence n An is measurable. �

�


Proposition 14. The (Lebesgue) measurable subsets of R form a collection, M, of the power set of R, including ∅ and R which is closed under complements, countable unions and countable intersections.

Proof. The countable intersection property follows from the others or directly by a similar argument to that above but with decreasing sequences. �

We have declared a set A which is measurable but not of finite measure to have infinite measure – for instance R is of infinite measure in this sense. Since the measure of a set is always non-negative (or undefined if it isn’t measurable) this does not cause any problems and in fact Lebesgue measure is countable additive provided we allow ∞ as a value of the measure:

(8.6) An ∈M, n ∈ N = ⇒ An ∈M and mu( An) ≤ µ(An) n n n

with equality if Aj ∩ Ak = ∅ for j =� k. It is a good exercise to prove this!

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Just to compensate for last week, I will make this problem set too short and easy!

Problem 4.1 Let H be a normed space in which the norm satisfies the parallelogram law:

(8.7) �u + v�2 + �u − v�2 = 2(�u�2 + �v�2) ∀ u, v ∈ H.

Show that the norm comes from a positive definite sesquilinear (i.e. Hermitian) inner product. Big Hint:- Try

1 � � (8.8) (u, v) =

4 �u + v�2 − �u − v�2 + i�u + iv�2 − i�u − iv�2 !

Problem 4.2 Let H be a finite dimensional (pre)Hilbert space. So, by definition H has a basis

{vi}ni=1, meaning that any element of H can be written

(8.9) v = civi

i

and there is no dependence relation between the vi’s – the presentation of v = 0 in the form (8.9) is unique. Show that H has an orthonormal basis, {ei}n satisfyingi=1 (ei, ej ) = δij (= 1 if i = j and 0 otherwise). Check that for the orthonormal basis the coefficients in (8.9) are ci = (v, ei) and that the map

(8.10) T : H � v �−→ ((v, ei)) ∈ Cn

is a linear isomorphism with the properties

(8.11) (u, v) = (Tu)i(Tv)i, �u�H = �Tu�Cn ∀ u, v ∈ H. i

Why is a finite dimensional preHilbert space a Hilbert space?

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This problem set is also intended to be a guide to what will be on the in-class test on March 5. In particular I will ask you to prove some of the properties of the Lebesgue integral, as below, plus one more abstract proof. Recall that equality a.e (almost everywhere) means equality on the complement of a set of measure zero.

Problem 3.1 If f and g ∈ L1(R) are Lebesgue integrable functions on the line show that

(1) If f(x) ≥ 0 a.e. then f ≥ 0. (2) If f(x) ≤ g(x) a.e. then f ≤ g. (3) If f is complex valued then its real part, Re f, is Lebesgue integrable and | Re f | ≤ |f |.

(4) For a general complex-valued Lebesgue integrable function

(8.12) | f | ≤ |f |.

Hint: You can look up a proof of this easily enough, but the usual trick is to choose θ ∈ [0, 2π) so that eiθ f = (eiθf) ≥ 0. Then apply the preceeding estimate to g = eiθf.

(5) Show that the integral is a continuous linear functional

(8.13) : L1(R) −→ C.

Solution:

(1) If f is real and fn is a real-valued absolutely summable series of step functions converging to f where it is absolutely convergent (if we only have a complex-valued sequence use part (3)). Then we know that

(8.14) g1 = |f1|, gj = |fj | − |fj−1|, f ≥ 1

is an absolutely convergent sequence converging to f | almost everywhere. 1

|It follows that f+ = 2 (|f | + f) = f, if f ≥ 0, is the limit almost everywhere of the series obtained by interlacing 1

2 gj and 12 fj :

1

(8.15) hn = 2 gk n = 2k − 1

fk n = 2k.

Thus f+ is Lebesgue integrable. Moreover we know that ⎛ ⎞ � � � k k

(8.16) f+ = lim hk = lim ⎝ fj + fj ⎠ k→∞

n≤2kk→∞

| j=1

| j=1

where each term is a non-negative step function, so f+ ≥ 0. (2) Apply the preceeding result to g − f which is integrable and satisfies

(8.17) g − f = (g − f) ≥ 0.

� � �

� � � | | �

� | |

� � � � � � � �

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(3) Arguing from first principles again, if fn is now complex valued and an absolutely summable series of step functions converging a.e. to f then define ⎧ ⎪⎨Re fk n = 3k − 2

(8.18) hn = ⎪Im fk n = 3k − 1 ⎩ − Im fk n = 3k.

This series of step functions is absolutely summable and

(8.19) |hn(x)| < ∞⇐⇒ |fn(x)| < ∞ = ⇒ hn(x) = Re f. n n n

Thus Re f is integrable. Since ± Re f ≤ f

(8.20) ± Re f ≤ |f | = ⇒ | Re f | ≤ |f |.

(4) For a complex-valued f proceed as suggested. Choose z ∈ C with z = 1 such that z f ∈ [0, ∞) which is possible by the properties of complex numbers. Then by the linearity of the integral

(8.21)

z f = (zf) = Re(zf) ≤ |z Re f | ≤ |f | = ⇒ | f | = z f ≤ |f |.

(where the second equality follows from the fact that the integral is equal to its real part).

(5) We know that the integral defines a linear map

(8.22) I : L1(R) � [f ] �−→ f ∈ C

since f = g if f = g a.e. are two representatives of the same class in L1(R). To say this is continuous is equivalent to it being bounded, which follows from the preceeding estimate

(8.23) |I([f ])| = | f | ≤ |f | = �[f ]�L1

(Note that writing [f ] instead of f ∈ L1(R) is correct but would normally be considered pedantic – at least after you are used to it!)

(6) I should have asked – and might do on the test: What is the norm of I as an element of the dual space of L1(R). It is 1 – better make sure that you can prove this.

Problem 3.2 If I ⊂ R is an interval, including possibly (−∞, a) or (a, ∞), we define Lebesgue integrability of a function f : I −→ C to mean the Lebesgue integrability of

(8.24) f(x) = f(x) x ∈ I

f : R −→ C, 0 x ∈ R \ I.

The integral of f on I is then defined to be

(8.25) f = f. ˜I

(1) Show that the space of such integrable functions on I is linear, denote it L1(I).

(2) Show that is f is integrable on I then so is |f |.

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(3) Show that if f is integrable on I and I |f | = 0 then f = 0 a.e. in the sense

that f(x) = 0 for all x ∈ I \ E where E ⊂ I is of measure zero as a subset of R.

(4) Show that the set of null functions as in the preceeding question is a linear space, denote it N (I).

(5) Show that I |f | defines a norm on L1(I) = L1(I)/N (I).

(6) Show that if f ∈ L1(R) then

(8.26) g(x) = f(x) x ∈ I

g : I −→ C, 0 x ∈ R \ I

is in L1(R) an hence that f is integrable on I. (7) Show that the preceeding construction gives a surjective and continuous

linear map ‘restriction to I’

(8.27) L1(R) −→ L1(I).

(Notice that these are the quotient spaces of integrable functions modulo equality a.e.)

Solution:

(1) If f and g are both integrable on I then setting h = f + g, h = f + g, directly from the definitions, so f + g is integrable on I if f and g are by the linearity of L1(R). Similarly if h = cf then h = cf is integrable for any constant c if f is integrable. Thus L1(I) is linear.

(2) Again from the definition, |f | = h if h = |f |. Thus f integrable on I implies f ∈ L1(R), which, as we know, implies that f 1(R). So in turn h ∈ L1(R) where h = |f |, so |f | ∈ L� 1(I). | | ∈ L

(3) If f ∈ L1(I) and �

I |f | = 0 then R |f | = 0 which implies that f = 0 on R \ E where E ⊂ R is of measure zero. Now, EI = E ∩ I ⊂ E is also of measure zero (as a subset of a set of measure zero) and f vanishes outside EI .

(4) If f, g : I −→ C are both of measure zero in this sense then f + g vanishes on I \ (Ef ∪ Eg) where Ef ⊂ I and Ef ⊂ I are of measure zero. The union of two sets of measure zero (in R) is of measure zero so this shows f + g is null. The same is true of cf + dg for constant c and d, so N (I) is a linear space.

(5) If f ∈ L1(I) and g ∈ N (I) then |f +g|−|f | ∈ N (I), since it vanishes where g vanishes. Thus � �

(8.28) I |f + g| =

I |f | ∀ f ∈ L1(I), g ∈ N (I).

Thus � (8.29) �[f ]�I =

I |f |

is a well-defined function on L1(I) = L1(R)/N (I) since it is constant on equivalence classes. Now, the norm properties follow from the same properties on the whole of R.

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(6) Suppose f ∈ L1(R) and g is defined in (8.26) above by restriction to I. We need to show that g ∈ L1(R). If fn is an absolutely summable series of step functions converging to f wherever, on R, it converges absolutely consider

fn(x) on I(8.30) gn(x) =

0 on R \ I

where I is I made half-open if it isn’t already – by adding the lower endpoint (if there is one) and removing the upper end-point (if there is one). Then gn is a step function (which is why we need I). Moreover,

� gn� | | ≤

|fn| so the series gn is absolutely summable and converges to gn outside I and at all points inside I where the series is absolutely convergent (since it is then the same as fn). Thus g is integrable, and since f differs from g by its values at two points, at most, it too is integrable so f is integrable on I by definition.

(7) First we check we do have a map. Namely if f ∈ N (R) then g in (8.26) is certainly an element of N (I). We have already seen that ‘restriction to I’ maps L1(R) into L1(I) and since this is clearly a linear map it defines (8.27) – the image only depends on the equivalence class of f. It is clearly linear and to see that it is surjective observe that if g ∈ L1(I) then extending it as zero outside I gives an element of L1(R) and the class of this function maps to [g] under (8.27).

Problem 3.3 Really continuing the previous one.

(1) Show that if I = [a, b) and f ∈ L1(I) then the restriction of f to Ix = [x, b) is an element of L1(Ix) for all a ≤ x < b.

(2) Show that the function

(8.31) F (x) = f : [a, b) −→ C Ix

is continuous. (3) Prove that the function x−1 cos(1/x) is not Lebesgue integrable on the

interval (0, 1]. Hint: Think about it a bit and use what you have shown above.

Solution:

(1) This follows from the previous question. If f ∈ L1([a, b)) with f � a representative then extending f � as zero outside the interval gives an element of L1(R), by defintion. As an element of L1(R) this does not depend on the choice of f � and then (8.27) gives the restriction to [x, b) as an element of L1([x, b)). This is a linear map.

(2) Using the discussion in the preceeding question, we now that if fn is an absolutely summable series converging to f � (a representative of f) where it converges absolutely, then for any a ≤ x ≤ b, we can define

(8.32) fn� = χ([a, x))fn, fn

�� = χ([x, b))fn

� � � � � � � �

� � � � � �

�

� � � � �

�


where χ([a, b)) is the characteristic function of the interval. It follows that fn� converges to fχ([a, x)) and fn

�� to fχ([x, b)) where they converge absolutely. Thus

(8.33) f = fχ([x, b)) = fn��, f = fχ([a, x)) = fn

� . [x,b) n [a,x) n

Now, for step functions, we know that fn = fn� + fn

�� so

(8.34) f = f + f [a,b) [a,x) [x,b)

as we have every right to expect. Thus it suffices to show (by moving the end point from a to a general point) that

(8.35) lim f = 0 x a→ [a,x)

for any f integrable on [a, b). Thus can be seen in terms of a defining absolutely summable sequence of step functions using the usual estimate that

(8.36) | [a,x)

f | ≤ [a,x) | n≤N

fn| + n>N [a,x)

|fn|.

The last sum can be made small, independent of x, by choosing N large enough. On the other hand as x a the first integral, for fixed N, tends→to zero by the definition for step functions. This proves (8.36) and hence the continuity of F.

(3) If the function x−1 cos(1/x) were Lebesgue integrable on the interval (0, 1] (on which it is defined) then it would be integrable on [0, 1) if we define it arbitrarily, say to be 0, at 0. The same would be true of the absolute value and Riemann integration shows us easily that � 1

(8.37) lim x cos(1/x) dx = ∞. t↓0 t

| |

This is contrary to the continuity of the integral as a function of the limits just shown.

Problem 3.4 [Harder but still doable] Suppose f ∈ L1(R). (1) Show that for each t ∈ R the translates

(8.38) ft(x) = f(x − t) : R −→ C

are elements of L1(R). (2) Show that

(8.39) tlim

0 |ft − f | = 0.

→

This is called ‘Continuity in the mean for integrable functions’. Hint: I will add one!

(3) Conclude that for each f ∈ L1(R) the map (it is a ‘curve’)

(8.40) R � t �−→ [ft] ∈ L1(R)

is continuous.Solution:

� � �

� � � � �

� � � � �

� �

� �

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(1) If fn is an absolutely summable series of step functions converging to f where it converges absolutely then fn(· − t) is such a series converging to f(· − t) for each t ∈ R. Thus, each of the f(x − t) is Lebesgue integrable, i.e. are elements of L1(R)

(2) Now, we know that if fn is a series converging to f as above then

(8.41) |f | ≤ |fn|. n

We can sum the first terms and then start the series again and so it follows that for any N,

(8.42) |f | ≤ | fn| + |fn|. n≤N n>N

Applying this to the series fn(· − t) − fn(·) we find that

(8.43) |ft − f | ≤ | fn(· − t) − fn(·)| + |fn(· − t) − fn(·)|n≤N n>N

The second sum here is bounded by 2 |fn|. Given δ > 0 we can choose n>N

N so large that this sum is bounded by δ/2, by the absolute convergence. So the result is reduce to proving that if |t| is small enough then

(8.44) | fn(· − t) − fn(·)| ≤ δ/2. n≤N

This however is a finite sum of step functions. So it suffices to show that

(8.45) | g(· − t) − g(·)| → 0 as t → 0

for each component, i.e. a constant, c, times the characteristic function of an interval [a, b) where it is bounded by 2|c||t|.

