Functional Analysis

July 12, 2007

Contents

1 Metric Spaces 3

1.1 Open Sets, Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Convergence, Cauchy Sequence, Completeness . . . . . . . . . . . . . . . . . 71.3 Completeness Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Completion of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Normed Spaces 13

2.1 Finite dimensional Normed Spaces or Subspaces . . . . . . . . . . . . . . . 152.2 Compactness and Finite Dimension . . . . . . . . . . . . . . . . . . . . . . . 182.3 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.4 Bounded and Continuous Linear Operators . . . . . . . . . . . . . . . . . . 242.5 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6 Finite Dimensional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.6.1 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.7 Dual Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3 Hilbert Spaces 39

3.1 Representation of Functionals on Hilbert Spaces . . . . . . . . . . . . . . . 493.2 Hilbert Adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4 Fundamental Theorems 59

4.1 Bounded, Linear Functionals on C[a, b] . . . . . . . . . . . . . . . . . . . . . 684.2 Adjoint Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.3 Reflexive Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.4 Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 784.5 Strong and Weak Convergence . . . . . . . . . . . . . . . . . . . . . . . . . 814.6 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 884.7 Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

1

5 Exam 1 93

6 Exam 2 95

7 Final Exam 96

2

1 Metric Spaces

Definition (Metric Space). The pair (X, d), where X is a set and the function

d : X × X → R

is called a metric space if

1. d(x, y) ≥ 0

2. d(x, y) = 0 iff x = y

3. d(x, y) = d(y, x)

4. d(x, y) ≤ d(x, z) + d(z, y)

Example 1.1 (Metric Spaces).

1. d(x, y) = |x − y| in R.

2. d(x, y) = [∑n

i=1(xi − yi)2]

12 in Rn.

3. d(x, y) = ‖x − y‖ in a normed space.

4. Let (X, ρ), (Y, σ) be metric spaces and define the Cartesian product X × Y ={(x, y)|x ∈ X, y ∈ Y }. Then the product measure τ((x1, y1), (x2, y2)) = [ρ(x1, x2)

2 +

σ(y1, y2)2]

12 .

5. (Subspace) (Y, d) of (X, d) if Y ⊂ X and d = d|Y ×Y .

6. l∞. Let X be the set of all bounded sequences of complex numbers, i.e., x =(ξi) and |ξi| ≤ cx, ∀ i. Then

d(x, y) = supi∈N

|ξi − ηi|

defines a metric on X.

7. X = C[a, b] andd(x, y) = max

t∈[a,b]|x(t) − y(t)|.

8. (Discrete metric)

d(x, y) =

{

1 x = y0 x 6= y

.

3

9. lp. x = (ξi) ∈ lp if∑∞

i=1 |ξi|p < ∞, (p ≥ 1, fixed),

d(x, y) =

( ∞∑

i=1

|ξi − ηi|p)

1p

.

Problem 1.

1. Show that d is a metric on C[a, b], where

d(x, y) =

∫ b

a

|x(t) − y(t)|dt.

2. Show that the discrete metric is a metric.

3. Sequence space s: set of all sequences of complex numbers with the metric

d(x, y) =

∞∑

i=1

1

2i

|ξi − ηi|1 + |ξi − ηi|

. (1)

Solution.

1. d(x, y) = 0 iff |x(t) − y(t)| = 0 for all t ∈ [a, b] because of the continuity. We haved(x, y) ≥ 0 and d(x, y) = d(y, x) trivially. We can argue the triangle inequality asfollows::

d(x, y) =

∫ b

a

|x(t)−y(t)|dt ≤∫ b

a

|x(t)−z(t)|dt+

∫ b

a

|z(t)−y(t)|dt = d(x, z)+ d(z, y).

2. Left as an exercise.

3. We show only the triangle inequality. Let a, b ∈ R. Then we have the inequalities

|a + b|1 + |a + b| ≤

|a| + |b|1 + |a| + |b| ≤

|a|1 + |a| +

|b|1 + |b| ,

where in the first step we have used the monotonicity of the function

f(x) =x

1 + x= 1 − 1

1 + x, for x > 0.

Substituting a = ξi − ζi and b = ζi − ηi, where x = (ξi), y = (ηi), and z = (ζi) we get

|ξi − ηi|1 + |ξi − ηi|

≤ |ξi − ζi|1 + |ξi − ζi|

+|ζi − ηi|

1 + |ζi − ηi|.

If we multiply both sides by 12i and sum over from i = 1 to ∞ we get the stated

result.

4

1.1 Open Sets, Closed Sets

Definition (Open Ball, Closed Ball, Sphere).

1. B(x0, r) = {x ∈ X|d(x, x0) < r}

2. B(x0, r) = {x ∈ X|d(x, x0) ≤ r}

3. S(x0, r) = {x ∈ X|d(x, x0) = r}

Definition (Open, Closed, Interior).

1. M is open if contains a ball about each of its points.

2. K ⊂ X is closed if Kc = X − K is open.

3. B(x0; ε) denotes the ε neighborhood of x0.

4. Int(M) denotes the interior of M .

Remark 1.2 (Induced Topology). Consider the set X with the collection τ of all opensubsets of X. Then we have

1. ∅ ∈ τ,X ∈ τ .

2. The union of any members of τ is a member of τ .

3. The finite intersection of members of τ is a member of τ .

We call the pair (X, τ) a topological space and τ a topology for X. It follows that a metricspace is a topological space.

Definition (Continuous). Let X = (X, d) and Y = (Y, d) be metric spaces. The mappingT : X → Y is continuous at x0 ∈ X if for every ε > 0 there is δ > 0 such that

d(Tx, Tx0) < ε , ∀ x such that d(x, x0) < δ.

Theorem 1.3 (Continuous Mapping). T : X → Y is continuous if and only if the inverseimage of any open subset of Y is an open subset of X.

Proof.

1. Suppose that T is continuous. Let S ⊂ Y be open S0 the inverse image of S. LetS0 6= ∅ and take x0 ∈ S0. We have Tx0 = y0 ∈ S. Since S is open there exists anε-neighborhood of y0, say N ⊂ S such that y0 ∈ N . The continuity of T impliesthat x0 has a δ-neighborhood N0 which is mapped into N . Since N ⊂ S we get thatN0 ⊂ S0, and it follows that S0 is open.

5

2. Assume that the inverse image of every open set in Y is an open set in X. Then∀ x0 ∈ X, and N (ε-neighborhood of Tx0) the inverse image N0 of N is open.Therefore N0 contains a δ-neighborhood of x0. Thus T is continuous.

Some more definitions:

Definition (Accumulation Point). x ∈ M is said to be an accumulation point of M if∃ (xn) ⊂ M s.t. xn → x.

Definition (Closure). M is the closure of M .

Definition (Dense Set). M ⊂ X is in X dense if M = X.

Definition (Separable Space). X is separable if there is a countable subset which is densein X.

Remark 1.4.

1. If M is dense, then every ball in X contains a point of M .

2. R, C are separable.

3. A discrete metric space is separable if and only if it is countable.

Theorem 1.5. ℓ∞ is not separable.

Proof. Let y = (ηi) where ηi = 0, 1. There are uncountably many y’s. If we put small ballswith radius 1

3 at the y’s they will not intersect. It follows that if M ⊂ l∞ is dense in l∞,then M is uncountable. Therefore l∞ is not separable.

Problem 2. Show that lp, 1 ≤ p < ∞ is separable.

Solution. Let M the set of all sequences of the form x = (ξ1, ξ2, ..., ξn, 0, 0, ...), where n isany positive integer and the ξis are rational. M is countable. We argue that M is dense inlp as follows. Let y = (ηi) ∈ lp be arbitrary. Then for every ε > 0 there is an n such that

∞∑

i=n+1

|ηi|p <εp

2.

Since the rationals are dense in R, for each ηi there is a rational ξi close to it. Hence thereis an x ∈ M such that

n∑

i=1

|ηi − ξi| <εp

2.

It follows that d(y, x) < ε.

6

1.2 Convergence, Cauchy Sequence, Completeness

Definition (Convergent Sequence). We say that the sequence (xn) is convergent in themetric space X = (X, d) if

limn→∞

d(xn, x) = 0 for some x ∈ X.

We shall also use the notation xn → x.

Definition (Diameter). For a set M ⊂ X we define the diameter δ(M) by

δ(M) = supx,y∈M

d(x, y).

M is said to be bounded if δ(M) is finite.

Lemma 1.6.

1. A convergent sequence (xn) in X is bounded and its limit x is unique.

2. If xn → x and yn → y in X, then d(xn, yn) → d(x, y).

Problem 3. Prove Lemma 1.6.

Solution.

1. Suppose that xn → x. Then, taking ε = 1 we can find an N such that d(xn, x) < 1for all n > N . By the triangle inequality we have

d(xn, x) < 1 + max d(x1, x), d(x2, x), ..., d(xN , x).

Therefore (xn) is bounded. If xn → x and xn → z, then

0 ≤ d(x, z) ≤ d(xn, x) + d(xn, z) → 0 as n → ∞

and uniqueness of the limit follows.

2. We haved(xn, yn) ≤ d(xn, x) + d(x, y) + d(y, yn),

and henced(xn, yn) − d(x, y) ≤ d(xn, x) + d(yn, y).

Interchanging xn and x, yn and y, and multiplying by −1 we get

d(x, y) − d(xn, yn) ≤ d(xn, x) + d(yn, y).

Combining the two inequalities we get

|d(xn, yn) − d(x, y)| ≤ d(xn, x) + d(yn, y) → 0 as n → ∞.

7

Definition (Cauchy Sequence). (xn) in (X, d) is a Cauchy sequence if ∀ ε > 0 ∃ N =N(ε) s.t.

m,n > N ⇒ d(xm, xn) < ε.

Definition (Complete). X is complete if every Cauchy sequence in X converges in X.

Theorem 1.7. R, C are complete metric spaces.

Theorem 1.8. Every convergent sequence in a metric space is a Cauchy sequence.

Theorem 1.9. Let M ⊂ X = (X, d), X is a metric space and let M denote the closure ofM in X. Then

1. x ∈ M iff ∃ (xn) ∈ M s.t. xn → x.

2. M is closed iff xn ∈ M and xn → x imply that x ∈ M .

Theorem 1.10. Let X be a complete metric space and M ⊂ X. M is complete iff M isclosed in X.

Theorem 1.11. Let T be a mapping from (X, d) into (Y, d). T : X → Y is continuous atx0 ∈ X iff xn → x0 implies Txn → Tx0.

Problem 4.

1. Prove Theorem 1.10

2. Prove Theorem 1.11

Solution.

1. Let M be complete. Then for every x ∈ M there is a sequence (xn) which convergesto x. Since (xn) is Cauchy and M is complete xn → x ∈ M , therefore M is closed.

Conversely, let M be closed and (xn) be Cauchy in M . Then xn → x ∈ X, whichimplies x ∈ M , and therefore x ∈ M . Thus M is complete.

2. Assume that T is continuous at x0. Then for ε > 0 ∃ δ > 0 such that

d(x, x0) < δ implies d(Tx, Tx0) < ε.

Take a sequence (xn) such that xn → x0. Then ∃ N s.t. ∀ n > N we haved(xn, x0) < δ and hence

∀ n > N , d(Txn, Tx)) < ε.

Conversely, assume that xn → x) ⇒ Txn → Tx0 and show that T is continuousat x0. Otherwise, ∃ ε > 0 such that ∀ δ > 0, ∃ x 6= x0) satisfying d(x, x0) < δbut d(Tx, Tx0) ≥ ε. Let δ = 1

n. Then there is an xn such that d(xn, x0) < 1

nbut

d(Txn, Tx0) ≥ ε contrary to our assumption.

8

1.3 Completeness Proofs

Theorem 1.12. Rn, Cn are complete.

Theorem 1.13. ℓ∞ is complete.

Proof. Let (xm) be a Cauchy sequence in ℓ∞.

⇒ d(xm, xn) = supi

|x(m)i − x

(n)i | < ε, for m,n > N(ε).

⇒ for fixed i, the sequence (ξ(m)i ) is Cauchy in R, and therefore we have that ξ

(m)i →

ξi, for i = 1, 2, 3, ... Let x = (ξi). It follows easily that x ∈ ℓ∞ and xm → x.

Theorem 1.14. The space of convergent sequences x = (ξi) of complex numbers with themetric induced from ℓ∞ is complete.

Theorem 1.15. ℓp is complete if 1 ≤ p < ∞.

Theorem 1.16. C[a, b] is complete.

Proof. Let (xm) be a Cauchy sequence in C[a, b]. Then

∀ ε > 0 ∃ N s.t. for m,n > N ⇒ d(xm, xn) = maxt∈[a,b]

|xm(t) − xn(t)| < ε. (2)

⇒ for any fixed t0 ∈ [a, b] we have that |xm(t0) − xn(t0)| < ε. It follows that xm(t0) isCauchy and therefore xm(t0) → x(t0). Show that x(t) ∈ C[a, b] and xm → x. From (2)with n → ∞ we get

maxt∈[a,b]

|xm(t) − x(t)| ≤ ε.

Therefore xm(t) converges uniformly to x(t) on [a, b]. Since the xm’s are continuous on[a, b] and the convergence is uniform x(t) is continuous, and thus belongs to C[a, b].

Remark 1.17. Note that in C[a, b] convergence is uniform convergence; we also use theterminology uniform metric for the metric generated by the “sup”-norm.

Example 1.18 (Incomplete Metric Spaces).

1. Q

2. The polynomials

3. C[a, b] with the ‖ · ‖2 norm

9

1.4 Completion of Metric Spaces

Definition (Isometry). Let X = (X, d) and X = (X, d) be metric spaces. The mappingT : X → X is an isometry if d(Tx, Ty) = d(x, y). The metric spaces X and X are isometricif there is a bijective isometry of X onto X

Theorem 1.19. Let X = (X, d) be a metric space. Then there exists a complete metricspace X = (X, d) which has a subspace W that is isometric with X and is dense in X. Xis unique except for isometries.

Proof. This proof is divided into steps.

1. Construction of X. Let (xn) and (x′

n) be equivalent Cauchy sequences, i.e.,

limn→∞

d(xn, x′

n) = 0.

Define X to be the set of all equivalence classes x of Cauchy sequences. Define now

d(x, y) = limn→∞

d(xn, yn) where (xn) ∈ x and (yn) ∈ y. (3)

We show that limit in (3) exists and independent of the particular choice of therepresentatives. We have using the triangle inequality

d(xn, yn) ≤ d(xn, xm) + d(xm, ym) + d(ym, yn).

It follows thatd(xn, yn) − d(xm, ym) ≤ d(xn, xm) + d(ym, yn).

Exchanging the role of n and m we get

d(xm, ym) − d(xn, yn) ≤ d(xn, xm) + d(ym, yn).

The two inequality together yield

|d(xn, yn) − d(xm, ym)| ≤ d(xn, xm) + d(ym, yn).

It follows that (d(xn, yn)) is a Cauchy sequence in R and therefore it converges, i.e.,the limit in (3) exists.

We show now that the limit in (3) is independent of the particular choice of repre-sentatives. Let (xn), (x

′

n) ∈ x and (yn), (y′

n) ∈ y. Then

|d(xn, yn) − d(x′

n, y′

n)| ≤ d(xn, x′

n) + d(yn, y′

n) → 0 as n → ∞,

which implieslim

n→∞d(xn, yn) = lim

n→∞d(x

′

n, y′

n).

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2. Show d is a metric on X . Left as an exercise.

3. Construction of an isometry T : X → W ∈ X . With each b ∈ X we associate theclass b ∈ X which contains the constant Cauchy sequence (b, b, b, ...). This definesthe mapping T : X → W onto the subspace W = T (X) ∈ X. The mapping T isgiven by b → b = Tb, where (b, b, ...) ∈ b. According to (3) d(b, c) = d(b, c), so T isan isometry. T is onto W since T (X) = W . ( W and X are isometric.)

Show that W is dense in X. Let x ∈ X be arbitrary and let (xn) ∈ x. For everyε > 0 there is an N such that d(xn, xN ) < ε

2 , for n > N . Let (xN , xN , ...) ∈ xN .Then xN ∈ W , and by (3)

d(x, xN ) = limn→∞

d(xn, xN ) ≤ ε

2< ε.

4. Completeness of X. Let (xn) be an arbitrary Cauchy sequence in X. Since W isdense in X for every xn there is a zn ∈ W such that

d(xn, zn) <1

n.

It follows using the triangle inequality and the Cauchyness of (xn) that (zn) is Cauchy.Then (zn), where zn = T−1zn is Cauchy in X. Let x ∈ X be the class to which (zn)belongs. We have

d(xn, x) ≤ d(xn, zn) + d(zn, x) <1

n+ d(zn, x) < ε

for sufficiently large n.

5. Uniqueness of X. If (X, d) is another complete metric space with a subspace W densein X and isometric with X, then W is isometric with W and the distances on X andX must be the same. Hence X and X are isometric.

Problem 5. Show now that d is a metric on X.

Solution. Clearly, d(x, y) ≥ 0 because of the definition (3). Also, d(x, x) = 0, and d(x, y) =d(y, x), because of the properties of d.

d(x, y) ≤ d(x, z) + d(z, y)

because we haved(xn, yn) ≤ d(xn, zn) + d(zn, yn) for all n.

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Problem 6. A homeomorphism is a continuous bijective mapping T : X → Y whose inverseis continuous. If such mapping exists, then X and Y are homeomorphic.

1. Show that is X and Y are isometric, then they are homeomorphic.

2. Illustrate with an example that a complete and an incomplete metric space may behomeomorphic.

Solution. 1. If f : (X, d) → (Y, d) is an isometry, then d(f(x), f(y)) = d(x, y) < ǫwhenever d(x, y) < δ, so f is continuous. If (X, d) and (Y, d) are isometric, thenthere exists a bijective isometry f : X → Y . It follows that f−1 : Y → X isalso an isometry, and that both f and f−1 are continuous. Therefore X and Y arehomeomorphic.

2. Let f : (−π2 , π

2 ) → R be given by f(x) = tan x. f is cintinuous and bijective.Moreover, f−1 : R → (−π

2 , π2 ) given by f−1(x) = tan−1(x) is also continuous. So, R

is homeomorphic to (−π2 , π

2 ).

Problem 7. If (X, d) is complete, show that (X, d), where d = d1+d

is complete.

Solution. It is easy to see that if (xn) is Cauchy in (X, d), then (xn) is Cauchy in (X, d).Since (X, d) is complete, xn → x, but then we also have xn → x in (X, d). It follows that(X, d) is complete.

12

2 Normed Spaces

Vector spaces, linear independence, finite- and infinite-dimensional vector spaces.

Definition (Basis). If X is any vector space, and B is a linearly independent subset of Xwhich spans X, then B is called a basis (or Hamel basis) for X.

Definition (Banach Space). A complete normed space.

Definition (norm). A real-valued function ‖·‖ : X → R satisfying

1. ‖x‖ ≥ 0

2. ‖x‖ = 0 iff x = 0

3. ‖αx‖ = |α|‖x‖

4. |x + y| ≤ |x| + |y|

Definition (Induced Metric). In a normed space (X, d) we can always define a metric byd(x, y) = ‖x − y‖.

Remark 2.1. All previous results about metric spaces apply to normed spaces with theinduced metric.

Problem 8. The norm is continuous, that is ‖·‖ : X → R by x 7→ ‖x‖ is a continuous.

Solution. The triangle inequality implies that

|‖x‖ − ‖y‖| ≤ ‖x − y‖.

Example 2.2 (Norm Spaces). Rn, Cn, ℓp, ℓ∞, C[a, b],L2[a, b],Lp[a, b].

Remark 2.3 (Induced Norm is Translation Invariant). A metric induced by a norm ona normed space is translation invariant, i.e., d(x + a, y + a) = d(x, y) and d(αx,αy) =|α|d(x, y). For example the metric (1) is not coming from a norm.

Theorem 2.4. A subspace Y of a Banach space X is complete iff the set Y is closed inX.

Definition (Convergent, Cauchy, Absolutely Convergent).

1. (xn) is convergent if ‖xn − x‖ → 0. (x is the limit of (xn)).

2. (xn) is Cauchy if ‖xm − xn‖ < ǫ for m,n > N(ǫ).

13

Definition (Infinite series). Let (xk) be a sequence and consider the partial sums

sn = x1 + x2 + ... + xn.

If sn converges, say sn → s, then∑∞

k=1 xk is convergent and

s =∞∑

k=1

.

If ‖x1‖ + ‖x2‖ + ... + ‖xn‖ + ... converges, then∑∞

k=1 xk is absolutely convergent.

Remark 2.5. Absolute convergence implies convergence iff X is complete.

Definition (Schauder basis). If X contains a sequence (en) such that for every x ∈ Xthere exists a unique sequence (αn) with the property that

‖x − (α1e1 + ... + αnen)‖ → 0 as n → ∞,

then (en) is called a Schauder Basis for X. The representation

x =∞∑

k=1

αkek

is called the expansion of x with respect to (en).

Remark 2.6. If X has a Schauder basis, then X is separable.

Theorem 2.7. Let X = (X, ‖ ·‖) be a normed space. Then there is a Banach space X andan isometry A from X onto a subspace W ⊂ X which is dense in X. X is unique, exceptfor isometries.

Theorem 2.8. Let X be a normed space, where the absolute convergence of any seriesalways implies convergence. Then X is complete.

Proof. Let (sn) be any Cauchy sequence in X. Then for every k ∈ N there exists nk

such that ‖sn − sm‖ < 2−k, (m,n > nk) and we can choose nk+1 > nk for all k. Then(snk

) is a subsequence of (sn) and is the sequence of the partial sums of∑

xk, wherex1 = sn1, ..., xk = snk

− snk−1, .... Hence

∑

‖xk‖ ≤ ‖x1‖ + ‖x2‖ +∑

2−k = ‖x1‖ + ‖x2‖ + 1.

It follows that∑

xk is absolutely convergent. By assumption,∑

xk converges, say, snk→

s ∈ X. Since (sn) is Cauchy, sn → s because

‖sn − s‖ ≤ ‖sn − snk‖ + ‖snk

− s‖.

Since (sn) was arbitrary, sn → s shows that X is complete.

14

2.1 Finite dimensional Normed Spaces or Subspaces

Theorem 2.9 (Linear Combinations). Let X be a normed space, and let

{x1, . . . , xn}

be a linearly independent set of vectors in X. Then ∃ c > 0 s.t.

‖α1x1 + · · · + αnxn‖ ≥ c (|α1| + · · · + |αn|) .

Proof. Lets = |α1| + · · · + |αn|.

If s = 0 we are done, so assume s > 0. Define

βi =αi

s.

Clearly we have thatn∑

i=1

|βi| = 1.

Then we can reduce the above inequality to∥

∥

∥

∥

∥

n∑

i=1

βixi

∥

∥

∥

∥

∥

≥ c. (4)

We will prove the result for Theorem 4.Suppose Theorem 4 is not true. Then ∃ (ym)

ym =

n∑

i=1

β(m)i xi

such that‖ym‖ → 0 as m → ∞.

Note that∑n

i=1 |β(m)i | = 1 ⇒ |β(m)

i | ≤ 1 ∀ i. Hence for each fixed i the sequence

(

β(m)i

)

=(

β(1)i , β

(2)i , . . .

)

is bounded. Therefore β(m)1 has a convergent subsequence (B-W). Let β1 be the limit, and

let (y1,m) be the corresponding subsequence of (ym)

(y1,m) = γ(m)1 x1 +

n∑

i=2

β(m)i xi

15

γ(m)1 → β1.

Then there is a there exist corresponding β2, (y2,m) where

β(m)2 → β2 as n → ∞

(y2,m) = γ(m)1 x1 + γ

(m)2 x2 +

n∑

i=3

β(m)i xi.

Since n is finite this process will terminate with

(yn,m) =

n∑

i=1

γ(m)i xi

with (yn,m) a subsequence of (ym). By construction,

γ(m)i → βi ∀ i

n∑

i=1

|βi| = 1.

Therefore (yn,m) has a limit, say

y =n∑

i=1

βixi.

Since ‖ym‖ → 0 by assumption we must also have ‖yn,m‖ → 0, We know y 6= 0 since∑n

i=1 |βi| = 1 and {xi} is a basis, a contradiction.

Theorem 2.10 (Completeness). Every f.d.s.s. Y of a n.s. X is complete. In particular,every f.d.n.s. is complete.

Proof. Let (ym) be a Cauchy sequence in Y . We must find a limit y such that

ym → y

y ∈ Y.

Let n = dimY , and let {e1, . . . , en} be a basis for Y . Then ∀ m we can represent ym as

ym =

n∑

i=1

α(m)i ei.

But then for each fixed i sequence (α(m)i ) is Cauchy in R. So each of these sequence

converges, say

α(m)i → αi.

16

Then define

y =

n∑

i=1

αiei.

Clearly, y ∈ Y and ym → y.

Theorem 2.11. Every f.d.s.s. Y of a closed n.s. X is closed in X.

Proof. By Theorem 2.10 Y must be complete. Therefore Y is closed in X.

Note that finiteness in the previous proof is essential. Infinite dimensional subspacesneed not be closed in X. For example consider

X = C[0, 1]

Y = span ({1, t, ..., tn, ..}) .

Then the sequence (ti) converges to

f(t) =

{

0 t < 11 t = 1

which is not in Y .

Definition (Equivalent Norms). ‖·‖1 and ‖·‖2 are equivalent if ∃ a, b > 0 s.t.

a ‖x‖2 ≤ ‖x‖1 ≤ b ‖x‖2 , ∀ x ∈ X.

Theorem 2.12 (Finite Dimesional ⇒ All Norms are Equivalent). On a f.d.n.s. X everytwo norms ‖·‖1, ‖·‖2 are equivalent.

Proof. Let n = dim X, let {e1, . . . , en} be a basis for X with ‖ei‖1 = 1, let x ∈ X berepresented as

x =n∑

i=1

αiei.

Then ∃ c > 0 s.t.

‖x‖1 ≥ cn∑

i=1

|αi|.

Also note

‖x‖2 ≤n∑

i=1

|αi| ‖ei‖2 ≤ k

n∑

i=1

|αi|

wherek = max {‖ei‖2} .

17

Combining these inequalities we obtain

‖x‖2 ≤ k

cc

n∑

i=1

|αi| ≤1

a‖x‖1 ,

wherea =

c

k

and we obtain the inequality involving a. A similar argument yields for the inequalityinvolving b.

Problem 9. What is the largest possible c > 0 in∥

∥

∥

∥

∥

n∑

i=1

αixi

∥

∥

∥

∥

∥

≥ c

n∑

i=1

|αi|

if

1. X = R2, x1 = (1, 0), x2 = (0, 1).

2. X = R3, x1 = (1, 0, 0), x2 = (0, 1, 0), x3 = (0, 0, 1).

Solution. Find min{‖∑ βixi‖ :∑ |βi| = 1} and obtain 1√

2, 1√

3, respectively.

Problem 10. Show that if ‖·‖1 , ‖·‖2 are equivalent norms on X then the Cauchy sequencesin (X, ‖·‖1), (X, ‖·‖2) are the same.

Solution. Suppose that (xn) is a Cauchy sequence in ‖ · ‖2. Then for every ǫ > 0 thereexists N such that ‖xn − xm‖2 < ǫ

bfor all m,n ≥ N , where b is a constant for which

‖x‖1 ≤ b‖x‖2. Then ‖xn − xm‖1 ≤ b‖xn − xm‖2 < ǫ for all m,n ≥ N . It follows that (xn)is Cauchy in ‖ · ‖1. A similar argument works in the other direction.

2.2 Compactness and Finite Dimension

Definition (Compactness). A metric space (X, d) is compact if every sequence in X hasa convergent subsequence. M ⊂ X is compact if M is compact as a subspace of X, i.e., ifevery sequence in M has a convergent subsequence which converges to an element in M .

Theorem 2.13. If M is compact then M is closed and bounded.

Proof. First we show closed. For all x ∈ M ∃ (xn) ⊂ M s.t. xn → x. M is compact ⇒x ∈ M , hence M is closed.

To show boundedness assume it is not true. Then fix b ∈ M and choose yn ∈M s.t. d(yn, b) > n. Then (yn) ⊂ M is a sequence with no convergent subsequence,contradicting the assumption that M is compact. Thus M is bounded.

18

Note the converse of this theorem is false. Consider M ⊂ ℓ2 as

M ={

xn ∈ ℓ2 s.t. xn = (en), (en)i = δni

}

.

Then M is bounded since ‖xn‖ = 1 and M is closed because there are no accumulationpoints

‖xn − xm‖ =√

2.

But the sequence (xn) has no convergent subsequence, so M is not compact. In some sense,there is too much room in the infinite dimensional setting.

Theorem 2.14 (Compactness). Let X be a f.d.n.s. . Then M ⊂ X is (sequentially)compact iff M is closed and bounded.

Proof. The “⇒” case was proved by the previous theorem. For the other direction, assumeM is closed and bounded.

For the other direction (“⇐”), assume n = dim X, {e1, . . . , en} is a basis for X, and(xm) ⊂ M is an arbitrary sequence. Then there is a representation

xm =

m∑

i=1

ξ(m)i ei, ∀ m.

Since M is bounded, (xm) is bounded so ∃ k ≥ 0 s.t.

‖xm‖ ≤ k.

So then

k ≥ ‖xm‖

=

∥

∥

∥

∥

∥

n∑

i=1

ξ(m)i ei

∥

∥

∥

∥

∥

≥ c

n∑

i=1

|ξ(m)i |

and therefore ∀ i, (ξ(m)i ) ⊂ R is bounded and therefore ∀ i, (ξ

(m)i ) has an accumulation

point z. Since M is closed, z ∈ M and there is a subsequence (zm) ⊂ (xm) s.t. zm → z,say,

z =n∑

i=1

ξiei.

Since (xm) was arbitrary, M is compact.

We shall need the following technical result to prove subsequent theorems.

19

Theorem 2.15 (Riesz). Let Y,Z be subspaces of X, and suppose Y is closed and is aproper subspace of Z. Then ∀ θ ∈ (0, 1) ∃ z s.t.

‖z‖ = 1

‖z − y‖ ≥ θ, ∀ y ∈ Y.

Proof. Let v ∈ Z − Y be arbitrary and let

a = infy∈Y

‖v − y‖ .

Note if a = 0 then v ∈ Y since Y is closed. Therefore a > 0.Choose θ ∈ (0, 1). Then ∃ y0 ∈ Y s.t.

a ≤ ‖v − y0‖ ≤ a

θ.

Let

z = c(v − y0) where c =1

‖v − y0‖.

By construction, ‖z‖ = 1. We want to show ‖z − y‖ ≥ θ, ∀ y ∈ Y . So

‖z − y‖ = ‖c(v − y0) − y‖= c

∥

∥

∥v − y0 −

y

c

∥

∥

∥

= c ‖v − y1‖

for some y1 ∈ Y,⇒ ‖v − y1‖ ≥ a ⇒

‖z − y‖ = ‖v − y1‖≥ ca

=a

‖v − y0‖≥ a

aθ

= θ.

Theorem 2.16 (Finite Dimension and Compactness). Let X be a normed space. Thenthe closed unit ball

M = {x s.t. ‖x‖ ≤ 1}is compact iff dimX < ∞.

20

Proof. “⇒”. Suppose M is compact, but dim X = ∞. Pick x1, ‖x1‖ = 1. Then x1

generates a closed, 1-dimensional subspace X1 $ X. Then also ∃ x2 ∈ X s.t. ‖x2‖ = 1and

‖x2 − x1‖ ≥ θ =1

2.

Then x1, x2 generate a closed, 2-dimensional subspace X2 $ X. Then ∃ x3 ∈ X s.t. ‖x3‖ =1 and

‖x3 − xi‖ ≥ θ =1

2, ∀ i = 1, 2.

In this way construct xn s.t.

‖xn − xi‖ ≥ θ =1

2, ∀ i = 1, . . . , n − 1.

Then (xn) ⊂ X is a sequence such that ‖xm − xn‖ ≥ 12 , ∀ m,n. Therefore (xn) has no

accumulation point, contradicting that M was compact.“⇐” was proved by Theorem 2.14.

Theorem 2.17 (Continuous Preserves Compact). Let T : X → Y be continuous. ThenM ⊂ X compact ⇒ T (M) ⊂ Y compact.

Proof. Let (yn) ⊂ T (M) be arbitrary. Then ∀ n ∃ xn s.t. yn = Txn. Then (xn) ⊂ Mhas a convergent subsequence (xnk

). Since T is continuous, the image (ynk) = T (xnk

) alsoconverges. Therefore T (M) is compact.

Corollary 2.18 (Max, Min). Let T : X → R be continuous and let M ⊂ X be compact.Then T obtains its max and min on M .

Proof. T (M) is compact ⇒ T (M) is closed and bounded. Therefore

inf T (M) ∈ T (M)

supT (M) ∈ T (M).

Problem 11. Show that a discrete metric space with infinitely many points is not compact.

Solution. The space contains an infinite sequence (xn), xn 6= xm (n 6= m) which cannothave a convergent subsequence since d(xn, xm) = 1.

Problem 12 (Local Compactness). A metric space X is said to be locally compact if everypoint x ∈ X has a compact neighborhood. Show that R, Rn, C, Cn are locally compact.

Solution. The closed ball B(x, ǫ) around an arbitrary point x is compact.

21

2.3 Linear Operators

Definition (Linear, Domain and Range). Let X,Y be vector spaces and let T : Xinto−→ Y .

Then

1. If ∀ x, y ∈ D (T ) and scalars α, β

T (αx + βy) = αTx + βTy

then T is said to be a linear operator.

2. The null space of T is N (T ) = {x ∈ D (T ) s.t. Tx = 0}

3. The domain of T , D (T ), is a vector space

4. The range of T , R (T ), lies in a vector space

Note that obviously D (T ) ⊂ X. It is customary to write

T : D (T )onto−→ R (T ) .

Also, T0 = T (0 + 0) = T0 + T0 ⇒ T0 = 0 ⇒ N (T ) 6= ∅.

Example 2.19 (Differentiation). Let X be the space of polynomials on [a, b]. Then define

T : Xonto−→ X

Tx(t) =d

dtx(t).

Example 2.20 (Integration). Let X = C[a, b]. Then define T : Xinto−→ X by

Tx(t) =

∫ t

a

x(s)ds.

Example 2.21 (Multiplication). Let X = C[a, b]. Then define T : Xinto−→ X by

Tx(t) = tx(t).

Example 2.22 (Matrices). Let A = (aij). Then define T : Rm → Rn by

Tx = Ax.

Theorem 2.23 (Range and Null Space). Let T be a linear operator.

1. R (T ) is a vector space.

2. If dimD (T ) = n < ∞ then dimR (T ) ≤ n.

22

3. N (T ) is a vector space.

Proof. 1. Let y1, y2 ∈ R (T ) and let α, β be scalars. Since y1, y2 ∈ R (T ) , ∃ x1, x2 ∈D (T ) s.t.

Tx1 = y1

Tx2 = y2.

Since D (T ) is vector space, αx1 + βx2 ∈ D (T ), so

T (αx1 + βx2) ∈ R (T ) .

ButT (αx1 + βx2) = αTx1 + βTx2 = αy1 + βy2

means R (T ) is a vector space.

2. Pick n + 1 elements arbitrarily in R (T ). Then we have

yi = Txi, ∀ i = 1, . . . , n + 1

for some {x1, . . . , xn+1} in D (T ). Since dimD (T ) = n, this set must be linearlydependent. Similarly, the points in the range are also linearly dependent. Therefore,dimR (T ) ≤ n.

3. Let x1, x2 ∈ N (T ) , i.e. , Tx1 = Tx2 = 0, and let α, β be scalars. But then

T (αx1 + βx2) = αTx1 + βTx2 = 0 ⇒ αx1 + βx2 ∈ N (T ) .

Therefore N (T ) is a vector space.

The next result shows linear operators preserve linear independence.

Definition (Injective (One-to-one)). T : D (T )into−→ Y is injective (or one-to-one) if

∀ x1, x2 ∈ D (T ) s.t. x1 6= x2

Tx1 6= Tx2.

Equivalently,Tx1 = Tx2 ⇒ x1 = x2.

So therefore if T : D (T )into−→ Y is injective then ∃ T−1 : R (T )

onto−→ D (T )

y0 7→ x0

y0 = Tx0

and thereforeT−1Tx = x, ∀ x ∈ D (T )

TT−1y = y, ∀ y ∈ R (T ) .

23

Theorem 2.24 (Inverses). Let X,Y be vector spaces and T : D (T )into−→ Y be linear with

D (T ) ⊂ X and R (T ) ⊂ Y.

1. T−1 : R (T )onto−→ D (T ) exists iff Tx = 0 ⇒ x = 0.

2. If T−1 exists then it is linear.

3. If dimD (T ) = n < ∞ and T−1 exists then dimR (T ) = dimD (T ).

Proof.

1. Suppose Tx = 0 ⇒ x = 0. Then

Tx1 = Tx2 ⇒ T (x1 − x2) = 0 ⇒ x1 = x2

So T is injective T−1 exists.

Conversely, if T−1 exists then Tx1 = Tx2 ⇒ x1 = x2. Therefore

Tx = T0 = 0 ⇒ x = 0.

2. We assume that T−1 exists and show that it is linear in a straightforward fashion.

3. It follows from the earlier result applied to both T and T−1.

Theorem 2.25 (Inverse of Product (Composition)). Let T : X → Y and S : Y → Z bebijections. Then (ST )−1 : Z → X exists and

(ST )−1 = T−1S−1.

2.4 Bounded and Continuous Linear Operators

Definition (Bounded Linear Operator). Let X,Y be normed spaces and T : D (T )into−→

Y,D (T ) ⊂ X. Then T is said to be bounded if ∃ k > 0 s.t.