(3) For the ‘curve’ ft which is a map

(8.46) R � t �−→ ft ∈ L1(R)

it follows that ft+s = (ft)s so we can apply the argument above to show that for each s,

(8.47) lim ft − fs = 0 = lim s �[ft] − [fs]�L1 = 0

t→s | | ⇒

t→

which proves continuity of the map (8.46).

Problem 3.5 In the last problem set you showed that a continuous function on a compact interval, extended to be zero outside, is Lebesgue integrable. Using this, and the fact that step functions are dense in L1(R) show that the linear space of continuous functions on R each of which vanishes outside a compact set (which depends on the function) form a dense subset of L1(R).

Solution: Since we know that step functions (really of course the equivalence classes of step functions) are dense in L1(R) we only need to show that any step function is the limit of a sequence of continuous functions each vanishing outside a

�

�


compact set, with respect to L1 . So, it suffices to prove this for the charactertistic function of an interval [a, b) and then multiply by constants and add. The sequence ⎧ ⎪⎪⎪⎪⎪⎪0 x < a − 1/n ⎨n(x − a + 1/n) a − 1/n ≤ x ≤ a

(8.48) gn(x) = 0 a < x < b ⎪⎪⎪⎪⎪⎪n(b + 1/n − x) b ≤ x ≤ b + 1/n ⎩0 x > b + 1/n

is clearly continuous and vanishes outside a compact set. Since � � 1 � b+1/n

(8.49) |gn − χ([a, b))| = a−1/n

gn + b

gn ≤ 2/n

it follows that [gn] [χ([a, b))] in L1(R). This proves the density of continuous→functions with compact support in L1(R).

Problem 3.6

(1) If g : R −→ C is bounded and continuous and f ∈ L1(R) show that gf ∈ L1(R) and that � �

(8.50) gf ≤ sup g f .| | R | | · | |

(2) Suppose now that G ∈ C([0, 1] × [0, 1]) is a continuous function (I use C(K) to denote the continuous functions on a compact metric space). Recall from the preceeding discussion that we have defined L1([0, 1]). Now, using the first part show that if f ∈ L1([0, 1]) then

(8.51) F (x) = G(x, )f( ) ∈ C· ·[0,1]

(where is the variable in which the integral is taken) is well-defined for· each x ∈ [0, 1].

(3) Show that for each f ∈ L1([0, 1]), F is a continuous function on [0, 1]. (4) Show that

(8.52) L1([0, 1]) � f �−→ F ∈ C([0, 1])

is a bounded (i.e. continuous) linear map into the Banach space of continuous functions, with supremum norm, on [0, 1].

Solution:

(1) Let’s first assume that f = 0 outside [−1, 1]. Applying a result form Problem set there exists a sequence of step functions gn such that for any R, gn g uniformly on [0, 1). By passing to a subsequence we can arrange→that sup[−1,1] |gn(x) − gn−1(x)| < 2−n . If fn is an absolutly summable series of step functions converging a.e. to f we can replace it by fnχ([−1, 1]) as discussed above, and still have the same conclusion. Thus, from the uniform convergence of gn,

n

(8.53) gn(x) fk(x) g(x)f(x) a.e. on R.→ k=1

�

� � � �

� �

� � � �

� �

�

�


So define h1 = g1f1, hn = gn(x) n

fk(x) − gn−1(x) n�−1

fk(x). This series k=1 k=1

of step functions converges to gf(x) almost everywhere and since (8.54) � � � |hn| ≤ A|fn(x)| + 2−n |fk(x)|, |hn| ≤ A |fn| + 2 |fn| < ∞

k<n n n n

it is absolutely summable. Here A is a bound for |gn| independent of n. Thus gf ∈ L1(R) under the assumption that f = 0 outside [0, 1) and

(8.55) |gf | ≤ sup |g| |f |

follows from the limiting argument. Now we can apply this argument to fp

which is the restriction of p to the interval [p, p + 1), for each p ∈ Z. Then we get gf as the limit a.e. of the absolutely summable series gfp where (8.55) provides the absolute summablitly since

(8.56) |gfp| ≤ sup |g| [p,p+1)

|f | < ∞. p p

Thus, gf ∈ L1(R) by a theorem in class and

(8.57) |gf | ≤ sup |g| |f |.

(2) If f ∈ L1[(0, 1]) has a representative f � then G(x, )f �( 1([0, 1)) so �· ·) ∈ L

(8.58) F (x) = G(x, )f( ) ∈ C· ·[0,1]

is well-defined, since it is indpendent of the choice of f �, changing by a null function if f � is changed by a null function.

(3) Now by the uniform continuity of continuous functions on a compact metric space such as S = [0, 1] × [0, 1] given δ > 0 there exist � > 0 such that

(8.59) sup G(x, y) − G(x�, y) < δ if x − x� < �. y∈[0,1]

| | | |

Then if |x − x�| < �, � � (8.60) |F (x) − F (x�)| = |

[0,1] (G(x, ·) − G(x�, ·))f(·)| ≤ δ |f |.

Thus F ∈ C([0, 1]) is a continuous function on [0, 1]. Moreover the map f �−→ F is linear and

(8.61) sup F ≤ sup G[0,1] | |

S | |

[0,1] ||f |

which is the desired boundedness, or continuity, of the map

(8.62) I : L1([0, 1]) −→ C([0, 1]), F (f)(x) = G(x, ·)f(·),

�I(f)�sup ≤ sup |G|�f�L1 .





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Lecture 9. Thursday, March 5

My attempts to distract you all during the test did not seem to work very well. Here is what I wrote up on the board, more or less. We will use Baire’s theorem later (it is also known as ‘Baire category theory’ although it has nothing to do with categories in the modern sense).

This is a theorem about complete metric spaces – it could be included in 18.100B but the main applications are in Functional Analysis.

Theorem 4 (Baire). If M is a non-empty complete metric space and Cn ⊂ M, n ∈ N, are closed subsets such that (9.1) M = Cn

n

then at least one of the Cn’s has an interior point.

Proof. We can assume that the first set C1 =� ∅ since they cannot all be empty and dropping some empty sets does no harm. Let’s assume the contrary of the desired conclusion, namely that each of the Cn’s has empty interior, hoping to arrive at a contradiction to (9.1) using the other properties. This means that an open ball B(p, �) around a point of M (so it isn’t empty) cannot be contained in any one of the Cn.

So, choose p ∈ C1. Now, there must be a point p1 ∈ B(p, 1/3) which is not in C1. Since C1 is closed there exists �1 > 0, and we can take �1 < 1/3, such that B(p1, �1) ∩ C1 = ∅. Continue in this way, choose p2 ∈ B(p1, �1/3) which is not in C2 and �2 > 0, �2 < �1/3 such that B(p2, �2) ∩ C2 = ∅. Here we use both the fact that C2 has empty interior and the fact that it is closed. So, inductively there is a sequence pi, i = 1, . . . , k and positive numbers 0 < �k < �k−1/3 < �k−2/32 < <· · · �1/3k−1 < 3−k such that pj ∈ B(pj−1, �j−1/3) and B(pj , �j ) ∩ Cj = ∅. Then we can add another pk+1 by using the properties of Ck – it has non-empty interior so there is some point in B(pk, �k/3) which is not in Ck+1 and then B(pk+1, �k+1) ∩ Ck+1 = ∅where �k+1 > 0 but �k+1 < �k/3. Thus, we have a sequence {pk} in M. Since d(pk+1, pk) < �k/3 this is a Cauchy sequence, in fact

(9.2) d(pk, pk+l) < �k/3 + + �k+l−1/3 < 3−k .· · ·

Since M is complete the sequence converges to a limit, q ∈ M. Notice however that pl ∈ B(pk, 2�k/3) for all k > l so d(pk, q) ≤ 2�k/3 which implies that q /∈ Ck for any k. This is the desired contradiction to (9.1).

Thus, at least one of the Cn must have non-empty interior. �

One application of this, which we will get to later, is the uniform boundedness principle (which is just a theorem).

Theorem 5. Let B be a Banach space and suppose that Tk is a sequence of bounded (i.e. continuous) linear operators Tn : B −→ V where V is a normed space. Suppose that for each b ∈ B the set {Tn(b)} ⊂ V is bounded (in norm of course) then supn �Tn� < ∞.

Proof. You can look it up, but it follows from an application of Baire’s theorem to the sets

(9.3) Sp = {b ∈ B, �b� < 1, �Tnb�V ≤ p ∀ n}, p ∈ N..


You can check that these are closed and that their union must be the closed ball of radius one around the origin in B (because of the assumption of ‘pointwise boundedness’) So, by Baire’s theorem one of them has non-empty interior. This means that for some p, some v ∈ Sp and some δ > 0,

(9.4) w ∈ B, �w�B ≤ δ =⇒ �Tn(v + w)�V ≤ p ∀ n. Using the triangle inequality, and the fact that �Tn(v)�V ≤ p this means

(9.5) w ∈ B, �w�B ≤ δ =⇒ �Tn(w)�V ≤ 2p =⇒ �Tn� ≤ 2p/δ

since the norm of the operator is sup{�Tw�V ; �w�B = 1� �

Why this should be useful we shall see!





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�

�

�

� �

�


Lecture 10. Tuesday, Mar 10

All of this is easy to find in the various reference notes and/or books so I will keep these notes very brief.

(1) Bessel’s inequality If in a preHilbert space H, ei, i = 1, . . . , N are orthonormal – so (ei, ej ) =

δij then for any element u ∈ H, set N

v = (u, ei)ei then i=1

N(10.1) � 2 2 2�v� = |(u, ei)| ≤ �u�H H ,

i=1

(u − v) ⊥ ei, i = 1, . . . , N.

The last statement follows immediately by computing (u, ej ) = (v, ej ) and similarly �v�2 can be computed directly. Then the inequality, which is Bessel’s inequality, follows from Cauchy’s inequality since from the last statement

(10.2) �v�2 = (v, v) = (v, u) + (v, v − u) = (v, u) = |(v, u)| ≤ �v��u� shows �v� ≤ �u�.

(2) Orthonormal bases: Since in the inequality in (10.1) the right side is independent of N it

follows that if {ei}∞ is a countable orthonormal set then i=1

∞

(10.3) |(u, ei)|2 ≤ �u�2 H .

i=1

From this it follows that the sequence n

(10.4) vn = (u, ei)ei

i=1

is Cauchy since if m > n, ∞

(10.5) �vn − vm�2 = �(u, ej )|2 ≤ �(u, ej )|2

n<j≤m j=n+1

and the right side is small if n is large, independent of m.

Lemma 5. If H is a Hilbert space – so now we assume completeness – and {ei}∞ is an orthonormal sequence then for each u ∈ H, i=1

∞

(10.6) v = (u, ej )ej ∈ H j=1

converges and (u − v) ⊥ ej for all j.

Proof. The limit exists since the sequence is Cauchy and the space is complete. The orthogonality follows from the fact that (u − vn, ej ) = 0 as soon as n ≥ j and

(10.7) (u − v, ej ) = lim (u − vn, ej ) = 0 n→∞

�

�

� �


by continuity of the inner product (which follows from Cauchy’s inequality).

Now, we say an orthonormal sequence is complete, or is and orthonormal basis of H if u ⊥ ej = 0 for all j implies u = 0. Then we see:

Proposition 15. If {ej �∞ is an orthonormal basis in a Hilbert space Hj=1 then

∞

(10.8) u = (u, ej )ej ∀ u ∈ H. j=1

Proof. From the lemma the series converges to v and (u − v) ⊥ ej for all j so by the assumed completeness, u = v which is (10.8). �

(3) Gram-Schmidt

Theorem 6. Every separable Hilbert space has an orthonormal basis.

Proof. Take a countable dense subset – which can be arranged as a sequence {vj } and the existence of which is the definition of separability – and orthonormalize it. Thus if v1 =� 0 set ei = v1/�v1�. Proceeding by induction we can suppose to have found for a given integer n elements ei, i = 1, . . . ,m, where m ≤ n, which are orthonormal and such that the linear span

(10.9) sp(e1, . . . , em) = sp(v1, . . . , vn).

To show the inductive step observe that if vn+1 is in the span(s) in (10.9) then the same ei work for n + 1. So it follows that

n� w(10.10) w = vn+1 − (vn+1, ej )ej = 0 so em+1 =

j=1

��w�

makes sense. Adding em+1 gives the equality of the spans for n + 1. Thus we may continue indefinitely. There are only two possibilities,

either we get a finite set of ei’s or an infinite sequence. In either case this must be an orthonormal basis. That is we claim

(10.11) H � u ⊥ ej ∀ j = ⇒ u = 0.

This uses the density of the vn’s. That is, there must exist a sequence wj

where each wj is a vn, such that wj → u in H. Now, each each vn, and hence each wj , is a finite linear combination of ek’s so, by Bessel’s inequality

(10.12) �wj �2 = |(wj , ek)|2 = |(u − wj , ek)|2 ≤ �u − wj �2

k k

where (u, ej ) = 0 for all j has been used. Thus �wj � → 0 and u = 0. �

(4) Isomorphism to l2

A finite dimensional Hilbert space is isomorphic to Cn with its standard inner product. Similarly from the result above

Proposition 16. Any infinite-dimensional separable Hilbert space (over the complex numbers) is isomorphic to l2 , that is there exists a linear map

(10.13) T : H −→ L2

�


which is 1-1, onto and satisfies (Tu, Tv)l2 = (u, v)H and �Tu�l2 = �u�H

for all u, v ∈ H.

Proof. Choose an orthonormal basis – which exists by the discussion above and set

(10.14) Tu = {(u, ej )�∞j=1.

This maps H into l2 by Bessel’s inequality. Moreover, it is linear since the entries in the sequence are linear in u. It is 1-1 since Tu = 0 implies (u, ej ) = 0 for all j implies u = 0 by the assumed completeness of the orthonormal basis. It is surjective since if {cj }∞ thenj=1

∞

(10.15) u = cj ej

j=1

converges in H. This is the same argument as above – the sequence of partial sums is Cauchy by Bessel’s inequality. Again by continuity of the inner product, Tu = {cj } so T is surjective.

The equality of the norms follows from equality of the inner products and the latter follows by computation for finite linear combinations of the ej and then in general by continuity. �

�

� �

�

� � � � � �



You should be thinking about using Lebesgue’s dominated convergence at several points below.

Problem 5.1 Let f : R −→ C be an element of L1(R). Define

(10.16) fL(x) = f(x) x ∈ [−L,L] 0 otherwise.

Show that fL ∈ L1(R) and that |fL − f | → 0 as L →∞. Problem 5.2 Consider a real-valued function f : R −→ R which is locally inte

grable in the sense that

(10.17) gL(x) = f(x) x ∈ [−L,L] 0 x ∈ R \ [−L,L]

is Lebesgue integrable of each L ∈ N. (1) Show that for each fixed L the function ⎧ ⎪⎨ ⎪⎩

gL(x) if gL(x) ∈ [−N,N ] N if gL(x) > N

−N if gL(x) < −N

(N )(10.18) gL (x) =

is Lebesgue integrable. (N)� |g − gL| → 0 as N →∞.

(3) Show that there is a sequence, hn, of step functions such that (2) Show that L

(10.19) hn(x) f(x) a.e. in R.→

(4) Defining ⎧⎪⎪⎪⎨ ⎪⎪⎪⎩

0 x �∈ [−L,L] hn(x) if hn(x) ∈ [−N,N ], x ∈ [−L,L](N)(10.20) hn,L = . N if hn(x) > N, x ∈ [−L,L] −N if hn(x) < −N, x ∈ [−L,L]

(N) (N) n,L − gShow that → 0 as n →∞.|h |L

Problem 5.3 Show that L2(R) is a Hilbert space. First working with real functions, define L2(R) as the set of functions f : R −→ R

which are locally integrable and such that |f |2 is integrable.

(1) For such f choose hn and define gL, g(N) and hn

(N) by (10.17), (10.18) andL (10.20).

(2) Show using the sequence h(N) (N) (N))2 arefor fixed N and L that g and (gn,L L L (N) (N)in L1(R) and that )2 )2(h − (g 0 as n →∞.| |

(gn,L L →

(N)(3) Show that (gL)2 ∈ L1(R) and that ||(gL)2 − f2 0 as L →∞.

(5) Show that f, g ∈ L2(R) then fg ∈ L

)2 − (gL)2 0 as N →∞.| →L (4) Show that | →

1(R) and that

(10.21) | fg| ≤ |fg| ≤ �f�L2 �g�L2 , �f�2 = |f |2 .L2

�

�

�

�


(6) Use these constructions to show that L2(R) is a linear space. (7) Conclude that the quotient space L2(R) = L2(R)/N , where N is the space

of null functions, is a real Hilbert space. (8) Extend the arguments to the case of complex-valued functions.

Problem 5.4 Consider the sequence space ⎧⎨

⎫⎬ (10.22) h2,1 = ⎩

c : N � j �−→ cj ∈ C; (1 + j2) cj 2| | < ∞⎭

. j

(1) Show that

(10.23) h2,1 × h2,1 � (c, d) �−→ �c, d� = (1 + j2)cj dj

j

is an Hermitian inner form which turns h2,1 into a Hilbert space. (2) Denoting the norm on this space by � · �2,1 and the norm on l2 by � · �2,

show that

(10.24) h2,1 ⊂ l2 , �c�2 ≤ �c�2,1 ∀ c ∈ h2,1 .

Problem 5.5 In the separable case, prove Riesz Representation Theorem directly. Choose an orthonormal basis {ei} of the separable Hilbert space H. Suppose

T : H −→ C is a bounded linear functional. Define a sequence

(10.25) wi = T (ei), i ∈ N. (1) Now, recall that |Tu| ≤ C�u�H for some constant C. Show that for every

finite N, N

(10.26) |wi|2 ≤ C2 . j=1

(2) Conclude that {wi} ∈ l2 and that

(10.27) w = wiei ∈ H. i

(3) Show that

(10.28) T (u) = �u,w�H ∀ u ∈ H and �T � = �w�H .

� �

� � �



Just to compensate for last week, I will make this problem set too short and easy!

Problem 4.1 Let H be a normed space in which the norm satisfies the parallelogram law:

(10.29) �u + v�2 + �u − v�2 = 2(�u�2 + �v�2) ∀ u, v ∈ H.

Show that the norm comes from a positive definite sesquilinear (i.e. Hermitian) inner product. Big Hint:- Try

1 � � (10.30) (u, v) =

4 �u + v�2 − �u − v�2 + i�u + iv�2 − i�u − iv�2 !

Solution: Setting u = v, even without the parallelogram law,

(10.31) (u, u) = 14 ��2u�2 + i�(1 + i)u�2 − i�(1 − i)u�2

� = �u�2 .

So the point is that the parallelogram law shows that (u, v) is indeed an Hermitian inner product. Taking complex conjugates and using properties of the norm, �u + iv� = �v − iu� etc

1 � � (10.32) (u, v) =

4 �v + u�2 − �v − u�2 − i�v − iu�2 + i�v + iu�2 = (v, u).

Thus we only need check the linearity in the first variable. This is a little tricky! First compute away. Directly from the identity (u, −v) = −(u, v) so (−u, v) = −(u, v) using (10.32). Now, (10.33)

(2u, v) = 41 � �u + (u + v)�2 − �u + (u − v)�2

+ i�u + (u + iv)�2 − i�u + (u − iv)�2

1 �=

2 �u + v�2 + �u�2 − �u − v�2 − �u�2

+ i�(u + iv)�2 + i�u�2 − i�u − iv�2 − i�u�2

− 41 � �u − (u + v)�2 − �u − (u − v)�2 + i�u − (u + iv)�2 − i�u − (u − iv)�2

� =2(u, v).

Using this and (10.32), for any u, u� and v,

1(u + u�, v) = (u + u�, 2v)

2

=1 1 � �(u + v) + (u� + v)�2 − �(u − v) + (u� − v)�2

2 4 + i�(u + iv) + (u − iv)�2 − i�(u − iv) + (u� − iv)�2

1 � (10.34) =

4 �u + v� + �u� + v�2 − �u − v� − �u� − v�2

+ i�(u + iv)�2 + i�u − iv�2 − i�u − iv� − i�u� − iv�2

− 21

41 � �(u + v) − (u� + v)�2 − �(u − v) − (u� − v)�2

+ i�(u + iv) − (u − iv)�2 − i�(u − iv) = (u� − iv)�2

= (u, v) + (u�, v).

�

�

�


Using the second identity to iterate the first it follows that (ku, v) = k(u, v) for any u and v and any positive integer k. Then setting nu� = u for any other positive integer and r = k/n, it follows that

(10.35) (ru, v) = (ku�, v) = k(u�, v) = rn(u�, v) = r(u, v)

where the identity is reversed. Thus it follows that (ru, v) = r(u, v) for any rational r. Now, from the definition both sides are continuous in the first element, with respect to the norm, so we can pass to the limit as r x in R. Also directly from the definition,

→

1 � � (10.36) (iu, v) =

4 �iu + v�2 − �iu − v�2 + i�iu + iv�2 − i�iu − iv�2 = i(u, v)

so now full linearity in the first variable follows and that is all we need. Problem 4.2 Let H be a finite dimensional (pre)Hilbert space. So, by definition H has a basis

{vi}ni=1, meaning that any element of H can be written

(10.37) v = civi

i

and there is no dependence relation between the vi’s – the presentation of v = 0 in the form (10.37) is unique. Show that H has an orthonormal basis, {ei}n satisi=1 fying (ei, ej ) = δij (= 1 if i = j and 0 otherwise). Check that for the orthonormal basis the coefficients in (10.37) are ci = (v, ei) and that the map

(10.38) T : H � v �−→ ((v, ei)) ∈ Cn

is a linear isomorphism with the properties

(10.39) (u, v) = (Tu)i(Tv)i, �u�H = �Tu�Cn ∀ u, v ∈ H. i

Why is a finite dimensional preHilbert space a Hilbert space? Solution: Since H is assumed to be finite dimensional, it has a basis vi, i =

1, . . . , n. This basis can be replaced by an orthonormal basis in n steps. First replace v1 by e1 = v1/�v1� where �v1� = 0 by the linear indepedence of the basis. �Then replace v2 by

(10.40) e2 = w2/�w2�, w2 = v2 − �v2, e1�e1.

Here w2 ⊥ e1 as follows by taking inner products; w2 cannot vanish since v2 and e1

must be linearly independent. Proceeding by finite induction we may assume that we have replaced v1, v2, . . . , vk, k < n, by e1, e2, . . . , ek which are orthonormal and span the same subspace as the vi’s i = 1, . . . , k. Then replace vk+1 by

k

(10.41) ek+1 = wk+1/�wk+1�, wk+1 = vk+1 − �vk+1, ei�ei. i=1

By taking inner products, wk+1 ⊥ ei, i = 1, . . . , k and wk+1 =� 0 by the linear independence of the vi’s. Thus the orthonormal set has been increased by one element preserving the same properties and hence the basis can be orthonormalized.

Now, for each u ∈ H set

(10.42) ci = �u, ei�.

�

�

�

�

� �


n

It follows that U = u − ciei is orthogonal to all the ei since i=1

(10.43) �u, ej � = �u, ej � − ci�ei, ej � = �u.ej � − cj = 0. i

This implies that U = 0 since writing U = diei it follows that di = �U, ei� = 0. i

Now, consider the map (10.38). We have just shown that this map is injective, since Tu = 0 implies ci = 0 for all i and hence u = 0. It is linear since the ci depend linearly on u by the linearity of the inner product in the first variable. Moreover it is surjective, since for any ci ∈ C, u = ciei reproduces the ci through (10.42).

i Thus T is a linear isomorphism and the first identity in (10.39) follows by direct computation:

n

(Tu)i(Tv)i = �u, ei� i=1 i

(10.44) � = �u, �v, ei�ei�

i

= �u, v�. Setting u = v shows that �Tu�Cn = �u�H .

Now, we know that Cn is complete with its standard norm. Since T is an isomorphism, it carries Cauchy sequences in H to Cauchy sequences in Cn and T −1

carries convergent sequences in Cn to convergent sequences in H, so every Cauchy sequence in H is convergent. Thus H is complete.





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�


Lecture 11. Thursday, 12 Mar

Quite a lot of new material, but all of it in the various notes and books. So, I will keep it brief.

(1) Convex sets and length minimizer The following result does not need the hypothesis of separability of the

Hilbert space and allows us to prove the subsequent results – especially Riesz’ theorem – in full generality.

Proposition 17. If C ⊂ H is a subset of a Hilbert space which is (a) Non-empty (b) Closed (c) Convex, in the sense that v1, v1 ∈ C implies 1 (v1 + v2) ∈ C2

then there exists a unique element v ∈ C closest to the origin, i.e. such that

(11.1) �v�H = inf u∈C �u�H .

Proof. By definition of inf there must exist a sequence {vn} in C such that �vn� → d = infu∈C �u�H . We show that vn converges and that the limit is the point we want. The parallelogram law can be written

(11.2) �vn − vm�2 = 2�vn�2 + 2�vm�2 − 4�(vn + vm)/2�2 .

Since �vn� → d, given � > 0 if N is large enough then n > N implies 2�vn�2 < 2d2 +�2/2. By convexity, (vn +vm)/2 ∈ C so �(vn +vm)/2�2 ≥ d2 . Combining these estimates gives

(11.3) n,m > N = ⇒u∈C �u�H ≤ 4d2 + �2 − 4d2

so {vn} is Cauchy. Since H is complete, vn v ∈ C since C is closed. →Moreover, the distance is continuous so �v�H = limn→∞ �vn� = d.

Thus v exists and uniqueness follows again from the parallelogram law. If v and v� are two points in C with �v� = �v�� = d then (v + v�)/2 ∈ C so

(11.4) �v − v��2 = 2�v�2 + 2�v��2 − 4�(v + v�)/2�2 ≤ 0 =⇒ v = v�.

(2) Orthocomplements

Proposition 18. If W ⊂ H is a linear subspace of a Hilbert space the

(11.5) W ⊥ = {u ∈ H; (u,w) = 0 ∀ w ∈ W } is also a linear subspace and W ∩ W ⊥ = {0}. If W is also closed then

(11.6) H = W ⊕ W ⊥

meaning that any u ∈ H has a unique decomposition u = w + w⊥ where w ∈ W and w⊥ ∈ W ⊥.

Proof. That W ⊥ defined by (11.5) is a linear subspace follows from the linearity of the condition defining it. If u ∈ W ⊥ and u ∈ W u ⊥ u by the definition so (u, u) = �u�2 = 0 and u = 0.

Now, suppose W is closed. If W = H then W ⊥ = {0} and there is nothing to show. So consider u ∈ H, u /∈ W. Consider

(11.7) C = u + W = {u� ∈ H; u� = u + w, w ∈ W }.