‖Tx‖ ≤ k ‖x‖ , ∀ x ∈ D (T ) .

Definition (Induced Operator Norm). Let X,Y be normed spaces and T : D (T )into−→

Y,D (T ) ⊂ X. Then the norms on X,Y induce a norm for operators

‖T‖ = supx∈D(T )

x 6=0

‖Tx‖‖x‖ .

Equivalently,‖T‖ = sup

x∈D(T )‖x‖=1

‖Tx‖ .

24

The preceding definition implies that

‖Tx‖ ≤ ‖T‖ ‖x‖ , ∀ x.

Example 2.26 (Differention is Unbounded). Let T : C[0, 1] → C[0, 1],be defined byTx(t) = x

′(t). Then T is linear but unbounded. Indeed, let xn(t) = tn. Then ‖xn‖ = 1

and ‖Txn‖ = n, i.e., T is not bounded.

Example 2.27 (Integration is Bounded). Let T : C[0, 1] → C[0, 1], k ∈ C[0, 1]2, y = Txvia

y(t) =

∫ 1

0k(t, τ)x(τ)dτ.

Since k is continuous, ∃ k0 > 0 s.t.

|k(t, τ)| ≤ k0.

So then

‖y‖ = ‖Tx‖

≤ maxt∈[0,1]

∫ 1

0|k(t, τ)| ‖x(t)‖ dτ

≤ k0 ‖x‖

Theorem 2.28 (Finite Dimension). Let X be normed and dimX = n < ∞. Then everylinear operator on X is bounded.

Proof. Let e1, . . . , en be a basis for X and let x ∈ X be arbitrary and represented as

x =n∑

i=1

αiei.

Then

‖Tx‖ =∥

∥

∥

∑

∥

∥

∥

n

i=1αiTei ≤ k

n∑

i=1

|αi|,

where k = maxi=1,...,n ‖Tei‖. Also,

n∑

i=1

|αi| ≤1

c‖x‖ .

It follows that T is bounded.

Next, we illustrate a general method for computing bounds on operators.

25

Example 2.29 (Calculating Bounds on Operators). Let

x =n∑

i=1

ξiei.

Then

‖Tx‖ =

∥

∥

∥

∥

∥

n∑

i=1

ξiTei

∥

∥

∥

∥

∥

≤n∑

i=1

|ξi| ‖Tei‖

≤ maxk=1,...,n

‖Tek‖n∑

i=1

|ξi| .

Then using Theorem 2.9 we can conclude

1

c‖x‖ =

1

c

∥

∥

∥

∥

∥

n∑

i=1

ξiei

∥

∥

∥

∥

∥

≥n∑

i=1

|ξi|

which implies

‖Tx‖ ≤ γ ‖x‖ where γ =1

cmax

k=1,...,n‖Tek‖ .

Definition (Continuity for Operators). Let T : D (T )into−→ Y . Then T is said to be

continuous at x0 ∈ D (T ) if

∀ ε > 0 ∃ δ > 0 s.t. ‖x − x0‖ < δ ⇒ ‖Tx − Tx0‖ < ε.

T is said to be continuous if it is continuous at every x0 ∈ D (T ).

Theorem 2.30 (Continuity and Boundedness). Let X,Y be vector spaces D (T ) ⊂ X, and

T : D (T )into−→ Y be linear.

1. T is continuous iff T is bounded

2. T is continuous at a single point ⇒ T is continuous everywhere

Proof.

1. The case T = 0 is trivial. Assume T 6= 0, then ‖T‖ 6= 0. Suppose T is bounded andlet x0 ∈ D (T ) , ε > 0 be arbitrary. Then because T is linear, ∀ x ∈ D (T ) s.t.

‖x − x0‖ < δ =ε

‖T‖

26

we have that

‖Tx − Tx0‖ = ‖T (x − x0)‖ ≤ ‖T‖ ‖x − x0‖ < ‖T‖ δ = ε.

Therefore T is continuous.

Conversely, suppose T is continuous at x0 ∈ D (T ). Then ∀ ε > 0 ∃ δ > 0 s.t.

‖x − x0‖ ≤ δ ⇒ ‖Tx − Tx0‖ ≤ ε.

So choose any y ∈ D (T ) , y 6= 0 and set

x = x0 + δy

‖y‖ .

Thenx − x0 = δ

y

‖y‖ ⇒ ‖x − x0‖ = δ,

so

‖Tx − Tx0‖ = ‖T (x − x0)‖

=

∥

∥

∥

∥

T

(

δ

‖y‖y

)∥

∥

∥

∥

=δ

‖y‖ ‖Ty‖≤ ε.

Therefore we have‖Ty‖ ≤ c ‖y‖ where c =

ε

δ.

2. Above we only used continuity at a single point to show boundedness. Boundednessimplies continuity.

Corollary 2.31. Let T : D (T )into−→ Y be a bounded, linear operator. Then

1. (xn) ⊂ D (T ) , x ∈ D (T ) and xn → x ⇒ Txn → Tx

2. N (T ) is closed.

Proof. 1. ‖Txn − Tx‖ = ‖T (xn − x)‖ ≤ ‖T‖ ‖xn − x‖ → 0

2. ∀ x ∈ N (T ) ∃ (xn) ⊂ N (T ) s.t.

xn → x ⇒ Txn → Tx.

But xn ∈ N (T ) ⇒ Txn = 0, ∀ n ⇒ Tx = 0 ⇒ x ∈ N (T ). Therefore N (T ) isclosed.

27

Compare to the notion of subadditivity (triangle inequality).

Remark 2.32 (Operator Norm is Submultiplicative).

1. ‖TS‖ ≤ ‖T‖ ‖S‖

2. ‖T n‖ ≤ ‖T‖n

Remark 2.33 (Equality of Operators). Two operators T, S are said to be equal and wewrite T = S if

D (T ) = D (S) and Tx = Sx, ∀ x ∈ D (T ) .

Definition (Rescriction). Let T : D (T )into−→ Y and let B ⊂ D (T ). We define the restric-

tion of T to B, denoted T |B : Binto−→ Y , as

T |Bx = Tx, ∀ x ∈ B.

Definition (Extension). Let T : D (T )into−→ Y and D (T ) ⊂ M . An extension of T is

T : Minto−→ Y s.t.

T |D (T ) = T

Tx = Tx, ∀ x ∈ D (T ) .

Theorem 2.34 (Bounded Linear Extension). Let D (T ) ⊂ X, X a normed space, Y a

Banach space, and T : D (T )into−→ Y be linear. Then there is a unique bounded, linear

extension T of T s.t.∥

∥

∥T∥

∥

∥= ‖T‖ .

Proof. Let x ∈ D (T ). Then ∃ (xn) ∈ D (T ) s.t. xn → x. Since T is linear and boundedwe have

‖Txn − Tx‖ = ‖T (xn − x)‖ ≤ ‖T‖ ‖xn − x‖ ,

so (Txn) is Cauchy. Since Y is Banach, (Txn) converges, say

Txn → y ∈ Y.

In this way, we defineT x = y.

Because limits are unique, T is uniquely defined ∀ x ∈ D (T ). Of course T inherits linearityand

T x = Tx, ∀ x ∈ D (T )

so T is a genuine extension.

28

Finally, we need to show T is bounded. Note

‖Txn‖ ≤ ‖T‖ ‖xn‖

and recallx 7→ ‖x‖

is continuous. Therefore Txn → y = T x and

∥

∥

∥T x∥

∥

∥≤ ‖x‖ ‖x‖ .

Therefore T is bounded and∥

∥

∥T∥

∥

∥≤ ‖T‖. By definition,

∥

∥

∥T∥

∥

∥≥ ‖T‖ so

∥

∥

∥T∥

∥

∥= ‖T‖ .

2.5 Linear Functionals

Definition (Linear Functional). A linear functional f is a linear operator with D (f) ⊂ Xwhere X is vector space and R (f) ⊂ K where K is the scalar field corresponding to X (Ror C). Then

f : D (f)into−→ K.

Definition (Bounded Linear Functional). Let f : D (f)into−→ K be a linear functional.

Then f is said to be bounded if ∃ c > 0 s.t.

|f(x)| ≤ c ‖x‖ , ∀ x ∈ D (f) .

Definition (Norm of a Linear Functional). Let f be a linear functional. Then

‖f‖ = supx∈D(f)

x 6=0

|f(x)|‖x‖ = sup

x∈D(f)‖x‖=1

|f(x)| ,

so|f(x)| ≤ ‖f‖ ‖x‖ .

Theorem 2.35 (Continuity and Boundedness). Let f be a linear functional. f is contin-uous iff f is bounded.

29

Example 2.36 (Dot Product). Let a ∈ R, a 6= 0 be fixed and let f : Rn → R be definedas

f(x) = x · a.

Then f is linear functional and|f(x)| ≤ ‖x‖ ‖a‖ .

On the other hand, if we take x = a

|f(x)| = ‖a‖2 ⇒ ‖f‖ ≥ ‖a‖ .

Example 2.37 (Integral). Define f : C[a, b] → R by

f(x) =

∫ b

a

x(t)dt.

Then|f(x)| ≤ (b − a) ‖x‖ .

Further, if x = 1|f(x)| = b − a ⇒ ‖f‖ = b − a.

Example 2.38 (Point Evaluation). Let t0 ∈ [a, b] be fixed and define f : C[a, b] → R by

f(x) = x(t0).

Choosing the sup-norm, we see

|f(x)| = |x(t0)| ≤ ‖x‖ ,

and since f(1) = 1 = ‖1‖ , we have‖f‖ = 1.

Example 2.39 (Point Evaluation in L2). Let f : L2[a, b] → R by

f(x) = x(a).

Choose the L2 norm, and let x ∈ L2[a, b] be such that x(a) = 1 but then x rapidly decreasesto 0. Then f(x) = 1 but

‖x‖ =

(∫ b

a

|x(t)|2 dt

)

12

and for any ε > 0 we can choose an x such that ‖x‖ < ε. Therefore there is no suchc > 0 s.t.

|f(x)| = 1 ≤ c ‖x‖ .

So point evaluation in this space with this norm is unbounded.

30

Example 2.40 (Inner Product in ℓ2). Fix a ∈ ℓ2 and define f : ℓ2 → R by

f(x) =

∞∑

i=1

xiai = 〈x, a〉 .

Then

|f(x)| ≤∞∑

i=1

|xiai|

≤

√

√

√

√

∞∑

i=1

x2i

√

√

√

√

∞∑

i=1

a2i

= ‖x‖ ‖a‖ .

Definition (Algebraic Dual). Let X be a vector space. Then we denote X∗ as the spaceof linear functionals on X. X∗∗ represents the dual of X∗, et cetera.

Space General Element Value at a Point

X x n/a

X∗ f f(x)

X∗∗ g g(f)

Definition (Isomorphism = Bijective Isometry). Let X,Y be vector spaces over the samescalar field K. Then an isomorphism T : X → Y is a bijective mapping which preservesthe two algebraic operations of vector spaces, i.e.

T (αx + βy) = αTx + βTy.

So T is a bijective linear operator and we say X,Y are isomorphic.If in addition X,Y are normed spaces, then T must also preserve the norms

‖x‖ = ‖Tx‖ .

We can obtain g ∈ X∗∗ picking a fixed x ∈ X and setting

g(f) = gx(f) = f(x), ∀ f ∈ X∗.

This g is linear, and x ∈ X ⇒ gx ∈ X∗∗.

Definition (Canonical Mapping/Embedding). Define C : X → X∗∗ by

x 7→ gx.

Then C is injective and linear. Therefore C is an isomorphism of X and R (C) ⊂ X∗∗.

31

An interesting and important problem is when R (C) = X∗∗, i.e., X,X∗∗ are isomor-phic.

Definition (Embeddable). Let X,Y be vector spaces. X is said to be embeddable in Y ifit is isomorphic with a subspace of Y .

From Definition 2.5 we can see X is always embeddable in X∗∗. C is also called thecanonical embedding of X into X∗∗.

Definition (Algebraic Reflexive). If the canonical mapping C is surjective (and hencebijective) then

R (C) = X∗∗

and X is said to be algebraically reflexive.

Problem 13. If Y is a subspace of a vector space X and f is a linear functional on X suchthat f(Y ) is not the whole scalar field K, show

f(y) = 0, ∀ y ∈ Y.

Solution. Suppose that f(y) = b 6= 0 for some y ∈ Y . For arbitrary a ∈ R we have aby ∈ Y

and f(aby) = a. It follows that f(Y ) = R; a contradiction.

Problem 14. Let f 6= 0 be a linear functional on a vector space X, and let x0 ∈ X −N (f)be fixed. Show that every x ∈ X has a unique representation

x = αx0 + y, some y ∈ N (f) .

Solution. Let f 6= 0 and consider x0 ∈ X − N(f). Then f(x0) = c 6= 0. Let a = f(x) =f(a

cx0). It follows that x − a

cx0 ∈ N(f). We can write

x =a

cx0 + y , where y = x − a

cx0 ∈ N(f).

2.6 Finite Dimensional Case

Let X,Y be finite dimensional vector spaces over the same field K, and let T : X → Y bea linear operator. Let E = {e1, . . . , en} , B = {b1, . . . bn} be bases for X,Y respectively.Then x ∈ X and y = Tx ∈ Y and Tek ∈ Y, ∀ k have the unique representations

x =

n∑

k=1

ξkek

Tek =

n∑

i=1

τikbi

y =n∑

i=1

ηibi.

32

Then calculate

y = Tx

= T

(

n∑

k=1

ξkek

)

=n∑

k=1

ξkTek

which implies

y =

n∑

i=1

ηibi =

n∑

k=1

ξk

(

n∑

i=1

τikbi

)

=

n∑

i=1

(

n∑

k=1

τikξk

)

bi

yielding

ηi =

n∑

k=1

τikξk.

So if we labelTEB = (τik), x = (ξk), y = (ηi)

theny = TEBx.

We say TEB represents T with respect to the bases E,B.

Remark 2.41 (Finite Dimensional Linear Operators are Matrices). The above argumentshows that every f.d.l.o. can be represented as a matrix.

2.6.1 Linear Functionals

Let X be a vector space, n = dim X, and let {e1, . . . , en} be a basis for X. Label

αi = f(ei).

Then

f(x) = f

(

n∑

i=1

ξiei

)

=

n∑

i=1

ξiαi.

(f is uniquely determined by its values αi at the n basis vectors of X).

33

Definition (Dual Basis). Define fk by

fk(ei) = δik, ∀ k = 1, . . . , n.

Then{f1, . . . , fn}

bij↔ {e1, . . . , en} .

Theorem 2.42 (Dimension of X∗). Let X be a vector space, n = dimX, and let E ={e1, . . . , en} be a basis for X. Then {f1, . . . , fk} is a basis for X∗ and dim X∗ = n. Sothen if f ∈ X∗ then we can represent

f =

n∑

i=1

αifi.

Proof. See Theorem 2.41.

Lemma 2.43 (Zero Vector). Let X be a fdvs. If x0 ∈ X has the property that f(x0) =0, ∀ f ∈ X∗ then x0 = 0.

Proof. Let {e1, . . . , en} be a basis for X and let

x0 =n∑

i=1

ξ0iei.

Then

f(x0) =n∑

i=1

ξ0iαi.

By assumption f(x0) = 0, ∀ f ∈ X∗ ⇒ ξ0i = 0, ∀ i = 1, . . . , n.

Theorem 2.44 (Algebraic Reflexivity). A fdvs X is algebraically reflexive, i.e., X isisomorphic to X∗∗.

Proof. By construction C : Xinto−→ X∗∗ is linear. Then

Cx0 = 0 ⇒ (Cx0)(f) = gx0(f) = f(x0) = 0, ∀ f ∈ X∗

and therefor x0 = 0 by the previous lemma. Therefore C has an inverse C−1 : R (C) → X.Therefore dimR (C) = dim X. Therefor C is an isomorphism and X is algebraicallyreflexive.

Problem 15. Find a basis for the null space of functional f : R3 → R given by

f(x) = α1ξ1 + α2ξ2 + α3ξ3

where α1 6= 0.

Solution. Let (α1, α2, α3) with α1 6= 0 be a fixed point in R3 and let f(x) =∑

aixi. Thevectors (−α2

α1, 1, 0) and (−α3

α2, 0, 1) form a basis for N(f).

34

2.7 Dual Space

Definition (B(X,Y )). Let X,Y be vector spaces. Then we define B(X,Y ) as the set ofall bounded linear operators from X to Y .

Remark 2.45 (B(X,Y ) is a Vector Space). (αS + βT )x = αSx + βTx

Theorem 2.46 (B(X,Y ) is a Normed Space). Let X,Y be normed spaces. The vectorspace B(X,Y ) is also a normed space with norm

‖T‖ = supx∈X

x 6=0

‖Tx‖‖x‖ = sup

x∈X

‖x‖=1

‖Tx‖ .

Theorem 2.47 (B(X,Y ) is a Banach Space). If in the assumptions of the previous proofY is Banach, then B(X,Y ) is also a Banach space.

Proof. Let (Tn) be an arbitrary Cauchy sequence in B(X,Y ). Then

∀ ε > 0 ∃ N s.t. ‖Tn − Tm‖ < ε, ∀ m,n > N.

Therefore ∀ x ∈ X, and m,n > N

‖Tnx − Tmx‖ = ‖(Tn − Tm)x‖≤ ‖Tn − Tm‖ ‖x‖< ε ‖x‖ .

So fix x ∈ X. We see ∀ x ∈ X that (Tnx) ⊂ Y is Cauchy. Y is complete so ∃ y ∈Y s.t. Tnx → y. We need to construct a limit for (Tn). Define T : X → Y by

Tx = y.

By construction, T is linear

limn→∞

Tn(αx + βy) = limn→∞

(αTnx + βTnx)

= α limn→∞

Tnx + β limn→∞

Tnx

= αTx + βTx.

The following

‖Tnx − Tx‖ =∥

∥

∥Tnx − lim

m→∞Tmx

∥

∥

∥

= limm→∞

‖Tnx − Tmx‖≤ ε ‖x‖

35

implies that Tn − T is bounded. Since Tn ∈ B(X,Y ) ⇒ Tn is bounded and

T = Tn − (Tn − T )

T is also bounded. Therefore T ∈ B(X,Y ).Finally,

‖Tnx − Tx‖‖x‖ ≤ ε

implies‖Tn − T‖ ≤ ε ⇒ ‖Tn − T‖ → 0.

Definition (Normed Dual Space X ′). Let X be a n.s.. Then the set of all bounded linearfunctionals on X constitutes a n.s. with norm

‖f‖ = supx∈Xx 6=0

|f(x)|‖x‖ = sup

x∈X‖x‖=1

|f(x)| .

This is the normed dual of X.

Theorem 2.48 (X ′ is a Banach Space). Let X be a n.s. and let X ′ be the dual of X.Then X ′ is a Banach space.

Proof. See Theorem 2.47.