Then C is closed, since a sequence in it is of the form u�n = u+wn where wn

is a sequence in W and u�n converges if and only if wn converges. Now, C is non-empty, since u ∈ C and it is convex since u� = u + w� and u�� = u + w��

in C implies (u� + u��)/2 = u + (w� + w��)/2 ∈ C. Thus the length minimization result above applies and there exists a

unique v ∈ C such that �v� = infu�∈C �u��. The claim is that this v is perpendicular to W – draw a picture in two real dimensions! To see this consider an aritrary point w ∈ W and λ ∈ C then v + λw ∈ C and

(11.8) �v + λw�2 = �v�2 + 2 Re(λ(v, w)) + |λ|2�w�2 .

Choose λ = teiθ where the phase is chosen so that eiθ(v, w) = |(v, w)| ≥ 0. Then the fact that �v� is minimal means that

(11.9) t(2|(v, w)| + t�w�2) ≥ 0 ∀ t ∈ R = ⇒ |(v, w)| = 0

which is what we wanted to show. Thus indeed, give u ∈/ W we have constructed v ∈ W ⊥ such that u =

v + w, w ∈ W. This is (11.6) with the uniqueness of the decomposition already shown since it reduces to 0 having only the decomposition 0 + 0 and this in turn is W ∩ W ⊥ = {0}. �

(3) Riesz’ theorem The most important application of these results is to prove Riesz’ rep

resentation theorem (for Hilbert space, there is another one to do with measures).

Theorem 7. If H is a Hilbert space then any continuous linear functional T : H −→ C there exists a unique element φ ∈ H such that

(11.10) T (u) = (u, φ) ∀ u ∈ H.

Proof. (a) Here is the proof I gave quickly in Lecture 10, not using the preceeding Lemma. If T is the zero functional then w = 0 satisfies (11.10). Otherwise there exists some u� ∈ H such that T (u�) = 0 and �then u ∈ H, namely u = u�/F (u�), such that F (u) = 1. Thus

(11.11) C = {u ∈ H; T (u) = 1} = T −1({1})

is non-empty. The continuity of T and the second form shows that C is closed, as the inverse image of a closed set under a continuous map. Moreover C is convex since

(11.12) T ((u + u�)/2) = (T (u) + T (u�))/2.

Thus, there exists an element v ∈ C of minimal length. As in the proof above, it follows that �v + λw�2 ≥ �v�2 for all w ∈ W and λ ∈ C this implies that v ∈ W ⊥. Now continue as in the proof below.

(b) Here is the proof I gave in Lecture 11 using the orthocomplement above. Since T is continuous the null space

(11.13) W = T −1({0}) = {u ∈ H; T (u) = 0}

is a closed linear subspace. Thus

(11.14) H = W ⊕ W ⊥

�


by Proposition 18 above. Now, if T = 0 is the zero functional then W = H and W ⊥ = {0} and w = 0 works in (11.10). Othewise, W ⊥ � v�, which is not in W, i.e. has T (v�) =� 0 and hence v ∈ W ⊥

with T (v) = 1. Then for any u ∈ H, (11.15)u−T (u)v satisfies T (u−T (u)v) = T (u)−T (u)T (v) = 0 −→ u = w+T (u)v, w ∈ W.

Then, (u, v) = T (u)�v�2 since (w, v) = 0. Thus if φ = v/�v�2 then

(11.16) u = w + (u, φ)v = T (u) = (u, φ)T (v) = (u, φ).⇒

(4) Adjoints of bounded operators. As an application of Riesz’ theorem I showed that any bounded linear operator on a Hilbert space

(11.17) A : H −→ H, �Hu�H ≤ C�u�H ∀ u ∈ H

has a unique adjoint operator. That is there exists a unique bounded linear operator A∗ : H −→ H such that

(11.18) (Au, v)H = (u,A∗v) ∀ u, v ∈ H.

To see the existence of A∗v we need to work out what A∗v should be for each fixed v ∈ H. So, fix v in the desired identity (11.18), which is to say consider

(11.19) H � u −→ (Au, v) ∈ C.

This is a linear map and it is clearly bounded, since

(11.20) |(Au, v)| ≤ �Au�H �v�H ≤ (C�v�H )�u�H .

Thus it is a continuous linear functional on H which depends on v. In fact it is just the composite of two continouos linear maps

(11.21) H u �−→Au

H w �−→(w,v) C.−→ −→

By Riesz theorem there exists an unique element in H, which we can denote A∗v (since it only depends on v) such that

(11.22) (Au, v) = (u,A∗v) ∀ u ∈ H.

Now this defines the map A∗H −→ H but we need to check that it is linear and continuous. Linearity follows from the uniqueness part of Riesz’ theorem. Thus if v1, v2 ∈ H and c1, c2 ∈ C then

(11.23) (Au, c1v1 + c2v2) = c1(Au, v1) + c2(Au, v2)

= c1(u,A∗v1) + c2(u,A∗v2) = (u, c1A∗v2 + c2A∗v2)

where we have used the definitions of A∗v1 and A∗v2 – by uniqueness we must have A∗(c1v1 + c2v2) = c1A∗v1 + c2A∗v2.

Since we know the optimality of Cauchy’s inequality

(11.24) �v�H = sup |(u, v)|�u�=1

(do we? If not set u = v/�v� to see it.) it follows that

(11.25) �A∗v� = sup (u,A∗v) = sup (Au, v)�u�=1

| | �u�=1

| | ≤ �A��v�.


So in fact

(11.26) �A∗� ≤ �A�. In fact it is immediately the case that (A∗)∗ = A so the reverse equality also holds and so

(11.27) �A∗� = �A�.





http://ocw.mit.edu

�

�

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Lecture 12. Tuesay, Mar 17: Compactness and weak convergence

A subset in a general metric space is one with the property that any sequence in it has a convergent subsequence, with its limit in the set. You will recall with pleasure no doubt the equivalence of this condition to the (more general since it makes good sense in an arbitrary topological space) equivalence of this with the covering condition, that any open cover of the set has a finite subcover. So, in a separable Hilbert space the notion of a compact set is already fixed. We want to characterize it – in the problems this week you will be asked to prove several characterizations.

A general result in a metric space is that any compact set is both closed and bounded, so this must be true in a Hilbert space. The Heine-Borel theorem gives a converse to this, Rn or Cn (and hence in any finite dimensional normed space) any closed and bounded set is compact. Also recall that the convergence of a sequence in Cn is equivalent to the convergence of the n sequences given by its components and this is what is used to pass first from R to C and then to Cn . All of this fails in infinite dimensions and we need some condition in addition to being bounded and closed for a set to be compact.

To see where this might come from, observe that a set, S, consisting of the points of a convergent sequence, s : N −→ M, together with its limit, s, in any metric space is always compact. The set here is the image of the sequence, thought of as a map from the integers into the metric space, together with the limit (which might of course already be in the image). Certainly this set is bounded, since the distance from the intial point is certainly bounded. Moreover it is closed, although you might need to think about this for a minute. A sequence in the set which is the image of another sequence consists of elements of the original sequence in any order and maybe repeated at will. Since the original sequence may itself have reapeated points, the labelling of points is by no means unique. However S is closed since M \ S is open – a point in p ∈ M \ S is at a finite no-zero distance, d(p, s) from the limit so B(p, d(p, s)/2) can contain only finitely many elements of S hence a smaller open ball does not meet it.

Lemma 6. The image of a convergent sequence in a Hilbert space is a set with equi-small tails with respect to any orthonormal sequence, i.e. if ek is an othonormal sequence and un u is a convergent sequence then given � > 0 there exists N such→that

(12.1) |(un, ek)|2 < �2 ∀ n. k>N

Proof. Bessel’s inequality shows that for any u ∈ H,

(12.2) |(u, ek)|2 ≤ �u�2 . k

The convergence of this series means that (12.1) can be arranged for any single element un or the limit u by choosing N large enough, thus given � > 0 we can choose N � so that

(12.3) |(u, ek)|2 < �2/2. k>N �

�

�

�


In fact, for any orthonormal sequence such as ek – whether complete or not,

(12.4) P : H � u �−→ Pu = (u, ek)ek ∈ H k

is continuous and in fact has norm at most one. Indeed from Bessel’s inequality, �Pu�2 ≤ �u�2 . Now, applying this to � (12.5) PN u = (u, ek)ek

k>N

the convergence un → u implies the convergence in norm �PN un� → �PN u� and so

(12.6) |(u, ek)|2 < �2 . k>N �

So, we have arranged (12.1) for n > n� with N = N �. Of course, this estimate remains valid if N is increased, and we may arrange it for n ≤ n� by chossing N large enough. Thus indeed (12.1) holds for all n if N is chosen large enough. �

This suggest one useful characterization of compact sets in a separable Hilbert space.

Proposition 19. A set K ⊂ H in a separabel Hilbert space is compact if and only if it is bounded, closed and has equi-small tails with respect to any one orthonormal basis.

Proof. We already know that a compact set is closed and bounded. Suppose the equi-smallness of tails condition fails with respect to some orthonormal basis ek. This means that for some � > 0 and all N there is an element uN ∈ K such that

(12.7) |(uN , ek)|2 ≥ �2 . k>N

Then the sequence {uN } can have no convergent subsequence, since this would contradict the Lemma we have just proved, hence K is not compact in this case.

Thus we have proved the equi-smallness of tails condition to be necessary for the compactness of a closed, bounded set. So, it remains to show that it is sufficient. So, suppose K is closed, bounded and satisfies the equi-small tails condition with respect to an orthonormal basis ek and {un} is a sequence in K. We only need show that {un} has a Cauchy subsequence, since this will converge (H being complete) and the limit will be in K (since it is closed). Now, consider each of the sequences of coefficients (un, ek) in C. Here k is fixed. This sequence is bounded:

(12.8) |(un, ek)| ≤ �un� ≤ C

by the boundedness of K. So, by the Heine-Borel theorem, there is a subsequence of unl such that (unl , ek) converges as l →∞.

We can apply this argument for each k = 1, 2, . . . . First extracting a subsequence of {un} so that the sequence (un, e1) converges ‘along this subsequence’. Then extract a subsequence of this subsequence so that (un, e2) also converges along this sparser subsequence, and continue inductively. Then pass to the ‘diagonal’ subsequence of {un} which has kth entry the kth term in the kth subsequence. It is ‘eventually’ a subsequence of each of the subsequences previously constructed – meaning it coincides with a subsequence for some point onward (namely the

� �

� �


kth term onward for the kth subsquence). Thus, for this subsequence each of the (unl , ek) converges.

Now, let’s relabel this subsequence vn for simplicity of notation and consider Bessel’s identity (the orthonormal set ek is complete by assumption) for the difference

�vn − vn+l�2 = (vn − vn+l, ek) 2 + (vn − vn+l, ek) 2 H | | | |

(12.9) � k≤N �

k>N � ≤ |(vn − vn+l, ek)|2 + 2 |(vn, ek)|2 + 2 |(vn+l, ek)|2

k≤N k>N k>N

where the parallelogram law on C has been used. To make this sum less than �2

we may choose N so large that the last two terms are less than �2/2 and this may be done for all n and l by the equi-smallness of the tails. Now, choose n so large that each of the terms in the first sum is less than �2/2N, for all l > 0 using the Cauchy condition on each of finite number of sequence (vn, ek). Thus, {vn} is a Cauchy subsequence of {un} and hence as already noted convergent in K. Thus K is indeed compact. �

It is convenient to formalize the idea that each of the (un, ek), the sequence of coefficients of the Fourier-Bessel series, should converge.

Definition 6. A sequence, {un}, in a Hilbert space, H, is said to converge weakly to an element u ∈ H if it is bounded in norm and (uj , v) (u, v) converges in C for each v ∈ H. This relationship is written

→

(12.10) un � u.

In fact as we shall see next time, the assumption that �un� is bounded and that u exists are both unnecessary. That is, a sequence converges weakly if and only if (un, v) converges in C for each v ∈ H. Conversely, there is no harm in assuming it is bounded and that the ‘weak limit’ u ∈ H exists. Note that the weak limit is unique since if u and u� both have this property then (u − u�, v) = limn→∞(un, v) −limn→∞(un, v) = 0 for all v ∈ H and setting v = u − u� it follows that u = u�.

Lemma 7. A (strongly) convergent sequence is weakly convergent with the same limit.

Proof. This is the continuity of the inner product. If un u then→

(12.11) |(un, v) − (u, v)| ≤ �un − u��v� → 0

for each v ∈ H shows weak convergence. �

Now, there is a couple of things I will prove here and leave some more to you for the homework.

Lemma 8. For a bounded sequence in a separable Hilbert space, weak convergence is equivalent to component convergence with respect to an orthonormal basis.

Proof. Let ek be an orthonormal basis. Then if un is weakly convergent it follows immediately that (un, ek) (u, ek) converges for each k. Conversely, suppose this →is true for a bounded sequence, just that (un, ek) ck in C for each k. The norm boundedness and Bessel’s inequality show that

→

(12.12) |ck|2 = lim |(un, ek)|2 ≤ C2 sup �un�2

n k≤p

n→∞ k≤p

�

�

�

�


for all p. Thus in fact {ck} ∈ l2 and hence

(12.13) u = wkek ∈ H k

by the completeness of H. Clearly (un, ek) → (u, ek) for each k. It remains to show thta (un, v) (u, v) for all v ∈ H. This is certainly true for any finite linear combination of the

→ ek and for a general v we can write

(12.14) (un, v) − (u, v) = (un, vp) − (u, vp) + (un, v − vp) − (u, v − vp) =⇒

|(un, v) − (u, v)| = |(un, vp) − (u, vp)| + 2C�v − vp� where vp = (v, ek)ek is a finite part of the Fourier-Bessel series for v and C is a

k≤p

bound for �un�. Now the convergence vp → v implies that the last term in (12.14) can be made small by choosing p large, independent of n. Then the second last term can be made small by choosing n large since vp is a finite linear combination of the ek. Thus indeed, (un, v) → (u, v) for all v ∈ H an it follows that un converges weakly to u. �

Proposition 20. Any bounded sequence {un} in a separable Hilbert space has a weakly convergent subsequence.