Example 2.49 (Rn). The dual of Rn is Rn.

Proof. We claim Rn∗ = Rn′ and ∀ f ∈ Rn∗ we have

f(x) =

n∑

k=1

ξkαk, αk = f(ek).

So then

|f(x)| ≤n∑

k=1

|ξkαk| ≤ ‖x‖

√

√

√

√

n∑

k=1

α2k,

which implies

‖f‖ ≤

√

√

√

√

n∑

k=1

α2k.

If we choose x = (αk) then

|f(x)| =

√

√

√

√

n∑

k=1

α2k ⇒ ‖f‖ =

√

√

√

√

n∑

k=1

α2k.

36

So take the onto mapping from Rn′ to Rn

f 7→ α = (αk) = (f(ek))

which is norm-preserving, linear, and a bijection (an isomorphism), which completes theproof.

Problem 16.

1. Show ℓ1′ = ℓ∞

2. Show ℓp′ = ℓq where 1 < p < ∞ and 1p

+ 1q

= 1

Solution.

1. A Schauder basis for ℓ1 is (ek), where ek = (δki). Then every x ∈ ℓ1 has a uniquerepresentation

x =

∞∑

k=1

ξkek.

We consider any f ∈ ℓ1′ , where ℓ1′ is the dual space of ℓ1. Since f is linear andbounded,

f(x) =

∞∑

k=1

ξkγk , where γk = f(ek).

It is easy to see that (γk) ∈ ℓ∞.On the other hand, for every b = (βk) ∈ ℓ∞,

g(x) =∞∑

k=1

ξkβk , where x = (ξk) ∈ ℓ1

is a bounded linear functional on ℓ1, i.e., g ∈ ℓ1′ .It is also easy see that ‖f‖ = ‖c‖∞, where c = (γk) ∈ ℓ∞. It follows that the bijectivelinear mapping of ℓ1′ onto ℓ∞ defined by f 7→ c is an isomorphism.

2. Take the same Schauder basis and representation for x and f as in 1. Let q be the

conjugate of p and consider xn = (ξ(n)k ) with

ξ(n)k =

|γk|qγk

, if k ≤ n and γk 6= 0,

andξ(n)k = 0 if k > n or γk = 0.

37

Then we have

f(xn) =∞∑

k=1

ξ(n)k γk =

n∑

k=1

|γk|q ≤ ‖f‖(n∑

k=1

|γk|q)1p .

Using 1 − 1p

= 1q

we get

(

n∑

k=1

|γk|q)1q ≤ ‖f‖.

Since n is arbitrary, letting n → ∞, we obtain

(∞∑

k=1

|γk|q)1q ≤ ‖f‖,

and therefore (γk) ∈ ℓq.Conversely, for any b = (βk) ∈ ℓq we may define g on ℓp by

g(x) =

∞∑

k=1

ξkβk , where x = (ξk) ∈ ℓp.

Then g is linear, and boundedness follows from the Holder inequality. Hence g ∈ ℓp′ .From the Holder inequality we get

|f(x)| ≤ ‖x‖(∞∑

k=1

|γk|q)1q .

It follows that ‖f‖ = ‖c‖q , where c = (γk) ∈ ℓq and γk = f(ek). The mapping ofℓp′ onto ℓq defined by f 7→ c is linear, bijective, and norm-preserving, so it is anisomorphism.

38

3 Hilbert Spaces

Definition (Inner Product). Let X be vector space with scalar field K. 〈·, ·〉 : X ×X → Kis called an inner product if

1. 〈x + y, z〉 = 〈x, z〉 + 〈y, z〉

2. 〈αx, y〉 = α 〈x, y〉

3. 〈x, y〉 = 〈y, x〉

4. 〈x, x〉 ≥ 0, 〈x, x〉 = 0 iff x = 0

Definition (Hilbert Space). A Hilbert space is a complete inner product space.

Remark 3.1.

1. A vector space X with inner product 〈·, ·〉 has the induced norm and metric:

‖x‖ = 〈x, x〉

d(x, y) = ‖x − y‖ .

2. If X is real then 〈x, y〉 = 〈y, x〉.

3. From the definition of inner product, we obtain

〈αx + βy, z〉 = α 〈x, z〉 + β 〈y, z〉〈x, αy〉 = α 〈x, y〉

〈x, αy + βz〉 = α 〈x, y〉 + β 〈x, z〉

So the inner product is linear in the first argument and conjugate linear in the secondargument.

4. Parallelogram Equality.

‖x + y‖2 + ‖x − y‖2 = 2(

‖x‖2 + ‖y‖2)

(5)

Any norm which is induced from an inner product must satisfy this equality.

5. Orthogonality. x, y ∈ X are said to be orthogonal if 〈x, y〉 = 0 and we write

x ⊥ y.

For subsets A,B ⊂ X we say A ⊥ B if

a ⊥ b ∀ a ∈ A, b ∈ B.

39

Example 3.2.

1. Rn with inner product 〈x, y〉 = xT y.

2. Cn with inner product 〈x, y〉 = xT y.

3. Real L2[a, b] with

〈x, y〉 =

∫ b

a

x(t)y(t)dt.

4. Complex L2[a, b] with

〈x, y〉 =

∫ b

a

x(t)y(t)dt.

5. ℓ2 with inner product

〈x, y〉 =

∞∑

i=1

xiyi.

Problem 17.

1. Show ℓp with p 6= 2 is not a Hilbert space.

2. Show C[a, b] is not a Hilbert space.

Solution.

1. Consider x = (1, 1, 0, ..., 0, ..) and y = (1,−1, 0, ..., 0, ...). Then x, y ∈ lp and ‖x‖p =

‖y‖p = 21p , and ‖x+y‖p = ‖x−y‖p = 2. It follows that if p 6= 2, then the paralelogram

equality is not satisfied.

2. Let x(t) = t−ab−a

and y(t) = 1− t−ab−a

. Then we have ‖x‖ = ‖y‖ = ‖x+y‖ = ‖x−y‖ = 1and the paralelogram equality is not satisfied.

Problem 18. Show that ‖x‖ =√

〈x, x〉 defines a norm.

Solution. It follows from the definition of the inner product that ‖x‖ ≥ 0 and ‖x‖ = 0 iffx = 0. Also, we have using the definiton of the inner product that

‖αx‖ = (〈αx,αx〉) 12 = (αα 〈x, x〉) 1

2 = (|α|2 〈x, x〉) 12 = |α|‖x‖.

Concerning the triangle inequality we get

‖x + y‖2 = 〈x + y, x + y〉 = 〈x, x〉 + 〈x, y〉 + 〈y, x〉 + 〈y, y〉 = ‖x‖2 + 2Re 〈x, y〉 + ‖y‖2

≤ ‖x‖2 + 2‖x‖‖y‖ + ‖y‖2 = (‖x‖ + ‖y‖)2,from which the result follows.

40

Lemma 3.3 (Continuity of Inner Product). If in an inner product space yn → y andxn → x then

〈xn, yn〉 → 〈x, y〉 .

Theorem 3.4 (Completion). For any inner product space X there is a Hilbert space Hand isomorphism A from X onto a dense subspace W ⊂ H. The space H is unique up toisomorphisms.

Theorem 3.5 (Subspace). Let Y ⊂ H with H a Hilbert space. Then

1. Y is complete iff Y is closed in H.

2. If Y is finite dimensional then Y is complete.

3. If H is separable then Y is separable.

Definition (Convex Subset). M ⊂ X is said to be convex if

αx + (1 − α)y ∈ M, ∀ α ∈ [0, 1], a, b ∈ M.

Theorem 3.6. Let X be an inner product space and M 6= ∅ a convex subset which iscomplete. Then ∀ x ∈ X ∃! y ∈ M s.t.

δ = infy∈M

‖x − y‖ = ‖x − y‖ .

Proof.

1. Existence. By definition of inf,

∃ (yn) s.t. δn → δ, where δn = ‖x − yn‖ .

We will show (yn) is Cauchy. Write vn = yn − x. Then ‖vn‖ = δn and

‖vn + vm‖ = ‖yn + ym − 2x‖

= 2

∥

∥

∥

∥

1

2(yn + ym) − x

∥

∥

∥

∥

≥ 2δ

Furthermore, yn − ym = vn − vm,⇒‖yn − ym‖2 = ‖vn − vm‖2

= −‖vn + vm‖2 + 2(‖vn‖2 + ‖vm‖)≤ −(2δ)2 + 2(δ2

n + δ2m).

Since M is complete, yn → y ∈ M . Then ‖x − y‖ ≥ δ. Also, we have

‖x − y‖ ≤ ‖x − yn‖ + ‖yn − y‖= δn + ‖yn − y‖→ δ + 0

therefore ‖x − y‖ = δ.

41

2. Uniqueness. Suppose that two such elements exist, y1, y2 ∈ M and from above,

‖x − y1‖ = δ = ‖x − y2‖ .

By (5) we have

‖y1 − y2‖2 = ‖(y1 − x) + (x − y2)‖2

= 2 ‖y1 − x‖2 + 2 ‖y2 − x‖2 − ‖(y1 − x) + (y2 − x)‖2

= 2δ2 + 2δ2 − 22

∥

∥

∥

∥

1

2(y1 + y2) − x

∥

∥

∥

∥

2

≤ 0

since

22

∥

∥

∥

∥

1

2(y1 + y2) − x

∥

∥

∥

∥

2

≥ 4δ2.

Therefore y1 = y2.

Lemma 3.7 (Orthogonality). With the setup of the previous theorem, z = x−y is orthog-onal to M .

Proof. Suppose z ⊥ M were false, then ∃ w ∈ M s.t.

〈z,w〉 = β 6= 0.

and so w 6= 0. For any α,

‖z − αw‖2 = 〈z − αw, z − αw〉= 〈z, z〉 − α 〈z,w〉 − α [〈w, z〉 − α 〈w,w〉]= ‖z‖2 − αβ − α [β − α 〈w,w〉] .

If we choose

α =β

〈w,w〉 ,

and continue the equality above,

‖z − αw‖2 = δ2 − |β|2〈w,w〉 − α [0]

< δ2

contradicting the minimality of y, δ. Therefore we must have z ⊥ M .

42

Definition (Direct Sum). A vector space X is said to be direct sum of subspaces Y andZ, written

X = Y ⊕ Z,

if ∀ x ∈ ∃! y ∈ Y, z ∈ Z s.t.x = y + z.

Definition (Orthogonal Complement). Let Y be a closed subspace of X. Then the or-thogonal complement of Y in X is

Y ⊥ = {z ∈ X s.t. z ⊥ Y } .

Theorem 3.8 (Direct Sum). Let Y be a closed subspace of H. Then

H = Y ⊕ Y ⊥.

Proof.

1. Existence. H is complete and Y is closed implies Y is complete. Further, we knowY is convex. Therefore ∀ x ∈ H ∃ y ∈ Y s.t.

x = y + z, where z ∈ Y ⊥.

2. Uniqueness. Assume x = y + z = y1 + z1. Then

y − y1 ∈ Y

z − z1 ∈ Y ⊥.

But Y ∩ Y ⊥ = {0} ⇒ y − y1 = z − z1 = 0.

Definition (Orthogonal Projection). Let Y be a closed subspace of H, so H = Y ⊕ Y ⊥,and let

x = y + z

y ∈ Y

z ∈ Y ⊥.

Then the orthogonal projection onto Y is P : H → Y given by

Px = y.

Clearly P is bounded. Further P is idempotent, that is, P 2 = P , which we call idempo-

tent.

43

Definition (Annihilator). Let X be an inner product space and let M be a nonemptysubset of X. Then the annihilator of M is

M⊥ = {x ∈ X s.t. x ⊥ M} = {x ∈ X s.t. 〈x, v〉 = 0 ∀ v ∈ M} .

Theorem 3.9. Let M,X be as the definition above and denote (M⊥)⊥ = M⊥⊥. Then

1. M⊥ is a vector space.

2. M⊥ is closed.

3. M ⊂ M⊥⊥.

Theorem 3.10. If M is a closed subspace of a Hilbert space H then

M⊥⊥ = M.

Lemma 3.11 (Dense Set). Let M 6= ∅ be a subset of a Hilbert space H. Then

span (M) = H iff M⊥ = {0} .

Proof.

1. ⇒. Assume x ∈ M⊥, V = span (M) is dense in H. Then x ∈ V = H ⇒ ∃ (xn) ⊂V s.t. xn → x. Now x ∈ M⊥ and M⊥ ⊥ V ⇒ 〈xn, x〉 = 0. But 〈·, ·〉 is continuous⇒〈xn, x〉 → 〈x, x〉 = 0 ⇒ x = 0 ⇒ M⊥ = {0}, since x ∈ M⊥ was arbitrary.

2. ⇐. Suppose M⊥ = {0} and write V = span (M). Then

x ⊥ V ⇒ x ⊥ M

⇒ x ∈ M⊥

⇒ x = 0

⇒ V ⊥ = {0}⇒ V = H.

Definition (Orthonormal Set/Sequence). M is said to be an orthogonal set if ∀ x, y ∈ M

x 6= y ⇒ 〈x, y〉 = 0.

M is said to be orthonormal if ∀ x, y ∈ M

〈x, y〉 = δxy =

{

0 x 6= y1 x = y

.

If M is countable and orthogonal (resp. orthonormal) then we can write M = (xn) and wesay (xn) is an orthogonal (resp. orthonormal) sequence.

44

Remark 3.12 (Pythagorean Relation, Linear Independence). Let M = {x1, . . . , xn} be anorthogonal set. Then

1.∥

∥

∥

∥

∥

n∑

i=1

xi

∥

∥

∥

∥

∥

2

=n∑

i=1

‖xi‖2 .

2. M is linearly independent.

Example 3.13 (Orthonormal Sequences).

1. {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1, 0), (0, . . . , 0, 1)} ⊂ R.

2. (en) ⊂ ℓ2 where en = δni.

3. Let X = C[0, 2π] with 〈x, y〉 =∫ 2π

0 x(t)y(t)dt. Then the functions

(sin nt), n = 1, 2, . . . , (cos nt), n = 0, 1, 2, . . .

form an orthogonal sequence.

Remark 3.14 (Unique Representation with Orthonormal Sequences). Let X be an innerproduct space and let (ek) be an orthonormal sequence in X, and suppose x ∈ span ({e1, . . . , en})where n is fixed. Then we can represent

x =n∑

k=1

αkek ⇒ 〈x, ei〉 =

⟨[

n∑

k=1

αkek

]

, ei

⟩

= αi

⇒ x =n∑

k=1

〈x, ek〉 ek.

Now let x ∈ X be arbitrary and take y ∈ span ({e1, . . . , en}), where

y =

n∑

k=1

〈x, ek〉 ek.

Define z by setting x = y + z ⇒ z ⊥ y because

〈z, y〉 = 〈x − y, y〉= 〈x, y〉 − 〈y, y〉=

⟨

x,[

∑

〈x, ek〉 ek

]⟩

− ‖y‖2

=∑

〈x, 〈x, ek〉 ek〉 − ‖y‖2

=∑

〈x, ek〉 〈x, ek〉 −∑

|〈x, ek〉|2

= 0.

45

Then

⇒ ‖x‖2 = ‖y‖2 + ‖z‖2

⇒ ‖z‖2 = ‖x‖2 −∑

|〈x, ek〉|2 ≥ 0

⇒ ∑nk=1 |〈x, ek〉|2 ≤ ‖x‖2

⇒ ∑∞k=1 |〈x, ek〉|2 ≤ ‖x‖2

Remark 3.15 (Bessel’s Inequaltiy). Let X be an inner product space and let (ek) be anorthonormal sequence in X. Then the last inequality in the above remark is called Bessel’s

Inequality:∞∑

k=1

|〈x, ek〉|2 ≤ ‖x‖2 .

Definition (Fourier Coefficients). The sequence (〈x, ek〉) is called the Fourier coefficientsof x w.r.t. (ek).

Problem 19 (Fourier Coefficients are Minimizers). Let {en, . . . , en} be an orthonormal setin an inner product space X (n is fixed). Let x ∈ X be an arbitrary, fixed element andlet y =

∑nk=1 βkek. Then ‖x − y‖ depends on β1, . . . , βn. Show by direct calculation that

‖x − y‖ is minimized iff βi = 〈x, ei〉 , ∀ i = 1, . . . , n.

Solution. Let γi = 〈x, ei〉, and y =∑

βiei. Then

‖x − y‖2 =⟨

x −∑

βiei, x −∑

βiei

⟩

= ‖x‖2 −∑

βiγi −∑

βiγi +∑

|βi|2

= ‖x‖2 −∑

|γi|2 +∑

|βi − γi|2

and this is minumum for given x and eis iff βi = γi.

Problem 20 (Gramm-Schmidt). Orthonormalize the first three terms of the sequence (1, t, t2, t3, . . . )on the interval [−1, 1] where

〈x, y〉 =

∫ 1

−1x(t)y(t)dt.

Solution. Let f1(t) = 1, then e1(t) = f1(t)‖f1(t)‖ = 1√

2. Let f2(t) = t. We have that

〈f2(·), e1(·)〉 = 0, so we just need to normalize f2(t) to get e2(t) =√

32t. Let f3(t) = t2.

Easy calculation shows that 〈f3(·), e1(·)〉 =√

23 and 〈f3(·), e2(·)〉 = 0. Then f3(t) =

〈f3(·), e2(·)〉 e2(t) = t2 − 13 . Normalizing this quantity we get e3(t) =

√

58(3t2 − 1).

46

Theorem 3.16 (Convergence). Let H be a Hilbert space and let (en) ⊂ H be an orthonor-mal sequence. Consider

∞∑

k=1

αkek. (6)

1. The series (6) converges in the induced norm of H iff

∞∑

k=1

|αk|2 < ∞.

2. If (6) converges to x, then αk = 〈x, ek〉 and

x =

∞∑

k=1

〈x, ek〉 ek.

3. For any x ∈ H the series (6) with αk = 〈x, ek〉 converges (in the norm of H).

Proof. Let

sn =n∑

k=1

αkek

σn =n∑

k=1

|αk|2 .

1. For n > m

‖sn − sm‖2 =

∥

∥

∥

∥

∥

n∑

k=m+1

αkek

∥

∥

∥

∥

∥

2

=n∑

k=m+1

|αk|2

= σn − σm.

Hence (sn) is Cauchy in H iff σn) is Cauchy in R.

2. Note 〈sn, ei〉 = αi, ∀ i = 1, . . . , k ≤ n. By assumption sn → x, so

⇒ αi = 〈sn, ei〉 → 〈x, ei〉 , for i ≤ k

⇒ αi = 〈x, ei〉 ∀ i = 1, 2, . . .

3. This follows from Bessel’s inequality and 1.

47

Definition (Total Set). Let X be an inner product space and let M ⊂ X.

1. span (M) = X ⇒ M is a total set.

2. If M is an orthonormal set then M is a total orthonormal set.

Remark 3.17.

1. A total orthonormal family in X is sometimes called an orthonormal basis for X.Note this is not equivalent to an algebraic basis unless X is finite dimensional.

2. In every nontrivial Hilbert space H 6= {0} there is a total orthonormal set.

Definition (Hilbert Dimension). The Hilbert dimension of X is the cardinality of the

smallest orthonormal set, i.e., if Λ ={

M s.t. span (M) = H}

then the Hilbert dimension

isinf

M∈Λ|M | .

Problem 21. Let X be an inner product space and let M ⊂ X. Then

1. If M is total in X thenx ⊥ M ⇒ x = 0 (7)

2. If X is complete, then (7) ⇒M is total in X.

Solution. 1. Let H be the completion of X. Then, X regarded as a subspace of H, isdense in H. By assumption, M is total in X, so spanM is dense in X, and dense inH. It follows that the orthogonal complement of M in H is {0}.

2. If x is a Hilbert space and M⊥ = {0}, then the Dense Set Lemma implies that M istotal in X.

Problem 22. Show that an orthonormal set M in a Hilbert space H is total iff∑

k

|〈x, ek〉|2 = ‖x‖2 , ∀ x ∈ M.

This is called Parseval’s equality.

Solution. If M is not total, by Problem 21 there is a nonzero x ⊥ M in H. Since x ⊥ Mwe have 0 on the left-hand side of Parseval’s Equality which is not equal to ‖x‖. Hence ifParceval’s Equality holds for all x ∈ H, then M must be total in H.

Conversely, assume M to be total in H. Consider any x ∈ H and its nonzero Fouriercoefficients arranged in a sequence, i.e., 〈x, e1〉 , 〈x, e2〉 , ...,. Define y =

∑ 〈x, ek〉 ek. Itfollows that x − y ∈ M⊥. Since M is total, then x = y.

48

Theorem 3.18 (Separable Hilbert Space). Let H be a Hilbert space.

1. If H is separable then every orthonormal set is countable.

2. If H contains an orthonormal sequence which is total then H is separable.

3.1 Representation of Functionals on Hilbert Spaces

Theorem 3.19 (Riesz). Let H be a Hilbert space. Every bounded, linear functional on Hcan be represented in terms of the inner product on H, i.e.,

f(x) = 〈x, z〉

where z is uniquely determined by f and

‖z‖ = ‖f‖ .

Proof.

1. Existence of z. If f = 0 take z = 0. Otherwise assume f 6= 0. Then N (f) 6= H andN (f) 6= H ⇒ N (f)⊥ 6= {0}. Let w ∈ N (f)⊥ s.t. w 6= 0 and set

v = f(x)w − f(w)x, x ∈ H.

Then

⇒ f(v) = f(x)f(w) − f(w)f(x) = 0

⇒ v ∈ N (f) .

Since w ⊥ N (f) we have

0 = 〈v,w〉= 〈f(x)w − f(w)x,w〉= f(x) 〈w,w〉 − f(w) 〈x,w〉= f(x) ‖w‖2 − f(w) 〈x,w〉 .

Then

⇒ f(x) =f(w)

‖w‖2 〈x,w〉

⇒ f(x) =

⟨

x,

(

f(w)

‖w‖2

)

w

⟩

.

Then

z =

(

f(w)

‖w‖2

)

w.

49

2. Uniqueness of z. Suppose f(x) = 〈x, z1〉 = 〈x, z2〉. Then

⇒ 〈x, z1 − z2〉 = 0, ∀ x ∈ H.

Choose x = z1 − z2. Then

⇒ 〈z1 − z2, z1 − z2〉 = 0

⇒ z1 = z2.

3. ‖f‖ = ‖z‖. If f = 0 then z = 0 and ‖f‖ = ‖z‖ = 0. For f 6= 0, z 6= 0. Note

f(z) = 〈z, z〉= ‖z‖2 , and

‖z‖2 ≤ ‖f‖‖z‖⇒ ‖z‖ ≤ ‖f‖ .

Also

|f(x)| = |〈x, z〉|≤ ‖x‖ ‖z‖

⇒ ‖f‖ ≤ ‖z‖ .

This yields‖f‖ = ‖z‖ .

Lemma 3.20 (Equality). Let X be an inner product space. Then

〈x,w〉 = 〈y,w〉 ∀ w ∈ X ⇒ x = y.

In particular,〈x,w〉 = 0 ∀ w ∈ X ⇒ x = 0.

Definition (Sesquilinear Form). Let X,Y be vector spaces over the same scalar field K(R or C). A sesquilinear form (or sesquilinear functional) h on X × Y is a mapping

h : X × Y → K

such that∀ x, x1, x2 ∈ X and y, y1, y2 ∈ Y and α, β ∈ K

1. h(x1 + x2, y) = h(x1, y) + h(x2, y)

50

2. h(x, y1 + y2) = h(x, y1) + h(x, y2)

3. h(αx, y) = αh(x, y)

4. h(x, βy) = βh(x, y)

Remark 3.21.

1. If X,Y are real (K = R), then

h(x, βy) = βh(x, y)

and h is said to be bilinear.

2. If X,Y are vector spaces and ∃ c ∈ R s.t.

|h(x, y)| ≤ c ‖x‖ ‖y‖ , ∀ x, y

then h is bounded and

‖h‖ = supx 6=0y 6=0

|h(x, y)|‖x‖ ‖y‖

= sup‖x‖=1‖y‖=1

|h(x, y)|

⇒ |h(x, y)| ≤ ‖h‖ ‖x‖ ‖y‖ .

Theorem 3.22 (Riesz Representation). Let H1,H2 be Hilbert spaces and

h : H1 × H2 → K

a bounded sesquilinear form. Then h has a representation

h(x, y) = 〈Sx, y〉

where S : H1 → H2 is a bounded, linear operator. S is uniquely determined by h and

‖S‖ = ‖h‖ .

Proof.

1. Existence of S. Consider h(x, y) which is linear in y. If we fix an x, then h(x, y) is abounded, linear functional and we can use the Riesz theorem to find z and represent

h(x, y) = 〈y, z〉 ,

51

henceh(x, y) = 〈z, y〉 . (8)

Note that z is unique, but it depends on x ∈ H1. This means that (8) with variablex defines an operator S : H1 → H2 given by

z = Sx,

and we writeh(x, y) = 〈z, y〉 = 〈Sx, y〉 .

We must show S is linear. Observe

〈S(αx1 + βx2), y〉 = h(αx1 + βx2, y)

= αh(x1, y) + βh(x2, y)

= α 〈Sx1, y〉 + β 〈Sx2, y〉 , ∀ y ∈ H2

⇒ S(αx1 + βx2) = αSx1 + βSx2.

2. Uniqueness of S. If h(x, y) = 〈Sx, y〉 = 〈Tx, y〉, then S = T by the equality lemma.

3. Boundedness of S. Note first

‖h‖ = supx 6=0y 6=0

|〈Sx, y〉|‖x‖ ‖y‖

≥ supx 6=0

Sx 6=0

|〈Sx, Sx〉|‖x‖ ‖Sx‖

= supx 6=0

‖Sx‖‖x‖

= ‖S‖⇒ ‖h‖ ≥ ‖S‖ .

So S is bounded. Also

‖h‖ = supx 6=0y 6=0

|〈Sx, y〉|‖x‖ ‖y‖

≤ supx 6=0y 6=0

‖Sx‖ ‖y‖‖x‖ ‖y‖

≤ supx 6=0

‖Sx‖‖x‖

= ‖S‖⇒ ‖h‖ ≤ ‖S‖ .

52

This yields‖S‖ = ‖h‖ .

Problem 23. Let H be a Hilbert space. Show that H ′ (the dual of H) is a Hilbert spacewith inner product 〈·, ·〉1 defined by

〈fz, fv〉1 = 〈z, v〉 = 〈v, z〉 ,

where

fz(x) = 〈x, z〉fv(x) = 〈x, v〉

Solution. It is easy to verify that 〈·, ·〉1 is an inner product on H′. Since ‖fz‖ = ‖z‖ =

(〈z, z〉) 12 = (〈fz, fz〉1)

12 the norm on H

′is induced by the inner product 〈·, ·〉1. We know

that the normed dual is always a Banach space. It follows that H′is complete, and therefore

a Hilbert space.

Problem 24. Show that any Hilbert space H is isomorphic with its second dual, i.e.,

H ∼= (H ′)′ = H ′′.

Solution. Let T : H → h′′

by z → Fz, where Fz : H′ → K is defined by Fz(f) = 〈f, fz〉1,

where 〈·, ·〉1 and fz are the same as in Problem 23. By repeated applications of the Rieszrepresentation theorem we have that F (f) = 〈f, g〉1 for some g ∈ h

′and g = fz for some

z ∈ H. It follows that T is surjective.Suppose that z,w ∈ H and Fz = Fw. Then 〈f, fz〉1 = 〈f, fw〉1 for all f ∈ H

′and we

get that fz = fw and z = w. Therefore, T is injective.It is easy to see that T is linear.Also, by the Riesz representation theorem we have that ‖Fz‖ = ‖fz‖ = ‖z‖, i.e., T

preserves norms, and inner products. It follows that T is an isomorphism.

3.2 Hilbert Adjoint

Definition (Hilbert Adjoint T ∗). Let H1,H2 be Hilbert spaces and let T : H1 → H2 be abounded, linear operator. Then the adjoint is T ∗ : H2 → H1 s.t.

〈Tx, y〉 = 〈x, T ∗y〉 , ∀ x ∈ H1 and x ∈ H2.

Theorem 3.23 (Existence of the Hilbert Adjoint).

1. T ∗ exists.

53

2. T ∗ is unique.

3. ‖T ∗‖ = ‖T‖.

Proof. Observe that h(y, x) = 〈y, Tx〉 is sesquilinear form on H2 × H1.

⇒ |h(x, y)| ≤ ‖y‖ ‖Tx‖≤ ‖T‖ ‖x‖ ‖y‖

⇒ ‖h‖ ≤ ‖T‖ .

Also

‖h‖ = supx 6=0y 6=0

|〈y, Tx〉|‖y‖ ‖x‖

≥ supx 6=0

Tx 6=0

|〈Tx, Tx〉|‖Tx‖ ‖x‖

= ‖T‖⇒ ‖h‖ = ‖T‖ .

Therefore h is a bounded sesquilinear form. Using a Riesz representation, h(x, y) = 〈T ∗y, x〉where T ∗ exists, is unique with norm

‖T ∗‖ = ‖h‖ = ‖T‖ .

Also

⇒ 〈y, Tx〉 = 〈T ∗y, x〉⇒ 〈Tx, y〉 = 〈x, T ∗y〉 .

Lemma 3.24 (Zero Operator). Let X,Y be inner product spaces and Q : X → Y bebounded, linear operators. Then

1. Q = 0 iff 〈Qx, y〉 = 0 ∀ x ∈ X and y ∈ Y

2. Let X be complex and Q : X → X. Then

〈Qx, x〉 = 0 ∀ x ∈ X ⇒ Q = 0.

Proof.

54

1. ⇒.

Q = 0 ⇒ Qx = 0 ∀ x ∈ X

⇒ 〈Qx, y〉 = 〈0, y〉 = 0 〈w, y〉 = 0.

⇐.

〈Qx, y〉 = 0 ∀ x, y ⇒ Qx = 0 ∀ x, y

⇒ Q = 0.

2. Let x, y ∈ X, then v = αx + y ∈ X. Then

〈Qv, v〉 = 〈Qx, x〉 = 〈Qy, y〉 = 0

⇒0 = 〈Q(αx + y), αx + y〉

= |α|2 〈Qx, x〉 + 〈Qy, y〉 + α 〈Qx, y〉 + α 〈Qy, x〉= α 〈Qx, y〉 + α 〈Qy, x〉 .

Now, take α = 1 then α = i to obtain the two relations

〈Qx, y〉 + 〈Qy, x〉 = 0

〈Qx, y〉 − 〈Qy, x〉 = 0.

Then 〈Qx, y〉 = 0 ⇒ Q = 0 by 1.

Remark 3.25. In the previous lemma, 2 is not necessarily true if X is real.

Theorem 3.26 (Properties of The Hilber Adjoint). Let H1,H2 be Hilbert spaces. LetS, T : H1 → H2 be bounded, linear operators and let α be a scalar. Then

1. 〈T ∗y, x〉 = 〈y, Tx〉

2. (S + T )∗ = S∗ + T ∗

3. (αT )∗ = αT ∗

4. (T ∗)∗ = T

5. ‖T ∗T‖ = ‖TT ∗‖ = ‖T‖2

6. T ∗T = 0 iff T = 0

55

7. (ST )∗ = T ∗S∗, assuming H1 = H2.

Definition (Self-Adjoint, Unitary, Normal). Let H be a Hilbert space and T : H → H.

1. T is self-adjoint (or Hermitian) if T ∗ = T

2. T is unitary if T is bijection and T ∗ = T−1

3. T is normal if TT ∗ = T ∗T

Remark 3.27.

1. If T is self-adjoint then 〈Tx, y〉 = 〈x, Ty〉.

2. If T is self-adjoint or normal then T is normal.

Example 3.28. Consider Cn with 〈x, y〉 = xT y. Let T : Cn → Cn. If we specify a basisfor Cn, we can represent T, T ∗ by matrices A,B. Then

〈Tx, y〉 = (Ax)T y

= xT AT y

〈x, T ∗y〉 = xT By.

Therefore

⇒ AT = B

⇒ B = AT.

If T is self-adjoint then A = AT.

Theorem 3.29 (Self-Adjointness). Let H be a Hilbert space and let T : H → H be abounded, linear operator. Then

1. If T is self-adjoint then 〈Tx, x〉 is real for all x ∈ H.

2. If H is complex and 〈Tx,X〉 is real for all x ∈ H then T is self-adjoint.

Proof.

1. If T is self-adjoint, then for all x ∈ X we have

〈Tx, x〉 = 〈x, Tx〉 = 〈Tx, x〉 .

56

2. If 〈Tx, x〉 is real for all x ∈ X, then

〈Tx, x〉 = 〈Tx, x〉 = 〈x, T ∗x〉 = 〈T ∗x, x〉 .

Hence

0 = 〈Tx, x〉 − 〈T ∗x, x〉= 〈(T − T ∗)x, x〉

⇒ T = T ∗

by the zero operator lemma, since H is complex.

Theorem 3.30. Let H be a Hilbert space and let (Tn) be a sequence of bounded, linear,self-adjoint operators Tn : H → H. Suppose that Tn → T in norm, that is ‖Tn − T‖ → 0,where ‖·‖ is the norm on the space B(H,H). Then T is a bounded, linear, self-adjointoperator on H.

Proof. We show T ∗ = T .

‖T − T ∗‖ ≤ ‖T − Tn‖ + ‖Tn − T ∗n‖ + ‖T ∗

n − T ∗‖= ‖T − Tn‖ + 0 + ‖T ∗

n − T ∗‖= 2 ‖T − Tn‖→ 0.

Theorem 3.31 (Unitary Operators). Let H be a Hilbert space and let U, V : H → H beunitary. Then

1. U is isometric, i.e., ‖Ux‖ = ‖x‖ ∀ x ∈ H

2. ‖U‖ = 1 provided H 6= {0}

3. U−1 = U∗ is unitary

4. UV is unitary

5. U is normal.

6. If T is a bounded, linear operator on H and H is complex, then T is unitary iff Tis isometric and surjective.

Proof.

57

1.

‖Ux‖2 = 〈Ux,Ux〉= 〈x,U∗Ux〉= 〈x, Ix〉= ‖x‖2

2. Follows from 1.

3. Since U is bijective, so is U−1 and

(U−1)∗ = U∗∗ = U = (U−1)−1.

4. UV is bijective and

(UV )∗ = V ∗U∗ = V −1U−1 = (UV )−1.

5. U−1 = U∗ and UU−1 = U−1U = I.

6. ⇒. Suppose that T is isometric and surjective. Isometry implies injectivity, so T isbijective. Need to show T ∗ = T−1. By isometry,

〈T ∗Tx, x〉 = 〈Tx, Tx〉= 〈x, x〉= 〈Ix, x〉

⇒ 〈(T ∗T − I)x, x〉 = 0

⇒ T ∗T = I.

Also,

TT ∗ = TT ∗(TT−1)

= T (T ∗T )T−1

= I.

Therefore T ∗ = T−1.

⇐. Conversely, T is isometric by 1 and surjective by definition.

58

4 Fundamental Theorems

Definition (Partially Ordered Set). Let M be a set with a relation ≤. This relation issaid to be a partial order if

1. a ≤ a ∀ a ∈ M

2. a ≤ b and b ≤ a ⇒ a = b

3. a ≤ b and b ≤ c ⇒ a ≤ c.

We say M is a partially ordered set.

Definition (Upper Bound). Let M be partially ordered set. If ∃ u ∈ M s.t. x ≤ u ∀ x ∈M , then u is said to be a upper bound. Such a u is not guaranteed to exist.

Definition (Maximal Element). Let M be partially ordered set. If ∃ m ∈ M s.t. m ≤x ⇒ m = x ∀ x ∈ M , then m is said to be a maximal elemtn. Such a m is not guaranteedto exist or be unique.

Example 4.1.

1. R with the usual ≤.

2. The power set P (X) is partially ordered by set inclusion. X is the unique maximalelement.

3. x = (xi), y = (yi) ∈ Rn ordered by

x ≤ y ⇔ xi ≤ yi ∀ i = 1, . . . , n.

4. The positive integers ordered by

x ≤ y ⇔ x|n.

Definition (Chain). Let M be a partially ordered set. A subset C ⊂ M is said to be achain if a, b ∈ C ⇒ a ≤ c or c ≤ a. I.e., all elements in C are comparable.

The next statement is equivalent to the axiom of choice. It retain’s its name (lemma)for historical reasons.

Lemma 4.2 (Zorn’s Lemma). Let M 6= ∅ be a partially ordered set. Suppose that everychain C ⊂ M has an upper bound. Then M has at least one upper bound.

If we decide to live with the the axiom of choice, we can use Zorn’s Lemma to provevarious important results.

59

Theorem 4.3 (Hemel Basis). Every vector space X 6= {0} has a Hemel basis.

Proof. Let M be the set of all linearly independent subsets of X. Note that M is notempty:

X 6= {0} ⇒ ∃ x ∈ X s.t. x 6= 0 ⇒ {x} ∈ M.

Set inclusion gives a partial order on M . Let C ⊂ M be a chain. Then

A =⋃

D∈C

D

is an upper bound for C.With this setup we invoke Zorn’s lemma to force the existence of a maximal element

B. Let Y = span (B) and so Y ⊂ X. We want to show Y = X. Assume not and letz ∈ X − Y . Then B ∪ {z} is linearly independent, contradicting the maximality of B.

Theorem 4.4 (Total Orthonormal Set). In ever Hilbert space H 6= {0}, there exists atotal orthonormal set.

Proof. Let M be the set of all orthonormal subsets of H. M is not empty since

X 6= {0} ⇒ ∃ x ∈ X s.t. x 6= 0 ⇒ { x

‖x‖} ∈ M.