This can be thought of as an analogue in infinite dimensions of the Heine-Borel theorem if you say ‘a bounded closed subset of a separable Hilbert space is weakly compact’.

Proof. Choose an orthonormal basis ek and apply the procedure in the proof of Proposition 19 to extract a subsequence of the given bounded sequence such that (unp , ek) converges for each k. Now apply the preceeding Lemma to conclude that this subsequence converges weakly. �

Lemma 9. For a weakly convergent sequence un � u

(12.15) �u� ≤ lim inf �un�.

Proof. Choose an orthonormal basis ek and observe that

(12.16) �u, ek�2 = lim 2 . n→∞

�un, ek�k≤p

Now the sequence on the right is bounded by �un�2 independently of p so

(12.17) �u, ek�2 ≤ lim inf �un�2

n k≤p

by the definition of lim inf . Now, take p →∞ to conclude that

(12.18) �u�2 ≤ lim inf �un�2

n

from which (12.15) follows. �


Problems 6: Due 11AM Tuesday, 31 Mar

Hint: Don’t pay too much attention to my hints, sometimes they are a little off-the-cuff and may not be very helpfult. An example being the old hint for Problem 6.2!

Problem 6.1 Let H be a separable Hilbert space. Show that K ⊂ H is compact if and only if it is closed, bounded and has the property that any sequence in K which is weakly convergent sequence in H is (strongly) convergent.

Hint:- In one direction use the result from class that any bounded sequence has a weakly convergent subsequence.

Problem 6.2 Show that, in a separable Hilbert space, a weakly convergent sequence {vn}, is (strongly) convergent if and only if the weak limit, v satisfies

(12.19) �v�H = lim n→∞

�vn�H .

Hint:- To show that this condition is sufficient, expand

(12.20) (vn − v, vn − v) = �vn�2 − 2 Re(vn, v) + �v�2 .

Problem 6.3 Show that a subset of a separable Hilbert space is compact if and only if it is closed and bounded and has the property of ‘finite dimensional approximation’ meaning that for any � > 0 there exists a linear subspace DN ⊂ H of finite dimension such that

(12.21) d(K,DN ) = sup inf {d(u, v)} ≤ �. u∈K v∈DN

Hint:- To prove necessity of this condition use the ‘equi-small tails’ property of compact sets with respect to an orthonormal basis. To use the finite dimensional approximation condition to show that any weakly convergent sequence in K is strongly convergent, use the convexity result from class to define the sequence {vn

� }in DN where vn

� is the closest point in DN to vn. Show that vn� is weakly, hence

strongly, convergent and hence deduce that {vn} is Cauchy. Problem 6.4 Suppose that A : H −→ H is a bounded linear operator with the

property that A(H) ⊂ H is finite dimensional. Show that if vn is weakly convergent in H then Avn is strongly convergent in H.

Problem 6.5 Suppose that H1 and H2 are two different Hilbert spaces and A : H1 −→ H2 is a bounded linear operator. Show that there is a unique bounded linear operator (the adjoint) A∗ : H2 −→ H1 with the property

(12.22) �Au1, u2�H2 = �u1, A∗u2�H1 ∀ u1 ∈ H1, u2 ∈ H2.

�

� � � �

�

�

�

�



You should be thinking about using Lebesgue’s dominated convergence at several points below.

Problem 5.1 Let f : R −→ C be an element of L1(R). Define

(12.23) fL(x) = f(x) x ∈ [−L,L] 0 otherwise.

χL

0 for each x as L →∞ and

1Show that ( ) and that 0 asRf f f L∈ L | − | → ∞L L .→Solution. If is the characteristic function of [ ] then Ifχ N,N f fχ=−L L L. is an absolutely summable series of step functions converging a.e.�

|

so by Lebesgue’s dominated convergence,

to f thenfn

is absolutely summable, since and converges a.e. to fL, sofn fnχL fn| | ≤ |1(R). Certainly |fL(x)−f(x)

f(x)fL(x)−f(x)f − fL

fL L | | | ≤0.

→ |fl(x)| f(x) ≤ 2

Problem 5.2|

Consider a real-valued function f : R −→ R which is locally inte+ | | | | | →

grable in the sense that

(12.24) gL(x) = f(x) x ∈ [−L,L] 0 x ∈ R \ [−L,L]

is Lebesgue integrable of each L ∈ N. (1) Show that for each fixed L the function ⎧ ⎪⎨ ⎪⎩

gL(x) if gL(x) ∈ [−N,N ] N if gL(x) > N

−N if gL(x) < −N

(N )(12.25) gL (x) =

is Lebesgue integrable. (N)(2) Show that 0 as N →∞.

(3) Show that there is a sequence, hn, of step functions such that|g − gL|L →

(12.26) hn(x) f(x) a.e. in R.→

(4) Defining ⎧⎪⎪⎪⎨ ⎪⎪⎪⎩

0 x �∈ [−L,L] hn(x) if hn(x) ∈ [−N,N ], x ∈ [−L,L](N)(12.27) hn,L = . N if hn(x) > N, x ∈ [−L,L] −N if hn(x) < −N, x ∈ [−L,L]

(N) (N) n,L − gShow that | → 0 as n →∞.|h L

Solution: (1) By definition g(N) = max(−NχL, min(NχL, gL)) where χL is the charac-L

teristic funciton of −[L,L], thus it is in L1(R). (N) (N)(2) Clearly g (x) gL(x) for every x and gL (x) ≤ gL(x) so by Dom|

, i.e. | | |L →

(N) (N)in L1inated Convergence, g 0 as N → ∞ |g(x)|.

to gL for ex

− gL|since the sequence converges to 0 pointwise and is bounded by 2

|gL gL L→ →

(3) Let SL,n be a sequence of step functions converging a.e. – ample the sequence of partial sums of an absolutely summable series of step functions converging to gL which exists by the assumed integrability.

�

� � �


Then replacing SL,n by SL,nχL we can assume that the elements all vanish outside [−N,N ] but still have convergence a.e. to gL. Now take the sequence

(12.28) hn(x) = Sk,n−k on [k, −k] \ [(k − 1), −(k − 1)], 1 ≤ k ≤ n,

0 on R \ [−n, n].

This is certainly a sequence of step functions – since it is a finite sum of step functions for each n – and on [−L,L] \ [−(L − 1), (L − 1)] for large integral L is just SL,n−L → gL. Thus hn(x) → f(x) outside a countable union of sets of measure zero, so also almost everywhere.

(N) (N)(4) This is repetition of the first problem, hn,L (x) gL almost everywhere � →

and h(N) ≤ NχL so g(N ) 1(R) and h(N) (N) 0 as n →∞.| n,L | L ∈ L | n,L − gL | →

Problem 5.3 Show that L2(R) is a Hilbert space – since it is rather central to the course I wanted you to go through the details carefully!

First working with real functions, define L2(R) as the set of functions f : R −→ R which are locally integrable and such that |f |2 is integrable.

(N) (N)(1) For such f choose hn and define gL, gL and hn by (12.24), (12.25) and (12.27).

(2) Show using the sequence h(N) for fixed N and L that g(N) and (g(N))2 are � n,L L L (N) (N)in L1(R) and that |(hn,L )

2 − (gL � )2| → 0 as n →∞. (N))2(3) Show that (�gL)2 ∈ L1(R) and that |(gL − (gL)2| → 0 as N →∞.

(4) Show that |(gL)2 − f2| → 0 as L →∞. (5) Show that f, g ∈ L2(R) then fg ∈ L1(R) and that

(12.29) | fg| ≤ |fg| ≤ �f�L2 �g�L2 , �f�L2 2 = |f |2 .

(6) Use these constructions to show that L2(R) is a linear space. (7) Conclude that the quotient space L2(R) = L2(R)/N , where N is the space

of null functions, is a real Hilbert space. (8) Extend the arguments to the case of complex-valued functions.

Solution:(N)(1) Done. I think it should have been hn,L .

(N)(2) We already checked that gL ∈ L1(R) and the same argument applies to (N)), (N))2 (N)(gL namely (hn,L gL almost everywhere and both are bounded →

by N2χL so by dominated convergence

(h(N))2 g(N))2 ≤ N2χL a.e. = g

(N))2 1(R) and n,L → L ⇒ L ∈ L

h(N ))2 (N))2 0 a.e. ,(12.30) | n,L − gL | �→

(N) (N) (N) (N)|hn,L )2 − gL )2| ≤ 2N2χL = ⇒ |hn,L )

2 − gL )2| → 0.

(3) Now, as N → ∞, (g(N))2 (gL)2 a.e. and (g(N ))2 (gL)2 so by L → � (N) L → ≤ f2

dominated convergence, (gL)2 ∈ L1 and |(g )2 −(gL)2| → 0 as N 2 →∞.L

(4) The same argument of dominated convergence shows now that g f2 � 2 1(R).

L →and |gL − f2| → 0 using the bound by f2 ∈ L

�

� �

�


(5) What this is all for is to show that fg ∈ L1(R) if f, F = g ∈ L2(R) (for easier notation). Approximate each of them by sequences of step functions as above, h(N ) for f and H(N) for g. Then the product sequence is in L1 – n,L n,L being a sequence of step functions – and

(12.31) h(N )(x)H(N)(x) g

(N)(x)G(N)(x)n,L n,L → L L

almost everywhere and with absolute value bounded by N2χL. Thus by dominated convergence g(N)

G(N) ∈ L1(R). Now, let N →∞; this sequence L L

converges almost everywhere to gL(x)GL(x) and we have the bound

(N ) (N) 1(f2 + F 2)(12.32) |gL (x)GL (x)| ≤ |f(x)F (x)|

2

so as always by dominated convergence, the limit gLGL ∈ L1 . Finally, letting L → ∞ the same argument shows that fF ∈ L1(R). Moreover, |fF | ∈ L1(R) and � �

(12.33) | fF | ≤ |fF | ≤ �f�L2 �F �L2

where the last inequality follows from Cauchy’s inequality – if you wish, first for the approximating sequences and then taking limits.

(6) So if f, g ∈ L2(R) are real-value, f + g is certainly locally integrable and

(12.34) (f + g)2 = f2 + 2fg + g 2 ∈ L1(R)

by the discussion above. For constants f ∈ L2(R) implies cf ∈ L2(R) is directly true. �

(7) The argument is the same as for L1 versus L1 . Namely f2 = 0 implies that f2 = 0 almost everywhere which is equivalent to f = 0 a@e. Then the norm is the same for all f + h where h is a null function since fh and h2

are null so (f + h)2 = f2 +2fh + h2 . The same is true for the inner product so it follows that the quotient by null functions

(12.35) L2(R) = L2(R)/N

is a preHilbert space. However, it remains to show completeness. Suppose {[fn]} is an ab

solutely summable series in L2(R) which means that �fn�L2 < ∞. It n

follows that the cut-off series fnχL is absolutely summable in the L1 sense since

(12.36) |fnχL| ≤ L 21 ( fn

2) 21

n

by Cauchy’s inequality. Thus if we set Fn = fk then Fn(x)χL converges k−1

almost everywhere for each L so in fact

(12.37) Fn(x) f(x) converges almost everywhere. →

We want to show that f ∈ L2(R) where it follows already that f is locally integrable by the completeness of L1 . Now consider the series

(12.38) g1 = F12 , gn = Fn

2 − Fn2 −1.

� � �

� � �

� � � �

� �

�


The elements are in L1(R) and by Cauchy’s inequality for n > 1, (12.39)

|gn| = |Fn 2 − Fn−1|2 ≤ �Fn − Fn−1�L2 �Fn + Fn−1�L2 ≤ �fn�L2 2 �fk�L2

k

where the triangle inequality has been used. Thus in fact the series gn is absolutely summable in L1

(12.40) |gn| ≤ 2( �fn�L2 )2 . n n

So indeed the sequence of partial sums, the Fn 2 converge to f2 ∈ L1(R).

Thus f ∈ L2(R) and moroever

(12.41) (Fn − f)2 = Fn 2 + f2 − 2 Fnf → 0 as n →∞.

Indeed the first term converges to f2 and, by Cauchys inequality, the series of products fnf is absulutely summable in L1 with limit f2 so the third term converges to −2 f2 . Thus in fact [Fn] [f ] in L2(R) and we have proved completeness.

→

(8) For the complex case we need to check linearity, assuming f is locally integrable and |f |2 ∈ L1(R). The real part of f is locally integrable and the approximation FL

(N) discussed above is square integrable with (FL (N))2 ≤

|f |2 so by dominated convergence, letting first N → ∞ and then L → ∞ the real part is in L2(R). Now linearity and completeness follow from the real case.

Problem 5.4 Consider the sequence space ⎧ ⎫ ⎨ � ⎬

(12.42) h2,1 = ⎩ c : N � j �−→ cj ∈ C; (1 + j2)|cj |2 < ∞⎭

. j

(1) Show that

(12.43) h2,1 × h2,1 � (c, d) �−→ �c, d� = (1 + j2)cj dj

j

is an Hermitian inner form which turns h2,1 into a Hilbert space. (2) Denoting the norm on this space by � · �2,1 and the norm on l2 by � · �2,

show that

(12.44) h2,1 ⊂ l2 , �c�2 ≤ �c�2,1 ∀ c ∈ h2,1 .

Solution: (1) The inner product is well defined since the series defining it converges ab

solutely by Cauchy’s inequality:

�c, d� = �

(1 + j2) 21 cj (1 + j2) 2

1 dj ,

(12.45) � 1

j � 1 � 1 |(1 + j2) 2 cj (1 + j2) 21 dj | ≤ ( (1 + j2)|cj |2) 2 ( (1 + j2)|dj |2) 2 .

j j j

�

� �

�

�

�

� � �

80

c

LECTURE NOTES FOR 18.102, SPRING 2009

It is sesquilinear and positive definite since

(12.46) �c�2,1 = ( (1 + j2)|cj |2)j

12

only vanishes if all cj vanish. Completeness follows as for l2 – if c(n) is a

Cauchy sequence then each component c(n) j converges, since (1 + j)

12

(n) j

is Cauchy. The limits cj define an element of h2,1 since the sequence is bounded and

N N1 2 |cj |2 = lim (1 + j2)

n→∞ j=1

|c(n) j |

2 ≤ A(1 + j2)(12.47) j=1

where A is a bound on the norms. Then from the Cauchy condition c(n) c in h2,1 by passing to the limit as m →∞ in �c(n) − c(m)�2,1 ≤ �.

→

(2) Clearly h2,2 ⊂ l2 since for any finite N

N N

(12.48) |cj |2 (1 + j)2|cj |2 ≤ �c�2 2,1

�

j=1 j=1

and we may pass to the limit as N →∞ to see that

�

(12.49) � � ≤ � �c 2 c 2 1.l ,

Problem 5.5 In the separable case, prove Riesz Representation Theorem directly.Choose an orthonormal basis {ei} of the separable Hilbert space H. Suppose

T : H −→ C is a bounded linear functional. Define a sequence

(12.50) wi = T (ei), i ∈ N.

(1) Now, recall that |Tu| ≤ C�u�H for some constant C. Show that for every finite N,

N

(12.51) |wi|2 ≤ C2 . j=1

(2) Conclude that {wi} ∈ l2 and that

(12.52) w = i

wiei ∈ H.

(3) Show that

(12.53)

Solution:

T (u) = �u, w�H ∀ u ∈ H and �T � = �w�H .

N

(1) The finite sum wN = wiei is an element of the Hilbert space with norm i=1

N� N

i=1

2 |wi|2 by Bessel’s identity. Expanding out =�wN �

N n N

(12.54) T (wN ) = T ( wiei) = wiT (ei) = |wi|2

i=1 i=1 i=1

and from the continuity of T,

(12.55) T (wN ) = 2 = 2 ≤ C2| | ≤ C�wN �H ⇒ �wN �H ≤ C�wN �H ⇒ �wN �

� �

�

�

�

�


which is the desired inequality. (2) Letting N →∞ it follows that the infinite sum converges and

(12.56) |wi|2 ≤ C2 = ⇒ w = wiei ∈ H i i

since �wN − w� ≤ |wi|2 tends to zero with N. j>N

N

(3) For any u ∈ H uN = �u, ei�ei by the completness of the {ei} so from the i=1

continuity of T

N

(12.57) T (u) = lim T (uN ) = lim �u, ei�T (ei) N→∞ N→∞

i=1

N

= lim lim N→∞

i=1

�u,wiei� = N→∞

�u,wN � = �u,w�

where the continuity of the inner product has been used. From this and Cauchy’s inequality it follows that �T � = sup�u�H =1 |T (u)| ≤ �w�. The converse follows from the fact that T (w) = �w�H

2 .





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Lecture 13. Thursday, Mar 19: Baire’s theorem

Note from Lecture 9, modified and considerably extended.

Theorem 8 (Baire). If M is a non-empty complete metric space and Cn ⊂ M, n ∈ N, are closed subsets such that (13.1) M = Cn

n

then at least one of the Cn’s has an interior point.

Proof. So, choose p1 ∈/ C1, which must exist since otherwise C1 contains an open ball. Since C1 is closed there exists �1 > 0 such that B(p1, �1)∩C1 = ∅. Next choose p2 ∈ B(p1, �1/3) which is not in C2, which is possible since otherwise B(p1, �1/3) ⊂C2, and �2 > 0, �2 < �1/3 such that B(p2, �2)∩C2 = ∅. So we have used both the fact that C2 has empty interior and the fact that it is closed. Now, proceed, inductively. Assume that there is a finite sequence pi, i = 1, . . . , k and positive numbers 0 < �k < �k−1/3 < �k−2/32 < < �1/3k−1 < 3−k such that pj ∈ B(pj−1, �j−1/3) and · · · B(pj , �j ) ∩ Cj = ∅. Then we can add another pk+1 by using the properties of Ck

– it has non-empty interior so there is some point in B(pk, �k/3) which is not in Ck+1 and then B(pk+1, �k+1) ∩ Ck+1 = ∅ where �k+1 > 0 but �k+1 < �k/3. Thus, we have constructe and infinite sequence {pk} in M. Since d(pk+1, pk) < �k/3 this is a Cauchy sequence. In fact

(13.2) d(pk, pk+l) < �k/3 + + �k+l−1/3 < 3−k < 2�k/3· · ·

for all l > 0, and this tends to zero as k →∞. Since M is complete this sequence converges. From (13.2) the limit, q ∈ M

must lie in the closure of B(pk, 2�k/3) for every k. Hence q /∈ Ck for any k which contradicts (13.1).

Thus, at least one of the Cn must have non-empty interior. �

One application of this is often called the uniform boundedness principle, I will just call it:

Theorem 9 (Uniform boundedness). Let B be a Banach space and suppose that Tn is a sequence of bounded (i.e. continuous) linear operators Tn : B −→ V where V is a normed space. Suppose that for each b ∈ B the set {Tn(b)} ⊂ V is bounded (in norm of course) then supn �Tn� < ∞.

Proof. This follows from a pretty direct application of Baire’s theorem to B. Consider the sets

(13.3) Sp = {b ∈ B, �b� ≤ 1, �Tnb�V ≤ p ∀ n}, p ∈ N.

Each Sp is closed because Tn is continuous, so if bk b is a convergent sequence →then �b� ≤ 1 and �Tn(p)� ≤ p. The union if the Sp is the whole of the closed ball of radius one around the origin in B : (13.4) {b ∈ B; d(b, 0) ≤ 1} = Sp

p

because of the assumption of ‘pointwise boundedness’ – each b with �b� ≤ 1 must be in one of the Sp’s.


So, by Baire’s theorem one of the sets Sp has non-empty interior. This means that for some p, some v ∈ Sp, and some δ > 0,

(13.5) w ∈ B, �w�B ≤ δ =⇒ �Tn(v + w)�V ≤ p ∀ n. Moving v to (1 − δ/2)v and halving δ as necessary it follows that this ball B(v, δ) is contained in the open ball around the origin of radius 1. Thus, using the triangle inequality, and the fact that �Tn(v)�V ≤ p this implies

(13.6) w ∈ B, �w�B ≤ δ =⇒ �Tn(w)�V ≤ 2p =⇒ �Tn� ≤ 2p/δ

since the norm of the operator is sup{�Tw�V ; �w�B = 1� it follows that the norms are uniformly bounded:

(13.7) �Tn� ≤ 2p/δ

as claimed. �

One immediate consequence of this is that, as I mentioned in last lecture, it is not necessary to assume that a weakly convergent sequence in a Hilbert space is norm bounded.

Corollary 2. If un ∈ H is a sequence in a Hilbert space and for all v ∈ H

(13.8) (un, v) F (v) converges in C→

then �un�H is bounded and there exists w ∈ H such that un � w (converges weakly).

Proof. Well, a corollary really should not need a proof but still I will give one since maybe it is a bit more than a corollary.

Apply the Uniform Boundedness Theorem to the continuous functionals

(13.9) Tn(u) = (u, un), Tn : H −→ C

where we reverse the order to make them linear rather than anti-linear. Thus, each set |Tn(u)| is bounded in C since it is convergent. It follows that there is a bound

(13.10) �Tn� ≤ C.

However, the norm is just �Tn� = �un�H so the sequence must be bounded in H. Define T : H −→ C as the limit for each u :

(13.11) T (u) = lim Tn(u) = lim (u, un). n→∞ n→∞

This exists for each u by hypothesis. It is a linear map an from (13.10) it is bounded, �T � ≤ C. Thus by the Riesz Representation theorem, there exists w ∈ H such that

(13.12) T (u) = (u,w) ∀ u ∈ H.

Thus (un, u) → (w, u) for all u ∈ H so un � w as claimed. �

The second major application of Baire’s theorem is to

Theorem 10 (Open Mapping). If T : B1 −→ B2 is a bounded and surjective linear map between two Banach spaces then T is open:

(13.13) T (O) ⊂ B2 is open if O ⊂ B1 is open.

This is ‘wrong way continuity’ and as such can be used to prove the continuity of inverse maps as we shall see. The proof uses Baire’s theorem but then another similar sort of argument is needed. I did not finish the second argument in the lecture.

�

� � �

�


Proof. (1) The first part, of the proof, using Baire’s theorem shows that the closure of the image, so in B2, of an open ball around the origin in B1, has 0 as an interior point – i.e. it contains an open ball around the origin in B2. To see this we apply Baire’s theorem to the sets

(13.14) Cp = clB2 T (B(0, p))

the closure of the image of the ball in B1 or radius p. We need to take the closure since the sets in Baire’s theorem are closed, but even before doing that we know that

(13.15) B2 = T (B(0, p)) p

since that is what surjectivity means – every point is the image of something. Thus one of the closed sets Cp has an interior point, v. Since T is surjective, v = Tu for some u ∈ B1. The sets Cp increase with p so we can take a larger p and v is still an interior point, from which it follows that 0 = v − Tu is an interior point as well. Thus indeed

(13.16) Cp ⊃ B(0, δ)

for some δ > 0. (2) Having applied Baire’s thereom, consider now what (13.16) means. It fol

lows that each v ∈ B2 with �v� < δ is the limit of a sequence Tun where �un� ≤ p. What we want to arrange is that this sequence converges. Note that we can scale the norm of v using the linearity of T. Thus, for a general v ∈ B2 we can apply (13.16) to v� = δv/2�v� to see that Tu� v�

where �u�n� ≤ p. Then un = �v�u�n/δ satisfies Tun → v, �un� ≤ 2pn �v→�/δ.

To simplifiy the arithmetic, let me replace T by cT where c = p/2δ. This means that for each v ∈ B2 there is a sequence un in B1 with �un� ≤ �v�and Tun v. →

Now, we can stop before we get to the limit of the sequence and get as close to v as we want. This means that

1(13.17) For each v ∈ B2, ∃ u ∈ B1, �u� < �v�, �v − Tu� ≤

2 �v�.

This in turn we can iterate to construct a better sequence. Fix w = w1 ∈ B1

with �w1� < 1 and choose u1 = u according to (13.17) for v = w = w1. Thus �u1� < 1 and w2 = w1 − Tu1 satisfies �w2� < 1

2 . Now proceed by induction, supposing that we have constructed a sequence uj , j < n, in B1

with �uj � ≤ 2−j+1 and �wj1� < 2−j+1 where wj = wj−1 − Tuj−1. Then we can choose un to extend the induction and so we get a sequence un such that for each n

n n n

(13.18) w − T ( ) = w1 − Tu1 − = w2 − Tu2 − = wn+1. j=1 j=2 j=3

The series with terms un is absolutely summabel, hence convergent since B1 is complete, and

(13.19) w = Tu, u = uj , �u� ≤ 2. j

�

��

��

�


So finally we have shown that each w ∈ B(0, 1) in B2 is in the image of B(0, 2) in B1. Going back to the unscaled T it follows that for some δ > 0,

(13.20) B(0, δ) ⊂ T (B(0, 1)).

(3) It follows of course that the image T (O) of any open set is open, since if w ∈ T (O) then w = Tu for some u ∈ O and hence B(w, �δ) is contained in the image of u + B(0, �) ⊂ O for � > 0 sufficiently small.

So, as I did not quite finish the proof in lecture. However at the very end I mention the two most important applications of of this ‘Open Mapping Theorem’. Namely:

Corollary 3. If T : B1 −→ B2 is a bounded linear map between Banach spaces which is 1-1 and onto, i.e. is a bijection, then it is a homeomorphism – meaning its inverse, which is necessarily linear, is also bounded.

Proof. The only confusing thing is the notation. Note that T −1 is used to denote the inverse maps on sets. So, the inverse of T, let’s call it S : B2 −→ B2 is certainly linear. If O ⊂ B1 is open then S−1(O) = T (O) is open by the Open Mapping theorem, so S is continuous. �

The second application is

Theorem 11 (Closed Graph). If T : B1 −→ B2 is a linear map between Banach spaces then it is bounded if and only if its graph

(13.21) Gr(T ) = {(u, v) ∈ B1 × B2; u2 = Tu1}is a closed subset of the Banach space B1 × B2.

Have we actually covered the product of Banach spaces explicitly? If not, think about it for a minute or two!

Proof. Suppose first that T is bounded, i.e. continuous. A sequence (un, vn) ∈B1 × B2 is in Gr(T ) if and only if vn = Tun. So, if it converges, then un → u and vn = Tun Tv by the continuity of T, so the limit is in Gr(T ) which is therefore→closed.

Conversely, suppose the graph is closed. Given the graph we can reconstruct the map it comes from (whether linear or not) from a little diagram. Form B1 × B2

consider the two projections, π1(u, v) = u and π2(u, v) = v. Both of them are continuous by inspection and we can restrict them to Gr(T ) ⊂ B1 × B2 to get

(13.22) Gr(T ) π1 � �π2

��

TB1

�� B2.