Set inclusion gives a partial order on M . Let C ⊂ M be a chain. Then

A =⋃

D∈C

D

is an upper bound for C.With this setup we invoke Zorn’s lemma to force the existence of a maximal element F .

We claim this F is total. To this end, suppose not. Then ∃ z ∈ H s.t. z 6= 0 and z ⊥ F .Hence F ∪ { z

‖z‖} is orthonormal, contradicting the maximality of F .

Definition (Sublinear Functional). Let X be a v.s. and let p : X → R. p is said to be asublinear functional if

1. p(x + y) ≤ p(x) + p(y) ∀ x, y ∈ X

2. p(αx) = αp(x) ∀ x ∈ X ∀ α ∈ R, α ≥ 0.

Remark 4.5. The norm on a n.s. is a sublinear functional.

Theorem 4.6 (Hanh-Banach (Extensions of Linear Functionals)). Let X be a real v.s.and let p be a sublinear functional on X. Let f be a linear functional defined on a subspaceZ ⊂ X s.t.

f(x) ≤ p(x) ∀ x ∈ Z.

60

Then f has a linear extension f : Z → X s.t.

f(x) ≤ p(x) ∀ x ∈ X

f(x) = f(x) ∀ x ∈ Z.

Proof.

1. Let E be the set of all linear extensions g of f which satisfy

g(x) ≤ p(x) ∀ x ∈ D (g) .

Then f ∈ E ⇒ E 6= ∅. We now define the following partial order on E: g ≤ h meansh is a linear extension of g, i.e.

(a) D (g) ⊂ D (h)

(b) h(x) = g(x) ∀ x ∈ D (g).

Let C be a chain in M . Then we define the maximal element g as follows

(a) D (g) =⋃

g∈C D (g)

(b) g(x) = g(x) ∀ g ∈ C s.t. x ∈ D (g).

Then g ≤ g ∀ g ∈ C ⇒ g is an upper bound for C. Since C ⊂ E was arbitrary,we invoke Zorn’s lemma to produce a maximal element f . Therefore f is a linearextension of f .

2. Now we want to show D(

f)

= X. In the manner above, assume otherwise and let

y1 ∈ x − D(

f)

and consider Y1 = span ({)D(

f)

∪ {y1}}. If x ∈ Y1 then we can

write x = y + αy1, y ∈ D(

f)

(and this representation is unique). Then let us define

a functional g1 : Y1 → R as

g1(y + αy1) = f(y) + αc

where c is any real constant. Then g1 is linear extension of f . If we can show

g1(x) ≤ p(x) ∀ x ∈ D (g1)

then g1 ∈ E, which would contradict the maximality of f and imply D(

f)

= X.

3. To this end, let us construct a suitable c. Let y, z ∈ D(

f)

. Then

f(y) − f(z) = f(y − z)

≤ p(y − z)

= p(y + y1 − y1 − z)

≤ p(y + y1) + p(−y1 − z)

61

and therefor−p(−y1 − z) − f(z) ≤ p(y + y1) − f(y).

Note

(a) y1 is fixed

(b) y does not appear on the RHS

(c) z does not appear on the LHS.

Then we can take supz∈D(f), inf

y∈D(f) and call these values m0,m1, respectively.

Then m0 ≤ m1. Let c be any value such that m0 ≤ c ≤ m1. Then

−p(−y1 − z) − f(z) ≤ c ∀ z ∈ D(

f)

c ≤ p(y + y1) − f(y) ∀ y ∈ D(

f)

.

Now let α ∈ R and consider the cases

(a) α < 0.

−p(−y1 −1

αy) − f(

1

αy) ≤ c

αp(−y1 −1

αy) + f(y) ≤ −αc

⇒

g1(x) = g(y + αy1)

= f(y) + αc

≤ −αp(−y1 −1

αy)

= p(αy1 + y) = p(x).

(b) α = 0. Then x ∈ D(

f)

so g1(x) = f(x) ≤ p(x).

(c) α > 0.

c ≤ p(1

αy + y1) − f(

1

αy)

αc ≤ αp(1

αy + y1) − f(y)

= p(x) − f(y)

62

⇒

g1(x) = f(y) + αc

≤ p(x).

Then g1 ∈ E and we have our contradiction. Therefore f is our bounded linearextension.

Problem 25.

1. Show that the norm on a v.s. X is a sublinear functional on X.

2. Show that a sublinear functional p satisfies p(0) = 0 and p(−x) ≥ −p(x).

Solution.

1. Let p(x) = ‖x‖. Then p(x + y) = ‖x + y‖ ≤ ‖x‖+ ‖y‖ = p(x) + p(y). Also, if α ≥ 0,then

p(αx) = ‖αx‖ = |α|‖x‖ = α‖x‖ = αp(x)

. It follows that p is a sublinear functional.

2. X is not empty, 0 ∈ X. Let x ∈ X, then p(0) = p(0x) = 0p(x) = 0. Also,0 = p(0) = p(x − x) ≤ p(x) + p(−x). It follows that p(−x) ≥ −p(x).

Problem 26. If p is a sublinear functional on a real v.s. X, show that there exists a linearfunctional f on X s.t.

−p(−x) ≤ f(x) ≤ p(x).

Solution. Let Z = {x ∈ X : x = αx + 0, α ∈ R} and define the linear functional f on Z by

f(x) = αp(x0).

Then f(x) ≤ p(x). By the Hahn-Banach Theorem there exists a f bounded linear func-tional defined on X such that f(x) ≤ p(x). Clearly, −f(x) = f(−x) ≤ p(x). It followsthat −p(−x) ≤ f(x) ≤ p(x).

Theorem 4.7 (Hanh-Banach Generalized). Let X be a v.s. over K (R or C) and let p bea real-valued functional X s.t.

1. p(x + y) ≤ p(x) + p(y) ∀ x, y ∈ X

2. p(αx) = |α| p(x) ∀ α ∈ K.

63

Let Z ⊂ X be a subspace and let f : Z → K be a functional s.t.

|f(x)| ≤ p(x) ∀ x ∈ Z.

Then f has a bounded linear extension f s.t.

1. D(

f)

= X

2. f(x) = f(x) ∀ x ∈ Z

3.∣

∣

∣f(x)

∣

∣

∣≤ p(x) ∀ x ∈ X.

Proof.

1. K = R. Then|f(x)| ≤ p(x) ⇒ f(x) ≤ p(x).

Then by the previous theorem ∃ f s.t. f(x) ≤ p(x) ∀ x ∈ X. Also,

−f(x) = f(−x) ≤ p(−x) = p(x)

so∣

∣

∣f(x)

∣

∣

∣≤ p(x).

2. K = C. X complex ⇒Z complex ⇒f complex ⇒

f = f1 + if2

where f1, f2 are real-valued. We introduce Xr, Zr the v.s.’s obtained by restricting Kto R. Now note that f linear on Z and f1, f2 real-valued means f1, f2 are real-valuedlinear functionals on Zr. Also,

f1(x) ≤ |f(x)| ⇒ f1(x) ≤ p(x) ∀ x ∈ Zr.

Then by the previous theorems ∃ f1 extension of f1 from Zr to Xr s.t.

f1(x) ≤ p(x) ∀ x ∈ Xr.

Going back to Z, we observe ∀ x ∈ Z

if(x) = i [f1(x) + if2(x)]

= f1(ix) + if2(ix)

= −f2(x) + if1(x)

⇒ f2(x) = −f1(ix)

64

So then ∀ x ∈ X we define

f(x) = f1(x) − if1(ix).

Notef(x) = f(x) ∀ x ∈ Z.

(a) We claim f is a linear functional on X.

f((a + ib)x) = f1(ax + ibx) − if1(iax − bx)

= af1(x) + bf1(ix) − i[

af1(ix) − bf1(x)]

= (a + ib)[

f1(x) − if1(ix)]

= (a + ib)f(x)

(b) We claim∣

∣

∣f(x)

∣

∣

∣≤ p(x) ∀ x ∈ X. Recall that p(x) ≥ 0. Then

f(x) = 0 ⇒∣

∣

∣f(x)

∣

∣

∣≤ p(x).

Assume f(x) 6= 0. We then use the exponential form of f

f(x) =∣

∣

∣f(x)

∣

∣

∣eiθ.

Then∣

∣

∣f(x)

∣

∣

∣= f(x)e−iθ

= f(e−iθx)

= f1(e−iθx) + if1(ie

−iθx)∣

∣

∣f(x)

∣

∣

∣∈ R ⇒∣

∣

∣f(x)

∣

∣

∣= f1(e

−iθx)

≤ p(e−iθx)

=∣

∣

∣e−iθ

∣

∣

∣p(x)

= p(x)

Then f is the bounded, linear extension.

65

Theorem 4.8 (Hanh-Banach (for Normed Spaces)). Let X be a n.s. and let Z ⊂ X bea subspace. Let f : Z → K be a bounded, linear functional. Then there exists a bounded,linear functional f which is an extension from Z to X s.t.

∥

∥

∥f∥

∥

∥

X= ‖f‖Z

where∥

∥

∥f∥

∥

∥

X= sup

x∈X‖x‖=1

∣

∣

∣f(x)

∣

∣

∣

‖f‖Z = supx∈Z

‖x‖=1

|f(x)| .

Proof. If Z = {0} then f = 0 and f = 0. Otherwise, let Z 6= {0}. Then ∀ x ∈ Z we have

|f(x)| ≤ ‖f‖Z ‖x‖ .

Then definep(x) ≤ ‖f‖Z ‖x‖ ∀ x ∈ X.

Notice

p(x + y) = ‖f‖Z ‖x + y‖≤ ‖f‖Z (‖x‖ + ‖y‖= p(x) + p(y)

p(αx) = ‖f‖Z ‖αx‖= |α| ‖f‖Z ‖x‖= |α| p(x).

Then we can use the Hanh-Banach (Generalized) theorem ⇒ ∃ f : X → K s.t.∣

∣

∣f(x)

∣

∣

∣≤ p(x) = ‖f‖Z ‖x‖ ∀ x ∈ X.

And so∥

∥

∥f∥

∥

∥

X= sup

x∈X‖x‖=1

∣

∣

∣f(x)

∣

∣

∣≤ ‖f‖Z .

Finally, since f is an extension of f∥

∥

∥f∥

∥

∥

X≤ ‖f‖Z .

Therefore,∥

∥

∥f∥

∥

∥

X= ‖f‖Z .

66

Remark 4.9. In the previous theorem if X = H is a Hilbert space and Z ⊂ H is a closedsubspace then we can represent f (for some fixed z ∈ Z)

f(x) = 〈x, z〉 ∀ x ∈ Z

‖f‖ = ‖z‖f(x) = 〈x, z〉 ∀ x ∈ X

Theorem 4.10 (Bounded Linear Functionals). Let X be a n.s. and let x0 ∈ X,x0 6= 0.Then ∃ f : X → K s.t.

∥

∥

∥f∥

∥

∥= 1

f(x0) = ‖x0‖

Proof. Let Z ⊂ X be the subspace defined as

Z = {x ∈ X|x = αx0}.

Then

f(x) = f(αx0)

= α ‖x0‖|f(x)| = |f(αx0)|

= |α| ‖x0‖= ‖αx0‖= ‖x‖

⇒ ‖f‖ = 1.

Therefore f has a linear extension f from Z to X with norm∥

∥

∥f∥

∥

∥= ‖f‖ = 1.

By definition of f and f , f(x0) = f(x0) = ‖x0‖.

Problem 27. To illustrate the Hanh-Banach (for Normed Spaces) consider a functionalf : R2 → R defined by

f(x) = α1x1 + α2x2 ∀ x = (x1, x2) ∈ R2.

Construct its linear extension to R3 and the corresponding norms.

Solution. Let f(x) = α1ξ1 + α2ξ2 + α3ξ3. Then ‖f‖(α21 + α2

2 + α23)

12 ≥ ‖f‖ and ‖f‖ = ‖f‖

iff α3 = 0.

67

Problem 28. Consider the n.s. R2 and let x0 ∈ R2, x0 6= 0. Find a bounded, linearfunctional f : R2 → R s.t.

1.∥

∥

∥f∥

∥

∥= 1

2. f(x0) = ‖x0‖Solution. let f = 〈x, x0

‖x0‖〉. Then ‖f‖‖ x0‖x0‖‖ = 1 and f(x0) = 〈x0,

x0‖x0‖ 〉 = ‖x0‖.

4.1 Bounded, Linear Functionals on C[a, b]

Definition (Variation). Let f : [a, b] → R. We define the variation of f as

Var f = sup

{

n∑

i=1

|f(ti) − f(ti−1)|}

wherea = t0 < t1 < · · · < tn = b

is a partition of [a, b].

Definition (Bounded Variation). We say f is of bounded variation and write f ∈ BV [a, b]if

Var f < ∞.

Note BV [a, b] is a vector space. If we define the norm

‖f‖ = |f(a)| + Var f, ∀ f ∈ BV [a, b]

then BV [a, b] is a normed space.

Definition (Riemann-Stieltjes Integral). Let x ∈ C[a, b] and f ∈ BV [a, b] and Pn be apartition of [a, b]. Define

ηPn = max {t1 − t0, . . . , tn − tn−1} .

Consider S defined by

S(Pn) =n∑

i=1

x(ti) [f(ti) − f(ti−1)] .

Then ∃ I ∈ R s.t. ∀ ε > 0 ∃ δ > 0 s.t.

ηPn < δ ⇒ |I − S(Pn)| ≤ ε.

We call I the Riemann-Stieltjes integral of x with respect to f and write

I =

∫ b

a

x(t)df(t).

68

Remark 4.11.

1. If f(t) = t then we have the usual integral.

2. If f has a integrable derivative then

∫ b

a

x(t)df(t) =

∫ b

a

x(t)f ′(t)dt.

3. We have∣

∣

∣

∣

∫ b

a

x(t)df(t)

∣

∣

∣

∣

≤ supt∈[a,b]

|x(t)|Var(f).

Theorem 4.12 (Reisz - B.L.F.’s on C[a, b]). Every b.l.f. g on C[a, b] can be representedby a Riemann-Stieltjes integral

g(x) =

∫ b

a

x(t)df(t)

and‖g‖ = Var(f).

Proof. By the Hanh-Banach theorem g has an extension from C[a, b] to B[a, b] the spaceof bounded functions with norm

‖x‖ =∑

t∈[a,b]

|x(t)|

and‖g‖ = ‖g‖ .

Let

xt(s) =

{

1 s ∈ [0, t]0 otherwise

,

then xt ∈ B[a, b]. Let f(a) = 0 and f(t) = g(xt) ∀ t ∈ (a, b]. We want to show f is of b.v.and Var F ≤ ‖g‖.

Use exponential form for a complex quantity z, z = |z| e(z) where

e(z) =

{

1 z = 0eiθ z 6= 0

.

Note that if z 6= 0 then |z| = ze−iθ ⇒

|z| = ze(z).

69

We shall use the notation

εi = e(f(ti) − f(ti−1)) and xti = xi.

Then

n∑

i=1

[f(ti) − f(ti−1)] = |g(x1)| +n∑

i=2

|g(xi) − g(xi−1)|

= ε1g(x1) +n∑

i=1

εi [g(xi) − g(xi−1)]

= g

(

ε1x1 +

n∑

i=2

εi [xi − xi−1]

)

≤ ‖g‖∥

∥

∥

∥

∥

ε1x1 +n∑

i=2

εi [xi − xi−1]

∥

∥

∥

∥

∥

= ‖g‖ · 1⇒ Var f ≤ ‖g‖

⇒ f ∈ BV [a, b]

Now, given P a partition of [a, b] define

zn = x(t0)x1 +n∑

i=2

x(ti−1 [xi − xi−1]

then zn ∈ BV [a, b]. Now we compute

g(zn) = x(t0)g(x1) +

n∑

i=2

x(ti−1) [g(xi) − g(xi−1)]

= x(t0)f(x1) +n∑

i=2

x(ti−1) [f(xi) − f(xi−1)]

=n∑

i=1

x(ti−1) [f(xi) − f(xi−1)]

Then taking a sequence of partitions (Pn) s.t. ηPn → 0, we obtain

∫ b

a

x(t)df(t).

We need to show g(zn) → g(x) = g(x) where x ∈ C[a, b]. Note zn(a) = x(a) · 1 ⇒ zn(z) =x(a) = 0. Note

ti−1 < t ≤ ti ⇒ |zn(t) − x(t)| = |x(ti−1) − x(t)| .

70

NoteηPn → 0 ⇒ ‖zn − x‖ → 0

since x ∈ C[a, b] ⇒ x is uniformly continuous (since [a, b] is compact). Then the continuityof g implies g(zn) → g(x) and g(x) = f(x).

Now we want to show Var f = ‖g‖. Computing,

|g(x)| ≤ maxt∈[a,b]

x(t)Var f = ‖x‖Var f

and so‖g‖ ≤ Var f.

Also Var f ≤ ‖g‖, so‖g‖ = Var f.

Remark 4.13. f is not unique in the above theorem. However, we can make f unique byrequiring:

1. f(a) = 0

2. f(t+) = f(t) (continuity from the right).

4.2 Adjoint Operator

Given a b.l.o. T : X → Y we are interested in constructing a new operator T× : Y ′ → X ′.We proceed as follows. Let g ∈ Y ′, x ∈ X. Setting y = Tx we obtain a function f on Xby defining

f(x) = g(Tx).

f is linear because g, T are linear. f is bounded because

|f(x)| = |g(Tx)| ≤ ‖g‖ ‖T‖ ‖x‖⇒ ‖f‖ ≤ ‖g‖ ‖T‖ .

Therefor f ∈ X ′. Thenf(x) = g(Tx) (9)

with variable g ∈ Y ′ defines an operator called the adjoint of T ,

T× : Y ′ → X ′.

Definition (Adjoint). Given a b.l.o. T : X → Y the adjoint T× : Y ′ → X ′ is given by

f(x) = (T×g)(x) = g(Tx).

71

Theorem 4.14. ‖T×‖ = ‖T‖.Proof. We know T× is linear and f = T×g. Then

‖T×g‖ = ‖f‖ ≤ ‖g‖ ‖T‖ ⇒ ‖T×‖ ≤ ‖T‖ .

Now we want to show ‖T×‖ ≥ ‖T‖. For any x0 ∈ X,x0 6= 0 ∃ g0 ∈ Y ′ s.t. ‖g0‖ = 1 and

g0(Tx0) = ‖Tx0‖ .

Hence

g0(Tx0) = (T×g0)(x0)

f0 = T×g0

⇒

‖Tx0‖ = g(Tx0)

= f0(x0)

≤ ‖f0‖ ‖x0‖= ‖T×g0‖ ‖x0‖≤ ‖T×‖ ‖g0‖ ‖x0‖ .

Recall ‖g0‖ = 1, then

‖Tx0‖ ≤ ‖T×‖ ‖x0‖ ⇒ ‖T×‖ ≥ ‖Tx0‖‖x0‖

.

But x0 6= 0 is arbitrary, and taking sup on the RHS yields

‖T×‖ ≥ ‖T‖ .

Example 4.15. Let T : Rn → Rn. Let E be a basis, and let x, y ∈ Rn, and let TE be therepresentation of T with respect to E.

E = {e1, . . . , en}x = (ξ1, . . . , ξn)

y = (η1, . . . , ηn)

TE = (τik)

y = TEx

ηi =n∑

k=1

τikξk.

72

Let F = {f1, . . . , fn} be the dual basis of E, i.e., a basis for Rn′. Then for g ∈ Rn′, we canrepresent

g =

n∑

i=1

αifi

fi(y) = fi

(

n∑

i=1

ηkek

)

= ηi

⇒ g(y) = g (TEx)

=n∑

i=1

αiηi

=

n∑

i=1

n∑

k=1

αiτikξk

⇒ g (TEx) =

n∑

k=1

βkξk where βk =

n∑

i=1

τikαi

⇒ f(x) = g (TEx)

=

n∑

k=1

βkξk

⇒ f = (TE)× g

βk =

n∑

i=1

τikαi.

Therefor, if T is represented by TE , then T× is represented by the transpose of TE .

Remark 4.16 (Properties of Adjoint).

1. (S + T )× = S× + T×

2. (αT )× = αT×

3. (ST )× = T×S×

4. If T ∈ B(X,Y ) and T−1 exists and T−1 ∈ B(Y,X), then (T×)−1 also exists, (T×)−1 ∈B(X ′, Y ′) and (T×)−1 = (T−1)×.

Remark 4.17 (Reltaion between T× and T ∗). Let T : X → Y,X = H1, Y = H2, T : H1 →H2, T

× : H ′2 → H ′

1, f ∈ H ′1g ∈ H ′

2

T×g = f

g(Tx) = f(x)

73

where H1,H2 are Hilbert spaces. Then let us represnet (Riesz)

f(x) = 〈x, x0〉 , x0 ∈ H1

g(y) = 〈y, y0〉 , y0 ∈ H2.

Then we can define A1 : H ′1 → H1 by A1f = x0 and A2 : H ′

2 → H2 by A2g = y0. ThenA1, A2 are bijective, isometric, and conjugate-linear. So we can define T ∗ : H2 → H1 by

T ∗y0 = A1T×A−1

2 y0 = x0.

⇒

〈Tx, y0〉 = g(Tx)

= f(x)

= 〈x, x0〉= 〈x, T ∗y0〉

Remark 4.18.

1. (αT )× = αT× but (αT )∗ = αT ∗

2. In the finite dimensional setting T ∗ is represented as the transpose and T ∗ is repre-sented by the conjugate transpose:

T× = T T

T ∗ = TT.

Problem 29.

1. Show (αT )× = αT×.

2. Show (T n)× = (T×)n.

Solution.

1. ((αT )xg)(x) = g((αT )x) = g(αTx) = αg(Tx) = α(txg)(x) = ((αT x)g)(x).

2. We have (ST )x = T xSx because ((ST )xg)(x) = g((ST )(x)) = g(S(T (x))) = (sxg)(Tx) =T x((Sxg)(x)) = (T x(Sxg))(x) = ((T xSx)g)(x). It follows that (T n)x = (T n−1)xT x =· · · = (tx)n.

74

4.3 Reflexive Spaces

Definition (Reflexive (Algebraic)). A vector space X is algebraically reflexive if the canon-ical map C : X → X∗∗ is surjective (and thus bijective). Recall C is the mapping

x 7→ gx where gx(f) = f(x), ∀ f ∈ X∗.

Definition (Reflexive (Dual)). A normed space X is called reflexive if X = X ′′, i.e., ifR (C) = X ′′.

Problem 30. Let X be a n.s.. Define gx : X ′ → K by fixing an x ∈ X and setting

gx(f) = f(x) ∀ f ∈ X ′.

1. Show ∀ fixed x ∈ X, gx is a b.l.f. on X ′ and

‖gx‖ = ‖x‖ .

2. C is an isomorphism Xonto−→ R (C).

3. If a n.s. X is reflexive (i.e., R (C) = X ′′) then X is complete.

Solution.

1. Let gx(f) = f(x), where x ∈ X is fixed. Then |gx(f)| = |f(x)| ≤ ‖f‖‖x‖ andtherefore g is bounded and ‖g‖ ≤ ‖x‖. Let f ∈ X

′such that ‖f‖ = 1 and f(x) = ‖x‖.

(We established the existence of such f .) Then ‖gx‖ ≥ |gx(f)|‖f‖ = ‖x‖. It follows that

‖gx‖ = ‖x‖.

2. Consider C : x → X′′, where x → gx. Then C is linear, because ∀f ∈ X

′we have

gαx+βy(f) = f(αx + βy) = αf(x) + βf(y) = αgx(f) + βgy(f).

Then C(αx + βy) = αgx + βgy = αC(x) + βC(y). Also, ‖C(x)‖ = ‖gx‖ = ‖x‖, andthen ‖gx − gy‖ = ‖gx−y‖ = ‖x − y‖ and C is isometric. Moreover, if x 6= y, thengx 6= gy and therefore C is injective. It follows that C is an isomorphism onto itsrange.

3. X′′

is complete being the dual of X′. By assumption X is reflexive, hence R(C) = X

′′.

Part (ii) implies the completeness of X via isomorphism.

Theorem 4.19. Every f.d.n.s. is reflexive.

Example 4.20.

75

1. ℓp,Lp[a, b], ∀ 1 < p < ∞ are reflexive.

2. C[a, b],L1[a, b], ℓ1 are not reflexive.

Theorem 4.21 (Hilbert Spaces are Reflexive). If H is a Hilbert space then H is reflexive.

Proof. The canonical map C : H → H ′′ given by x 7→ gx is injective. We want to show Cis surjective, i.e.,

∀ g ∈ H ′′ ∃ x ∈ H s.t. g = Cx.

Define A : H ′ → H by by Af = z where z is the Riesz representation of f :

f(x) = 〈x, z〉 .

Then A is bijective, an isometry and conjugate-linear. We view H ′ as the Hilbert spacewith inner product

〈f1, f2〉1 = 〈Af2, Af1〉 .

Let g ∈ H ′′ be arbitrary. Then

g(f) = 〈f, f0〉1 = 〈Af0, Af〉 .

Writing f(x) = 〈x, z〉 , z = Af,Af0 = z we have

〈Af0, Af〉 = 〈x, z〉 = f(x).

⇒ g(f) = f(x) ⇒ g = Cx. Then H is reflexive.

Lemma 4.22 (Existence of a Functional). Let Y be a n.s. and let Y ⊂ X be a proper,closed subspace. Let x0 ∈ X − Y be arbitrary. Calculate

δ = infy∈Y

‖y − x0‖ .

Note Y closed ⇒δ > 0. Then ∃ f ∈ X ′ s.t.

∥

∥

∥f∥

∥

∥= 1 and f(y) = 0 ∀ y ∈ Y and f(x0) = δ. (10)

Proof. Consider a subspace Z ⊂ X spanned by Y and x0 and define a sublinear functionalf on Z by

f(z) = f(y + αx0) = αδ.

Note δ 6= 0 ⇒ f 6= 0. Then f satisfies (10). Now we use the Hanh-Banach theorem toextend f to X. In a slight abuse of notation we refer to the extension of f as f . Then fis bounded. It is clear that α = 0 ⇒ f(z) = 0.

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For the case α 6= 0

|f(z)| = |α| δ= |α| inf

y∈Y‖y − x0‖

≤ |α| ‖−α−1y − x0‖= ‖y + αx0‖

⇒ |f(z)| ≤ ‖z‖⇒ ‖f‖ ≤ 1

Now to show ‖f‖ ≥ 1. ∃ (yn) ⊂ Y s.t. ‖yn − x0‖ → δ. Let zn = yn−x0, then f(zn) = −δ.Then

‖f‖ = supz∈Z

z 6=0

|f(z)|‖z‖

≥ |f(zn)|‖zn‖

=δ

‖zn‖→ 1,

⇒ ‖f‖ = 1.

Theorem 4.23 (Separability). X ′ separable ⇒X separable.

Proof. Assume X ′ is separable. Then

U ′ = {f | ‖f‖ = 1} ⊂ X ′

contains a countable, dense subset, say (fn) ⊂ U ′. Since fn ∈ U ′ ∀ n,

‖fn‖ = sup‖x‖=1

|f(x)| = 1.

By the definition of sup we can find a sequence (xn) ∈ X s.t. ‖xn‖ = 1 ∀ n s.t.

|fn(xn)| ≥ 1

2.

Let Y = span ((xn)), then Y is separable. We want to show Y = X (because then (xn) isa countable dense subset of X).

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Suppose Y ( X. Then ∃ f ∈ X ′ s.t.∥

∥

∥f∥

∥

∥= 1, f(y) = 0 ∀ y ∈ Y . Since (xn) ⊂ Y we

have f(xn) = 0 ∀ n. Then

1

2≤ |fn(xn)|

=∣

∣

∣fn(xn) = f(xn)

∣

∣

∣

=∣

∣

∣(fn − f)(xn)

∣

∣

∣

≤∥

∥

∥fn − f

∥

∥

∥‖xn‖

=∥

∥

∥fn − f

∥

∥

∥.

Then∥

∥

∥fn − f

∥

∥

∥≥ 1

2 and∥

∥

∥f∥

∥

∥= 1 contradicts that (fn) ⊂ U ′ is dense.

Corollary 4.24. X seprable and X ′ not separable ⇒X not reflexive.

Proof. X reflexive ⇒X = X ′′ ⇒X ′′ separable ⇒X ′ separable, which contradicts our as-sumptions.

Example 4.25. ℓ1 is not reflexive. This is because (ℓ1)′ = ℓ∞ and ℓ∞ is not separable.

4.4 Baire Category Theorem

Definition (Category). Let X be a metric space and M ⊂ X.

1. M is said to be nowhere dense in X if M has no interior points.

2. M is said to be of first category or meager if X is the countable union of nowheredense sets.

3. M is of the second category if it is not of first category.

Theorem 4.26 (Baire’s Theorem). If X 6= ∅ then X is of the second category (i.e., X isnot meager).

Proof. Suppose X 6= ∅ and X is meager. Then

X

∞⋃

k=1

Mk

with Mk nowhere dense in X ∀ k. By assumption, M1 does not contain a nonempty openset ⇒ M1 6= X, ⇒ M1

c= X − M1 is not empty and is open. Pick p1 ∈ M1

cand an open

ball containing p1, say B1 = B(p1, ε1) ⊂ M1c, with ε1 ≤ 1

2 .

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Similarly, M2 does not contain a nonempty open set ⇒ B(p1,ε12 ) ( M2. Then M2

c ∩B(p1,

ε12 ) 6= ∅. Then

∃ B2 = B(p2, ε2) ⊂(

M2c ∩ B(p1,

ε1

2))

with ε2 <ε1

2.

Continuing inductively, we can find Bk = B(pk, εk) with εk < 12k s.t.

Bk ∩ Mk 6= ∅Bk+1 ⊂ B(pk,

εk

2) ⊂ Bk.

Since ε < 12k we find (pn) is Cauchy. Then because X is complete ∃ p ∈ X s.t. pn → p.

Also ∀ n > m

d(pm, p) ≤ d(pm, pn) + d(pn, p)

<εm

2+ d(pn, p)

→ εm

2

and therefore p ∈ Bm ∀ m. Since Bm ⊂ Mmc, we have p 6∈ Mm ∀ m. Therefore p 6∈ ∪Mm =

X, a contradiction.

Theorem 4.27 (Uniform Boundedness). Let (Tn) ⊂ B(X,Y ) with X a Banach space andY a normed space. Suppose ∀ x ∈ X ∃ cx s.t.

‖Tnx‖ ≤ cx ∀ n.

Then (‖Tn‖) is bounded.

Proof. For each k defineAk = {x ∈ X| ‖Tnx‖ ≤ k ∀ n} .

Then we can see Ak is closed:

1. Let x ∈ Ak. Then ∃ (xi) ⊂ Ak s.t. xi → x.

2. For fixed n, ‖Tnxi‖ ≤ k ⇒ ‖Tnx‖ ≤ k by continuity of ‖·‖.

3. Then x ∈ Ak.

By assumption ∀ x ∈ X ∃ k s.t. x ∈ Ak, so

X =

∞⋃

k=1

Ak.

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Then since X is complete we invoke Baire to obtain Ak0, a set which contains an open ball,say

B0 = B(x0, v) ⊂ Ak0.

Let x ∈ X s.t. x 6= 0 be arbitrary. Set z = x0 + γx, γ = v2‖x‖ . Then

‖z − x0‖ < r ⇒ z ∈ B0

⇒ ‖Tnz‖ ≤ k ∀ n

Also, x0 ∈ B0 ⇒ ‖Tnx0‖ ≤ k0. Then

‖Tnx‖ = γ−1 ‖Tn(z − x0)‖≤ γ−1 (‖Tnz‖ + ‖Tnx0‖)≤ 4

v‖x‖ k0.

We conclude

‖Tn‖ = sup‖x‖=1

‖Tnx‖ ≤ 4

vk0 = c,

and c is independent of x.

Problem 31. Let X be the normed space of all polynomials with norm

‖x‖ = maxi

|αi| , where x(t) = αktk + · · · + α0.

Show X is not complete.

Solution. Let X be the normed space of all polynomials with norm ‖x‖ = maxi |αi| (theαi’s are the coefficients). Define the linear functional fn by fn(x) = α0 + α1 + ... + αn−1.Then |fn(x)| ≤ n‖x‖. Also for each fixed x we have |fn(x)| ≤ cx. On the other hand, for

x(t) = 1 + t + t2 + ... + tn, we have ‖x‖ = 1 and fn(x) = n‖x‖. Hence ‖fn‖ ≥ |fn(x)|‖x‖ = n.

It follows that the sequence (‖fn‖) is unbounded. The Uniform Boundedness Theoremimplies that X is not complete.

Problem 32. If X,Y are Banach and (Tn) ⊂ B(X,Y ) is a sequence, show TFAE:

1. (‖Tn‖) is bounded.

2. (‖Tnx‖) is bounded ∀ x ∈ X.

3. (|g(Tnx)| is bounded ∀ x ∈ X ∀ g ∈ Y ′.

Solution. Recall that in a Banach space X if a sequence (xn) is such that (f(xn)) is boundedfor all f ∈ X

′, then (‖xn‖) is bounded and therefore 3 implies 2. Also, 2 implies 1 by the

Uniform Boundedness Theorem. Finally, 1 implies 3 since |g(Tnx)| ≤ ‖g‖‖Tn‖‖x‖.

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4.5 Strong and Weak Convergence

Definition (Strong and Weak Convergence). Let X be a normed space and let (xn) ⊂ Xbe a sequence.

1. (xn) is strongly convergent if ∃ x ∈ X s.t.

limn→∞

‖xn − x‖ = 0.

We denote this property with

limn→∞

xn = x or xn → x.

2. (xn) is weakly convergent if ∃ x ∈ X s.t. ∀ f ∈ X ′

limn→∞

f(xn) = f(x).

We denote this property withxn

w−→ x.

Lemma 4.28 (Weak Convergence). Let xnw−→ x. Then

1. The the weak limit x is unique.

2. xnk

w−→ x ∀ subsequences (xnk) ⊂ (xn).

3. (‖xn‖) is bounded.

Proof.

1. Suppose xnw−→ x and xn

w−→ y. Then f(xn) → f(x) and f(xn) → f(y) ⇒ f(x) =f(y). Then 0 = f(x) − f(y) = f(x − y) ∀ f ∈ X ′. Then by a previous lemma x = y.

2. (f(xn)) ⊂ R ⇒ all subsequences of (f(xn)) converge and have the same limit.

3. (f(xn)) converges ⇒|f(xn)| ≤ cf ∀ n. Define gn ∈ X ′′ by

gn(f) = f(xn).

Then |gn(f)| = |f(xn)| ≤ cf , ⇒(|gn(f)|) is bounded ∀ f ∈ X ′. Then X ′ complete⇒(‖gn‖) bounded by the UBT. Now ‖gn‖ = ‖x‖ and the proof is complete.

Theorem 4.29 (Strong and Weak Convergence). Let X be a normed space and (xn) ⊂ X.

1. xn → x ⇒ xnw−→ x. The limits are the same.

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2. xnw−→ x 6⇒ xn → x in general.

3. dim X < ∞ ⇒xn

w−→ x ⇒ xn → x.

Proof.

1. ‖f(xn) − f(x)‖ = ‖f(xn − x)‖ ≤ ‖f‖ ‖xn − x‖. Therefore xn → x ⇒ xnw−→ x.

2. We show a counterexample. Let H be a Hilbert space and (en) ⊂ H and orthonormalsequence. Then ∀ f ∈ H ′, f(x) = 〈x, z〉. In particular, f(en) = 〈en, z〉 and by Bessel’sinequality

∞∑

n=1

|〈en, z〉|2 ≤ ‖z‖2 .

Therefore f(en) = 〈en, z〉 → 0 ∀ f ∈ H ′ ⇒ enw−→ 0. But (en) does not converge

strongly because for n 6= m

‖em − en‖2 = 〈em − en, em − en〉 = 2.

3. Suppose dimX = k < ∞ and xnw−→ x. Let {e1, . . . , ek} be a basis for X. Then we

can represent

xn =

k∑

i=1

α(n)i ei

x =

k∑

i=1

αiei.

By assumption f(xn) → f(x) ∀ f ∈ X ′. Take the dual basis {f1, . . . , fk} defined by

fk(ei) = δki.

Then

fi(xn) = α(n)i

fi(x) = αi.

Hencefi(xn) → fi(x) ⇒ α

(n)i → αi.

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Then

‖xn − x‖ =

∥

∥

∥

∥

∥

k∑

i=1

(α(n)i − αi)ei

∥

∥

∥

∥

∥

≤k∑

i=1

∣

∣

∣α

(n)i − αi

∣

∣

∣‖ei‖

→ 0 as n → ∞.

Remark 4.30.

1. In ℓ1 strong and weak convergence are equivalent.

2. In a Hilbert space H,

xnw−→ x iff 〈xn, z〉 → 〈x, z〉 ∀ z ∈ H.

Lemma 4.31 (Weak Convergence). Let X be a normed space and let (xn) ⊂ X. Thenxn

w−→ x iff both:

1. (‖xn‖) is bounded.

2. f(xn) → f(x) ∀ f ∈ M ⊂ X ′ ∀ M total .

Proof. 1. “⇒”. Weak convergence implies 1,2 by previous results.

2. “⇐”. Suppose 1,2 hold. Let f ∈ X ′ be arbitrary and show f(xn) → f(x). By 1 wecan find c > 0 s.t.

‖xn‖ ≤ c

‖x‖ ≤ c.

Since M ⊂ X ′ is total, ∀ f ∈ X ′ ∃ (fn) ⊂ span (M) s.t. fn → f . Hence ∀ ε >0 ∃ i s.t.