This little diagram commutes. Indeed there are two ways to map a point (u, v) ∈Gr(T ) to B2, either directly, sending it to v or first sending it u ∈ B1 and then to Tu. Since v = Tu these are the same.

Now, Gr(T ) ⊂ B1 × B2 is a closed subspace, so it too is a Banach space and π1

and π2 remain continuous when restricted to it. The map π1 is 1-1 and onto, because each u occurs as the first element of precisely one pair, namely (u, Tu) ∈ Gr(T ).


Thus the Corollary above applies to π1 to show that its inverse, S is continuous. But then T = π2 ◦ S, from the commutativity, is also continuous proving the theorem. �





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�

�


Lecture 14. Tuesday, March 31: Fourier series and L2(0, 2π).

Fourier series. Let us now try applying our knowledge of Hilbert space to a concrete Hilbert space such as L2(a, b) for a finite interval (a, b) ⊂ R. You showed that this is indeed a Hilbert space. One of the reasons for developing Hilbert space techniques originally was precisely the following result.

Theorem 12. If u ∈ L2(0, 2π) then the Fourier series of u,

(14.1)1 �

cke ikx , ck =

� u(x)e−ikxdx

2π k∈Z (0,2π)

converges in L2(0, 2π) to u.

Notice that this does not say the series converges pointwise, or pointwise almost everywhere since this need not be true – depending on u. We are just claiming that

1 � (14.2) lim |u(x) −

2πcke

ikx|2 = 0 n→∞

|k|≤n

for any u ∈ L2(0, 2π). First let’s see that our abstract Hilbert space theory has put us quite close to

proving this. First observe that if e�k(x) = exp(ikx) then these elements of L2(0, 2π) satisfy � � 2π

� 0 if k = j

(14.3) e�ke�j = exp(i(k − j)x) =

�0 2π if k = j.

Thus the functions e�k 1 ikx (14.4) ek = = e �e�k�

√2π

form an orthonormal set in L2(0, 2π). It follows that (14.1) is just the Fourier-Bessel series for u with respect to this orthonormal set:

(14.5) ck = √

2π�u, ek� =1 cke

ikx = �u, ek�ek.⇒ 2π

So, we alreay know that this series converges in L2(0, 2π) thanks to Bessel’s identity. So ‘all’ we need to show is

Proposition 21. The ek, k ∈ Z, form an orthonormal basis of L2(0, 2π), i.e. are complete:

(14.6) ue ikx = 0 ∀ k = ⇒ u = 0 in L2(0, 2π).

This however, is not so trivial to prove. An equivalent statement is that the finite linear span of the ek is dense in L2(0, 2π). I will prove this using Fejer’s method. In this approach, we check that any continuous function on [0, 2π] satisfying the additional condition that u(0) = u(2π) is the uniform limit on [0, 2π] of a sequence in the finite span of the ek. Since uniform convergence of continuous functions certainly implies convergence in L2(0, 2π) and we already know that the continuous functions which vanish near 0 and 2π are dense in L2(0, 2π) (I will recall why later) this is enough to prove Proposition 21. However the proof is a serious piece of analysis, at least it is to me!

�

�

� �

�


So, the problem is to find the sequence in the span of the ek. Of course the trick is to use the Fourier expansion that we want to check. The idea of Cesaro is to make this Fourier expansion ‘converge faster’, or maybe better. For the moment we can work with a general function u ∈ L2(0, 2π) – or think of it as continuous if you prefer. So the truncated Fourier series is

1 � � u(t)e−iktdt)e ikx (14.7) Un(x) = (

2π (0,2π)|k|≤n

where I have just inserted the definition of the ck’s into the sum. This is just a finite sum so we can treat x as a parameter and use the linearity of the integral to write this as

(14.8) Un(x) = Dn(x − t)u(t), Dn(s) = 21 π

� e iks .

(0,2π) |k|≤n

Now this sum can be written as an explicit quotient, since, by telescoping, 12

12 )s(14.9) (2π)Dn(s)(eis/2 is/2) = e i(n+− e )s − e−i(n+ .

So in fact, at least where s = 0, 12

12i(n+ )s − e−i(n+

2π(eis/2 − e−is/2)

)se(14.10) Dn(s) =

and of course the limit as s 0 exists just fine. →As I said, Cesaro’s idea is to speed up the convergence by replacing Un by its

average n1 �

(14.11) Vn(x) = Ul. n + 1

l=0

Again plugging in the definitions of the Ul’s and using the linearity of the integral we see that � n1 � (14.12) Vn(x) = Sn(x − t)u(t), Sn(s) =

n + 1 Dl(s).

(0,2π) l=0

So again we want to compute a more useful form for Sn(s) – which is the Fejer kernel. Since the denominators in (14.10) are all the same,

n n1 2

1 2 )sis/2 − e−is/2)Sn

i(n+(s) = e )s e−i(n+(14.13) 2π(n + 1)(e − . l=0 l=0

Using the same trick again, n

1 2 )s i(n+1)s= e − 1is/2 − e−is/2) i(n+e(14.14) (e

l=0

so

(14.15) 2π(n + 1)(eis/2 − e−is/2)2Sn(s) = e i(n+1)s + e−i(n+1)s − 2 =⇒

sin2( (n+1)1 2 s)Sn(s) = .

n + 1 2π sin2( 2 s )

�

�

�

�


Now, what can we say about this function? One thing we know immediately is that if we plug u = 1 into the disucssion above, we get Un = 1 for n ≥ 0 and hence Vn = 1 as well. Thus in fact

(14.16) Sn(x − ·) = 1. (0,2π)

Now looking directly at (14.15) the first thing to notice is that Sn(s) ≥ 0. Also, we can see that the denominator only vanishes when s = 0 or s = 2π in [0, 2π]. Thus if we stay away from there, say s ∈ (δ, 2π − δ) for some δ > 0 then – since sin is a bounded function

(14.17) |Sn(s)| ≤ (n + 1)−1Cδ on (δ, 2π − δ).

Now, we are interested in how close Vn(x) is to the given u(x) in supremum norm, where now we will take u to be continuous. Because of (14.16) we can write

(14.18) u(x) = Sn(x − t)u(x) (0,2π)

where t denotes the variable of integration (and x is fixed in [0, 2π]). This ‘trick’ means that the difference is

(14.19) Vn(x) − u(x) = Sx(x − t)(u(t) − u(x)). (0,2π)

For each x we split this integral into two parts, the set Γ(x) where x − t ∈ [0, δ] or x − t ∈ [2π − δ, 2π] and the remainder. So (14.20) � � |Vn(x) − u(x)| ≤

Γ(x) Sx(x − t)|u(t) − u(x)| +

(0,2π)\Γ(x) Sx(x − t)|u(t) − u(x)|.

Now on Γ(x) either |t − x| ≤ δ – the points are close together – or t is close to 0 and x to 2π so 2π − x + t ≤ δ or conversely, x is close to 0 and t to 2π so 2π − t + x ≤ δ. In any case, by assuming that u(0) = u(2π) and using the uniform continuity of a continuous function on [0, 2π], given � > 0 we can choose δ so small that

(14.21) |u(x) − u(t)| ≤ �/2 on Γ(x).

On the complement of Γ(x) we have (14.17) and since u is bounded we get the estimate

(14.22) |Vn(x)−u(x)| ≤ �/2 Γ(x)

Sn(x−t)+(n+1)−1C �(δ) ≤ �/2+(n+1)−1C �(δ).

Here the fact that Sn is non-negative and has integral one has been used again to estimate the integral of Sn(x − t) over Γ(x) by 1. Thus, having chosen δ to make the first term small, we can choose n large to make the second term small and it follows that

(14.23) Vn(x) → u(x) uniformly on [0, 2π] as n →∞

under the assumption that u ∈ C([0, 2π]) satisfies u(0) = u(2π). So this proves Proposition 21 subject to the density in L2(0, 2π) of the continuous

functions which vanish near (but not of course in a fixed neighbourhood) of the ends.

��

� �

�


In fact we know that the L2 functions which vanish near the ends are dense since we can chop of and use the fact that

(14.24) δlim

0 (0,δ) |f |2 +

(2π−δ,2π) |f |2 = 0.

→

The L2 functions which vanish near the ends are in the closure of the span of the step functions which vanish near the ends. Each such step function can be approximated in L2((0, 2π)) by a continuous function which vanishes near the ends so we are done as far as density is concerned. So we have proved Theorem 12.

Problem set 7, Due 11AM Tuesday 7 Apr.

I will put up some practice problems for the test next Thursday when I get a chance.

Problem 7.1 Question:- Is it possible to show the completeness of the Fourier basis

exp(ikx)/√

2π

by computation? Maybe, see what you think. These questions are also intended to get you to say things clearly.

(1) Work out the Fourier coefficients ck(t) = (0,2π) fte

−ikx of the step function

(14.25) ft(x) = 1 0 ≤ x < t

0 t ≤ x ≤ 2π

for each fixed t ∈ (0, 2π). (2) Explain why this Fourier series converges to ft in L2(0, 2π) if and only if

(14.26) 2 |ck(t)|2 = 2πt − t2 , t ∈ (0, 2π). k>0

(3) Write this condition out as a Fourier series and apply the argument again to show that the completeness of the Fourier basis implies identities for the sum of k−2 and k−4 .

(4) Can you explain how reversing the argument, that knowledge of the sums of these two series should imply the completeness of the Fourier basis? There is a serious subtlety in this argument, and you get full marks for spotting it, without going ahead a using it to prove completeness.

Problem 7.2 Prove that for appropriate constants dk, the functions dk sin(kx/2), k ∈ N, form an orthonormal basis for L2(0, 2π).

Hint: The usual method is to use the basic result from class plus translation and rescaling to show that d� exp(ikx/2) k ∈ Z form an orthonormal basis of k L2(−2π, 2π). Then extend functions as odd from (0, 2π) to (−2π, 2π).

Problem 7.3 Let ek, k ∈ N, be an orthonormal basis in a separable Hilbert space, H. Show that there is a uniquely defined bounded linear operator S : H −→ H, satisfying

(14.27) Sej = ej+1 ∀ j ∈ N. Show that if B : H −→ H is a bounded linear operator then S +�B is not invertible if � < �0 for some �0 > 0.

Hint:- Consider the linear functional L : H −→ C, Lu = (Bu, e1). Show that B�u = Bu − (Lu)e1 is a bounded linear operator from H to the Hilbert space


H1 = {u ∈ H; (u, e1) = 0}. Conclude that S + �B� is invertible as a linear map from H to H1 for small �. Use this to argue that S + �B cannot be an isomorphism from H to H by showing that either e1 is not in the range or else there is a non-trivial element in the null space.

Problem 7.4 Show that the product of bounded operators on a Hilbert space is strong continuous, in the sense that if An and Bn are strong convergent sequences of bounded operators on H with limits A and B then the product AnBn is strongly convergent with limit AB.

Hint: Be careful! Use the result in class which was deduced from the Uniform Boundedness Theorem.


Solutions to Problems 6

Hint: Don’t pay too much attention to my hints, sometimes they are a little off-the-cuff and may not be very helpfult. An example being the old hint for Problem 6.2!

Problem 6.1 Let H be a separable Hilbert space. Show that K ⊂ H is compact if and only if it is closed, bounded and has the property that any sequence in K which is weakly convergent sequence in H is (strongly) convergent.

Hint:- In one direction use the result from class that any bounded sequence has a weakly convergent subsequence.

Problem 6.2 Show that, in a separable Hilbert space, a weakly convergent sequence {vn}, is (strongly) convergent if and only if the weak limit, v satisfies

(14.28) �v�H = lim n→∞

�vn�H .

Hint:- To show that this condition is sufficient, expand

(14.29) (vn − v, vn − v) = �vn�2 − 2 Re(vn, v) + �v�2 .

Problem 6.3 Show that a subset of a separable Hilbert space is compact if and only if it is closed and bounded and has the property of ‘finite dimensional approximation’ meaning that for any � > 0 there exists a linear subspace DN ⊂ H of finite dimension such that

(14.30) d(K,DN ) = sup inf {d(u, v)} ≤ �. u∈K v∈DN

Hint:- To prove necessity of this condition use the ‘equi-small tails’ property of compact sets with respect to an orthonormal basis. To use the finite dimensional approximation condition to show that any weakly convergent sequence in K is strongly convergent, use the convexity result from class to define the sequence {vn

� }in DN where vn

� is the closest point in DN to vn. Show that vn� is weakly, hence

strongly, convergent and hence deduce that {vn} is Cauchy. Problem 6.4 Suppose that A : H −→ H is a bounded linear operator with the

property that A(H) ⊂ H is finite dimensional. Show that if vn is weakly convergent in H then Avn is strongly convergent in H.

Problem 6.5 Suppose that H1 and H2 are two different Hilbert spaces and A : H1 −→ H2 is a bounded linear operator. Show that there is a unique bounded linear operator (the adjoint) A∗ : H2 −→ H1 with the property

(14.31) �Au1, u2�H2 = �u1, A∗u2�H1 ∀ u1 ∈ H1, u2 ∈ H2.





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�


Lecture 15. Thursday, April 2

I recalled the basic properties of the Banach space, and algebra, of bounded operators B(H) on a separable Hilbert space H. In particular that it is a Banach space with respect to the norm

(15.1) �A� = sup �u�H=1

�Au�H

and that the norm satisfies

(15.2) �AB� ≤ �A��b�.

Restatated and went through the proof again of the

Theorem 13 (Open Mapping). If A : B1 −→ B2 is a bounded linear operator between Banach spaces and A(B1) = B2, i.e. A is surjective, then it is open:

(15.3) A(O) ⊂ B2 is open ∀ O ⊂ B1 open.

Proof in Lecture 13, also the two consequences of it: If A : B1 −→ B2 is bounded, 1-1 and onto (so it is a bijection) then its inverse is also bounded. Secondly the closed graph theorem. All this is in the notes for Lecture 13.

As a second example of the Uniform Boundedness Theorem I also talked about strong convergence of operators. Thus a sequence of bounded operators (on a separable Hilbert space) An ∈ B(H) is said to converge strongly if for each u ∈ H Anu converges. It follows that the limit is a bounded linear operator – or you can include this in the definition if you prefer. The Uniform Boundedness Theorem shows that if An is strongly convergent then it is bounded, supn �An� < ∞. You will need this for the problems this week.

I also talked about the shift operator S : l2 −→ l2 defined by �∞ ∞

(15.4) S( cj ej ) = cj ej+1

j=1 j=1

defined by moving each element of the sequence ‘up one’ and starting with zero. This is an example of a bounded linear operator, with �S� = 1 clearly enough, which is 1-1, since Au = 0 implies u = 0, but which is not surjective. Indeed the range of S is exactly the subspace

(15.5) H1 = {u ∈ L2; (u, e1) = 0}.

Using the open mapping theorem (or directly) it is easy to see that S is invertible as a bounded linear map from H to H1, but not on H. In fact as you should show in the problem set this week, it cannot be made invertible by a small perturbation. This shows in particular that the set of invertible elements of B(H) is not dense, which is quite different from the finite dimensional case.

Finally I started to talk about the set of invertible elements:

(15.6) GL(H) = {A ∈ B(H); ∃ B ∈ H(H), BA = AB = Id}.

Note that this is equivalent to saying A is 1-1 and onto in view of the discussion above.

Lemma 10. If A ∈ B(H) and �A� < 1 then

(15.7) Id −A ∈ GL(H).

�

�

� �


Proof. Neumann series. If �A� < 1 then �Aj � ≤ �A�j and it follows that the Neumann series

(15.8) B = Aj

j

1

is absolutely summable in B(H) sicne �Aj � converges. Thus the sum converges. j=0

Moreover by the continuity of the product with respect to the norm n n+1

(15.9) AB = A lim Aj = lim Aj = B − Id n→∞

j=0 n→∞

j=1

an similarly BA = B − Id . Thus (Id −A)B = B(Id −A) = Id shows that B is a (and hence the) 2-sided inverse of Id −A. �

Proposition 22. The group of invertible elements GL(H) ⊂ B(H) is open (but not dense if H is infinite-dimensional).

Proof. I will do the proof next time. �





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Lecture 16. Tuesday, April 7: partially reconstructed

From last time

Proposition 23. The invertible elements form an open subset GL(H) ⊂ B(H).

Proof. Recall that we showed using the convergence of the Neumann series that if B ∈ B(H) and �B� < 1 then Id −B is invertible, meaning it has a two-sided inverse in B(H) (which we know, from the open mapping Theorem to be equivalent to it being a bijection).

So, suppose G ∈ GL(H), meaning it has a two-sided (and unique) inverse G−1 ∈ B(H) :

(16.1) G−1G = GG−1 = Id .

Then we wish to show that B(G; �) ⊂ GL(H) for some � > 0. In fact we shall see that we can take � = �G−1�−1 . The idea is that we wish to show that G + B is a bijection, and hence invertible. To do so set

(16.2) E = G−1B = G + B = G−1(Id +G−1B).⇒

This is injective if Id +G−1B is injective, and surjective if Id +G−1B is surjective, since G−1 is a bijection. From last time we know that

(16.3) �G−1B� < 1 =⇒ Id +G−1B is invertible.

Since �G−1B� ≤ �G−1��B� this follows if �B� < �G−1�−1 as anticipated. �

Thus GL(H) ⊂ B(H), the set of invertible elements, is open. It is also a group – since the inverse of G1G2 if G1, G2 ∈ GL(H) is G−1G−1 .2 1

This group of invertible elements has a smaller subgroup, U(H), the unitary group, defined by

(16.4) U(H) = {U ∈ GL(H); U−1 = U∗}.

The unitary group consists of the linear isometric isomorphisms of H onto itself – thus

(16.5) (Uu,Uv) = (u, v), �Uu� = �u� ∀ u, v ∈ H, U ∈ U(H).

This is an important object and we will use it a little bit later on. The unitary group on a separable Hilbert space may seem very similar to the

familiar unitary group of n × n matrices, U(n). It is, of course it is much bigger for one thing. In fact there are some other important differences which I will describe a little later on (or get you to do some of it in the problems). On important fact that you should know, even though I will not prove it, is that U(H) is contractible as a metric space – it has no significant topology. This is to be constrasted with the U(n) which have a lot of topology, and not at all simple spaces – especially for large n. One upshot of this is that U(H) does not look much like the limit of the U(n) as n →∞.

Now, for the rest of today I will talk about the opposite of the ‘big’ operators such as the elements of GL(H).

Definition 7. An operator T ∈ B(H) is of finite rank if its range has finite dimension (and that dimension is called the rank of T ); the set of finite rank operators is denoted R(B).

�

�

� �


Why not F(B)? Because we want to use this for the Fredholm operators. Clearly the sum of two operators of finite rank has finite rank, since the range

is contained in the sum of the ranges (but is often smaller):

(16.6) (T1 + T2u) ∈ Ran(T1) + Ran(T2).

Since the range of a constant multiple of T is contained in the range of T it follows that the finite rank operators form a linear subspace of B(H).

It is also clear that

(16.7) B ∈ B(H) and T ∈ R(B) then BT ∈ R(B).

Indeed, the range of BT is the range of B restricted to the range of T and this is certainly finite dimensional since it is spanned by the image of a basis of Ran(T ). Similalry TB ∈ R(H) since the range of TB is contained in the range of T. Thus we have in fact proved most of

Proposition 24. The finite rank operators form a ∗-closed ideal in B(H), which is to say a linear subspace such that

(16.8) B1, B2 ∈ B(H), T ∈ R(H) =⇒ B1TB2, T ∗ ∈ R(H).

Proof. In fact it is only the fact that T ∗ is of finite rank if T is which remains to be checked. To do this let us find an explicit representation for an operator of finite rank. First, since Ran(T ) is finite dimensional, we can choose a basis, fi

i = 1, . . . , N, for it. Then for any element u ∈ H, N

(16.9) Tu = cifi. i=1

The constants ci are determined, since the fi are a basis, and so define linear functionals u �−→ ci. These are continuous. In fact we can simply choose the fi to be orthonormal and then, pairing (16.9) with fj we see that

(16.10) cj = (Tu, fj ) = (u, T ∗fj ).

In particular there are elements (really by Riesz’ theorem) ei = T ∗fi ∈ H sucht that

N

(16.11) Tu = (u, ei)fi. i=1

Conversely, if T can be written in the form (16.11) then it is of finite rank, since its range is contained in the span of the fi.

From (16.11) it follows that T ∗ is also of finite rank since

N N

(16.12) (T ∗v, u) = (v, Tu) = (v, fi)(ei, u) ∀ u ∈ H = ⇒ T ∗v = (u, fi)ei. j=1 i=1

The roles of the fi and ei are simply interchanged. �

Next time I will show that the closure of the ideal R(H) in B(H) is the ideal of compact operators. Of course this closure is certainly closed(!) Moreover it is a ∗-closed ideal, since Tn → K in norm and B1, B2 ∈ B(H) implies

(16.13) B1TnB2 → B1KB2, Tn ∗ → K∗.

�

� �

�


So, once we prove that the compact operators are the closure of the finite rank operators we will know that they form a closed, ∗-ideal.

Notice that the importance of the ideal condition – it is the analogue of the normal condition for a subgroup – is that the quotient B/I of the algebra by an ideal is again an algebra. The quotient by the ideal, K(H), of compact operators is a Banach space since K is closed. It is called the Calkin algebra.

Lemma 11 (Row rank=Colum rank). For any finite rank operator on a Hilbert space, the dimension of the range of T is equal to the dimension of the ranfe of T ∗.

Proof. We showed that a finite rank operator T always takes the form (16.11). If the fi are taken to be a basis for the range of T, so N = dim Ran(T ), then the ei

must be linearly independent. Indeed, if not then one of the ei can be replaced by a linear combination ei = cj ej . Inserting this into (16.11) shows that

j=i

(16.14) Tu = (u, ej )(fj + cj fi) j=i

from which it follows that the range has dimension at most N −1 – which contradicts the choice of the fi.

Since the ei are independent it follows from (16.12) that the range of T ∗ has dimension N (since the fi are independent) – if you like just say dim Ran(T ∗) ≤ N for all finite rank T and then use the fact that (T ∗)∗ = T to deduce equality. �

�


Problem set 8, Due 11AM Tuesday 14 April.

Okay, I forgot to put the problems up. So, here are three problems that should be reasonably quick. If anyone is seriously inconvenienced by the limited time they have to work on them, just let me know and I will give you a couple of days.

Problem 8.1 Show that a continuous function K : [0, 1] −→ L2(0, 2π) has the property that the Fourier series of K(x) ∈ L2(0, 2π), for x ∈ [0, 1], converges uniformly in the sense that if Kn(x) is the sum of the Fourier series over kthen Kn : [0, 1] −→ L2(0, 2π) is also continuous and

| | ≤ n

(16.15) sup �K(x) − Kn(x)�L2(0,2π) → 0. x∈[0,1]

Hint. Use one of the properties of compactness in a Hilbert space that you proved earlier.

Problem 8.2 Consider an integral operator acting on L2(0, 1) with a kernel which is continuous

– K ∈ C([0, 1]2). Thus, the operator is

(16.16) Tu(x) = K(x, y)u(y). (0,1)

Show that T is bounded on L2 (I think we did this before) and that it is in the norm closure of the finite rank operators.

Hint. Use the previous problem! Show that a continuous function such as K in this Problem defines a continuous map [0, 1] � x �−→ K(x, ) ∈ C([0, 1]) and hence ·a continuous function K : [0, 1] −→ L2(0, 1) then apply the previous problem with the interval rescaled.

Here is an even more expanded version of the hint: You can think of K(x, y) as a continuous function of x with values in L2(0, 1). Let Kn(x, y) be the continuous function of x and y given by the previous problem, by truncating the Fourier series (in y) at some point n. Check that this defines a finite rank operator on L2(0, 1) – yes it maps into continuous functions but that is fine, they are Lebesgue square integrable. Now, the idea is the difference K − Kn defines a bounded operator with small norm as n becomes large. It might actually be clearer to do this the other way round, exchanging the roles of x and y.

Problem 8.3 Although we have concentrated on the Lebesgue integral in one variable, you proved at some point the covering lemma in dimension 2 and that is pretty much all that was needed to extend the discussion to 2 dimensions. Let’s just assume you have assiduously checked everything and so you know that L2((0, 2π)2) is a Hilbert space. Sketch a proof – noting anything that you are not sure of – that the functions exp(ikx + ily)/2π, k, l ∈ Z, form a complete orthonormal basis.

�

�



Problem 7.1 Question:- Is it possible to show the completeness of the Fourier basis

exp(ikx)/√

2π

by computation? Maybe, see what you think. These questions are also intended to get you to say things clearly.

(1) Work out the Fourier coefficients ck(t) = (0,2π) fte

−ikx of the step function

(2) Explain why this Fourier series converges to ft in L2(0, 2π) if and only if

�

(16.17) ft(x) = 1

0

0 ≤ x < t

t ≤ x ≤ 2π

for each fixed t ∈ (0, 2π).

(16.18) 2 |ck(t)|2 = 2πt − t2 , t ∈ (0, 2π). k>0

(3) Write this condition out as a Fourier series and apply the argument again to show that the completeness of the Fourier basis implies identities for the sum of k−2 and k−4 .

(4) Can you explain how reversing the argument, that knowledge of the sums of these two series should imply the completeness of the Fourier basis? There is a serious subtlety in this argument, and you get full marks for spotting it, without going ahead a using it to prove completeness.

Problem 7.2 Prove that for appropriate constants dk, the functions dk sin(kx/2), k ∈ N, form an orthonormal basis for L2(0, 2π).

Hint: The usual method is to use the basic result from class plus translation and rescaling to show that d� exp(ikx/2) k ∈ Z form an orthonormal basis of k L2(−2π, 2π). Then extend functions as odd from (0, 2π) to (−2π, 2π).

Problem 7.3 Let ek, k ∈ N, be an orthonormal basis in a separable Hilbert space, H. Show that there is a uniquely defined bounded linear operator S : H −→ H, satisfying

(16.19) Sej = ej+1 ∀ j ∈ N. Show that if B : H −→ H is a bounded linear operator then S +�B is not invertible if � < �0 for some �0 > 0.

Hint:- Consider the linear functional L : H −→ C, Lu = (Bu, e1). Show that B�u = Bu − (Lu)e1 is a bounded linear operator from H to the Hilbert space H1 = {u ∈ H; (u, e1) = 0}. Conclude that S + �B� is invertible as a linear map from H to H1 for small �. Use this to argue that S + �B cannot be an isomorphism from H to H by showing that either e1 is not in the range or else there is a non-trivial element in the null space.

Problem 7.4 Show that the product of bounded operators on a Hilbert space is strong continuous, in the sense that if An and Bn are strong convergent sequences of bounded operators on H with limits A and B then the product AnBn is strongly convergent with limit AB.

Hint: Be careful! Use the result in class which was deduced from the Uniform Boundedness Theorem.

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