‖fi − f‖ <ε

3c.

Moreover, since fi ∈ span (M) (by2) ∃ N s.t.

|fi(xn) − fi(x)| <ε

3, ∀ n > N.

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Then

|f(xn) − f(x)| ≤ |f(xn) − fi(xn)| + |fi(xn) − fi(x)| + |fi(x) − f(x)|< ‖f − fi‖ ‖xn‖ +

ε

3+ ‖fi − f‖ ‖x‖

<ε

3cc +

ε

3+

ε

3cc

= ε.

Therefore xnw−→ x.

Definition (Weak Cauchy, Weak Complete).

1. In a normed space X, a sequence (xn) ⊂ X is said to be weak Cauchy if ∀ f ∈ X ′

the sequence (f(xn)) is Cauchy.

2. A normed space X is said to be weakly complete if every weak Cauchy sequenceconverges weakly in X.

Problem 33. Show X reflexive ⇒X weakly complete.

Solution. Let (xn) be any weak Cauchy sequence in X. Then (f(xn)) converges for everyf ∈ X

′. For xn ∈ X there is a gxn ∈ X

′′such that f(xn) = gxn(f). Hence (gxn(f))

converges, say, gxn(f) → g(f). Weak Cauchyness of (xn) implies the boundedness of (xn)and then since ‖gxn‖ = ‖xn‖ we have that g is bounded. Also, g is linear and thereforeg ∈ X

′′. Since X is reflexive, there is an x such that g(f) = f(x). Hence f(xn) → f(x).

Since f ∈ X′

was arbitrary, this shows that (xn) converges to x weakly. Since (xn) wasany weak Cauchy sequence, X is weakly complete.

Definition (Convergence of Operators). Let X,Y be normed spaces, and let (Tn) ⊂B(X,Y ) be a sequence. We say (Tn) is:

1. uniformly operator convergent if (Tn) converges in the norm of B(X,Y ).

2. strongly operator convergent if ∀ x ∈ X the sequence (Tnx) converges stronglyin Y .

3. weakly operator convergent if ∀ x ∈ X the sequence (Tnx) converges weakly inY .

Remark 4.32. Uniform ⇒strong ⇒weak:

1. ‖Tn − T‖ → 0 ⇒ ‖(Tn − T )x‖ ≤ ‖Tn − T‖ ‖x‖ → 0.

2. ‖(Tn − T )x‖ → 0 ⇒ ‖f(Tn − T )x‖ ≤ ‖f‖‖(Tn − T )x‖ → 0

84

Example 4.33 (Strong 6⇒ Uniform). Let X = Y = ℓ2. Let Tn be “zero-overwrite”operator, i.e., let (xi) ∈ ℓ2

Tn(xi) = (0, . . . , 0, xn+1, xn+2, . . . ).

Then Tn(xi) → 0 ∀ (xi) ∈ ℓ2 so (Tn) is strongly operator convergent to the zero operator.However given any n, we can construct x′ ∈ ℓ2 s.t. Tnx = x, i.e., by making the first nterms of x′ zero. Then

‖Tn‖ = supx 6=0

‖Tnx‖‖x‖ ≥ 1.

So (Tn) does not converge to 0 ∈ B(X,Y ).

Example 4.34 (Weak 6⇒ Strong). Let X = Y = ℓ2. Let Tn be “zero-shift” operator, i.e.,let (xi) ∈ ℓ2

Tn(xi) = (0, . . . , 0, x1, x2, . . . ), n zeros .

Let (zi) ∈ ℓ2 be another element. Define f ∈ ℓ2′ by

f(x) = 〈x, z〉 =∞∑

i=1

xizi.

Then

f(Tnx) = 〈Tnx, z〉

=

∞∑

k=1

xkzk+n.

Compute

|f(Tnx)|2 = |〈Tnx, z〉|2

≤∞∑

k=1

|xk|2∞∑

m=n+1

|zm|2

∞∑

m=n+1

|zm|2 → 0.

Then f(Tnx) → 0 = f(0x), i.e. (Tn) converges weakly to zero. But consider the sequencex = (1, 0, 0, . . . ). Then

‖Tnx − Tmx‖ =√

2, n 6= m .

So (Tn) does not converge strongly to zero.

Definition (Strong, Weak and Weak* Convergence of Functionals). Let X be a normedspace and let (fn) ∈ X ′ be a sequence.

85

1. (fn) is strongly convergent to f ∈ X ′ if ‖fn − f‖ → 0 and we write

fn → f.

2. (fn) is weakly convergent to f ∈ X ′ if

g(fn) → g(f) ∀ g ∈ X ′′.

3. (fn) is weak* convergent to f ∈ X ′ if

fn(x) → f(x) ∀ x ∈ X.

We writefn

w∗

−→ f.

Remark 4.35.

1. Weak ⇒weak*.

2. Limit operators. Let (Tn) ∈ B(X,Y ) be a sequence.

(a) If Tn → T (uniform) then T ∈ B(X,Y ).

(b) If convergence is strong or weak, it is possible that the limit T is unbounded (notcontinuous) if X is not complete. This is the why we use Banach spaces.

Example 4.36. Let X be the subspace of ℓ2 of sequences with finitely many nonzeroterms. Then X is not complete. Define Tn via

Tn(xi) = (x1, 2x2, . . . , nxn, xn+1, xn+2, . . . ).

Then (Tn) ⊂ B(X,X) converges strong to the unbounded operator T

T (xi) = (ixi).

Lemma 4.37 (Strong Operator Convergence). Let X be a Banach space, Y a normedspace, and (Tn) ⊂ B(X,Y ) a sequence. (Tn) strongly operator convergent ⇒T ∈ B(X,Y ).

Proof. Note

1. T is linear.

2. Tnx → Tx ∀ x ∈ X ⇒ Tnx is bounded ∀ x ∈ X ⇒‖Tn‖ ≤ c, for some c ∈ R by theuniform boundedness theorem.

86

Now,

‖Tnx‖ ≤ ‖Tn‖ ‖x‖≤ c ‖x‖ .

So ‖Tnx‖ ≤ c ‖x‖. Taking lim on the LHS yields

‖Tx‖ ≤ c ‖x‖ ,

therefore T ∈ B(X,Y ).

Theorem 4.38 (Strong Operator Convergence). Let X,Y be Banach spaces, and let (Tn) ⊂B(X,Y ) be a sequence. (Tn) is strongly operator convergent iff

1. (‖Tn‖) is bounded.

2. (Tnx) is Cauchy ∀ x ∈ M ∀ M total in X.

Proof.

1. ⇒. If Tnx → Tx ∀ x ∈ X then 1 follows by the UBT and 2 follows trivially.

2. ⇐. Suppose ‖Tn‖ ≤ c ∀ n. Let x ∈ X be arbitrary and show (Tnx) converges stronglyin Y .

Let ε > 0 be given. Since X = span (M), then ∃ y ∈ span (M) s.t.

‖x − y‖ <ε

3c.

Also, (Tny) Cauchy ⇒ ∃ N s.t.

‖Tny − Tmy‖ <ε

3.

Then

‖Tnx − Tmx‖ ≤ ‖Tnx − Tny‖ + ‖Tny − Tmy‖ + ‖Tmy − Tmx‖< c

ε

3c+

ε

3+ c

ε

3c= ε.

Then (Tn) is Cauchy, and since Y is complete, (Tnx) converges in Y .

Corollary 4.39. Let X be a Banach space and (fn) ⊂ X ′ a sequence which is weak*convergent to f ,

fnw∗

−→ f.

Then f ∈ X ′ iff

1. (‖fn‖) is bounded.

2. (fnx) is Cauchy ∀ x ∈ M ∀ M total in X.

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4.6 The Open Mapping Theorem

Definition (Open Mapping). Let X,Y be metric spaces, T : D (T ) → Y , D (T ) ⊂ X.Then T is said to be an open mapping if B ⊂ D (T ) open ⇒T (B) ⊂ Y open.

Theorem 4.40 (Open mapping). Let X,Y be Banach spaces and T ∈ B(X,Y ) s.t. T :

Xonto−→ Y . Then T is an open mapping. Hence, if T is injective then T is bijective and so

T−1 is continuous and so T−1 ∈ B(X,Y ).

Lemma 4.41 (Open Unit Ball). Let X,Y be Banach spaces and T ∈ B(X,Y ) s.t. T :

Xonto−→ Y and B0 = B(0, 1) ⊂ X. Then T (B0) ⊂ Y is open and 0 ∈ T (B0).

Proof. Let A ⊂ X. Define

1. αA = {x ∈ X|x = αa, a ∈ A}

2. A + w = {x ∈ X|x = a + w, a ∈ A}.Consider B1 = B(0, 1

2 ⊂ X, and let x ∈ X be arbitrary. Then choose k s.t. x ∈ kB1, e.g.,k > 2 ‖x‖. Then

X =

∞⋃

k=1

kB1.

Since T is surjective and linear,

Y = T (X)

= T

( ∞⋃

k=1

kB1

)

=∞⋃

k=1

kT (B1)

=∞⋃

k=1

kT (B1).

Then by the category theory, Y complete ⇒Y not meager. Then at least one of kT (B1)contains an open ball ⇒T (B1) contains an open ball, say

B∗ = B(y0, ε) ⊂ T (B1).

ThenB∗ − y0 = B(0, ε) ⊂ T (B1) − y0.

With these sets constructed, we want to show

T (B1) − y0 ⊂ T (B0).

Let y ∈ T (B1) − Y0. Then y + y0 ∈ T (B1).

88

Problem 34. Let X,Y be Banach spaces and T ∈ B(X,Y ) injective. Show that T−1 :R (T ) → X is bounded iff R (T ) is closed in Y .

Solution.

1. If R(T ) is closed in Y , it is complete, and boundedness follows from the Open MappingTheorem.

2. Assume T−1 to be bounded, y ∈ R(T ) ⊂ Y , (yn) in R(T ) such that yn → y, andxn = T−1yn. Since T−1 is continuous and X is complete, (xn) converges, say, xn → x.Since T is continuous, yn = Txn → Tx. Hence y = Tx ∈ R(T ), so that R(T ) is closedbecause y ∈ R(T ) was arbitrary.

4.7 Closed Graph Theorem

Definition (Closed Linear Operator). Let X,Y be normed spaces and T : D (T ) → Y alinear operator with D (T ) ⊂ X. T is said to be closed if the graph of T , denoted ΓT

ΓT = {(x, y)|x ∈ D (T ) , y = Tx}is closed in X × Y .

Theorem 4.42 (Closed Graph Theorem). Let X,Y be Banach spaces and let T : D (T ) →Y be a closed linear operator. If D (T ) is closed in X then T is bounded.

Proof. Note that X × Y is normed space with norm

‖(x, y)‖ = ‖x‖ + ‖y‖ (11)

First we show X × Y is complete with norm (11). Let zn = (xn, yn) and (zn) ∈ X × YCauchy. Let ε > 0 be given. ∃ N s.t.

‖zn − zm‖ = ‖xn − xm‖ + ‖yn − ym‖ < ε ∀ n,m > N.

Then (xn), (yn) are Cauchy in X,Y respectively. Since X,Y are Banach, these sequencesconverge, say

xn → x ∈ X, yn → y ∈ Y.

Then setting z = (x, y),zn → z ∈ X × Y.

Since (zn) was arbitrary, X × Y is complete.Now, by assumption ΓT is closed in X×Y and D (T ) is closed in X. Therefor ΓT,D (T )

are complete. Consider P : ΓT → D (T ) given by the map

(x, Tx) 7→ x.

Then

89

1. P is linear. Obvious.

2. P is bounded.

‖P (x, Tx)‖ = ‖x‖≤ ‖x‖ + ‖Tx‖= ‖(x, Tx)‖ .

3. P is bijective with inverse P−1 : D (T ) → ΓT given by the map

x 7→ (x, Tx).

Since ΓT and D (T ) are complete, P−1 is bounded, say

‖(x, Tx)‖ ≤ b ‖x‖ .

Then

‖Tx‖ ≤ ‖Tx‖ + ‖x‖= ‖(x, Tx)‖≤ b ‖x‖ ,

∀ x ∈ D (T ).

Then T is bounded, as required.

Theorem 4.43 (Closed Linear Operator). Let X,Y be normed spaces, T : D (T ) → Y alinear operator, and D (T ) ⊂ X. Then T is closed iff

1. Given (xn) ⊂ D (T ).

2. If xn → x and Txn → y then

x ∈ D (T ) and Tx = y.

Proof. z ∈ ΓT iff ∃ sequence (zn) ⊂ ΓT

zn = (xn, Txn)

s.t.zn → z.

Hence, xn → x and Txn → y. Then z = (x, y) ∈ ΓT iff

x ∈ D (T ) and y = Tx.

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Example 4.44 (Differential Operator). Let X = C[0, 1] and let T : D (T ) → X be givenby the map

x 7→ x′.

Note D (T ) is the space of continuously differentiable functions. Then

1. T is bounded. Obvious.

2. T is closed. Suppose xn → x and Txn = x′n → y. Then

∫ 1

0y(τ)dτ =

∫ 1

0lim

n→∞x′

n(τ)dτ

= limn→∞

∫ 1

0x′

n(τ)dτ

= x(t) − x(0).

⇒x(t) = x(0) +

∫ 1

0y(τ)dτ.

⇒

x ∈ D (T )

x′ = y

Remark 4.45. Boundedness does not imply closedness. To see this, let T : D (T ) →D (T ) ⊂ X be the identity operator where D (T ) is proper, dense subset of X. Then T islinear and bounded.

However T is not closed. Take x ∈ X −D (T ) and let (xn) ⊂ D (T ) be s.t.

xn → x.

ThenTxn = xn → x 6∈ D (T ) .

Lemma 4.46 (Closed Operator). Let X,Y be normed spaces, let T ∈ B(D (T ) , Y ) withD (T ) ⊂ X.

1. If D (T ) is a closed subset of X then T is close.

2. If T is closed and Y is complete then D (T ) is a closed subset of X.

Proof.

1. If (xn) ⊂ D (T ) and xn → x and (Txn) converges then

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(a) x ∈ D (T ) = D (T ) since D (T ) is closed.

(b) Txn → Tx since T is closed.

⇒T is closed.

2. For x ∈ D (T ) ∃ (xn) ⊂ D (T ) s.t.

xn → x.

Since T is bounded,‖Txn − Tx‖ ≤ ‖T‖ ‖xn − xm‖ .

Therefore (Txn) is Cauchy, and since Y is complete

Txn → y ∈ Y.

Since T is closed we have x ∈ D (T ) and Tx = y. Hence D (T ) is closed becausey ∈ D (T ) was arbitrary.

Problem 35. Let X and Y be normed spaces and let T : X → Y be a closed, linear operator.Show the following.

1. The image B of a compact subset C ⊂ X is closed in Y .

2. The inverse image A of compact subset K ⊂ Y is closed in X.

Solution. 1. Consider any a ∈ A. Let an → a, where an ∈ A. Let cn ∈ C be such thatan = Tcn. Since C is compact, (cn) has a subsequence (cnk

) which converges, say,cnk

→ c ∈ C. Also Tcnk→ a, and Tc = a ∈ A because T is closed by assumption.

2. Consider any b ∈ B. Let bn → b, where bn ∈ B. Let kn = Tbn. Since K is compact,(kn) has a subsequence (kni

) which converges, say, kni→ k ∈ K. Also bni

→ b, andTb = k ∈ K = T (B) by the closedness of T , so that b ∈ B and B is closed.

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5 Exam 1

Problem 1 (4 Points). Let M ⊂ l∞ be the subspace consisting of all sequences x = (ξi)with at most finitely many nonzero terms. Is M complete?

Solution. Let M ⊂ l∞, and x = (1, 1/2, 1/3, ....) = (ξi). Consider the sequence (xn), wherexn = (1, 1/2, ...., 1/n, 0, ...). Then xn ∈ M and d(xn, x) = 1/(n + 1), i.e., we have thatxn → x, but x 6∈ M . It follows that M is not closed.

Problem 2 (4 Points). Does

d(x, y) =

∫ b

a

|x(t) − y(t)|dt

define a metric or pseudometric on X if X is

1. the set of all real-valued continuous functions on [a, b]

2. the set of all real-valued Riemann integrable functions on [a, b]?

Solution. d is a metric on C[a, b], because if the integral is 0, then |x(t) − y(t)| = 0 for allt ∈ [a, b]. On the other hand d is a pseudo-metric on R[a, b], for example if x(a) = 1 andx(t) = 0 everywhere else on [a, b], then the integral of |x(t)| is 0.

Problem 3 (4 Points). If X is a compact metric space and M ⊂ X is closed, show thatM is compact.

Solution. Let (xn) ⊂ M ⊂ X. Since X is compact (xn) has a convergent subsequence,say (xnk

) ⊂ M , and xn → x. M is closed, so we must have x ∈ M . It follows that M iscompact.

Problem 4 (4 Points). Let T : X → Y be a linear operator and dimX = dimY = n < ∞.Show that R(T ) = Y iff T−1 exists.

Solution.

1. Suppose that T−1 exists and let {e1, e2, ..., en} a basis for X. We show that {Te1, T e2, ..., T en}are linearly independent (and therefore R(T ) = Y ).

Suppose that α1Te1 + ... + αnTen = T (α1e1 + ... + αnen) = 0. T is bijective byassumption, so we have that α1e1 + ... + αnen = 0, and then α1 = ... = αn = 0. Itfollows that {Te1, ..., T en} are linearly independent.

2. Suppose that R(T ) = Y , and (y1, y2, ..., yn) is a basis for Y . Then there existx1, x2, ..., xn ∈ X such that Txi = yi, for i = 1, 2, ..., n.

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We can show that {x1, x2, ..., xn} are linearly independent using the same argumentas above. It follows that {x1, x2, ..., xn} is a basis for X.

Suppose that Tx = 0. Then Tx = T (α1x1 + ... + αnxn) = α1y1 + ... + αnyn = 0, andα1 = ... = αn = 0. It follows that x = 0 and T is injective. Also, T is surjective byassumption. It follows that T−1 exists.

Problem 5 (4 Points). Show that the operator T : l∞ → l∞ defined by

y = (ηi) = Tx, ηi =ξi

i, x(ξi),

is linear and bounded What can you say about T−1?.

Solution. Let T : l∞ → l∞, x = (ξi) ∈ l∞, Tx = (ξi/i). Then T is bounded, linear,injective, but not surjective, e.g., y = (1, 1, 1, ..) ∈ l∞ is not in R(T ). T−1 is only definedon R(T ) ⊂ l∞. T−1 is not bounded, because for xn = (ξni, where ξni = 1 if i = n and 0otherwise we get

‖t−1xn‖‖xn‖

= n.

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6 Exam 2

Problem 1 (4 Points). If x and y are different vectors in a finite dimensional vector spaceX, show that there is a linear functional f on X such that f(x) 6= f(y).

Solution. Let x, y ∈ X, x 6= y, dimX = n.

1. Suppose that x 6= αy, for any α. Then x and y are linearly independent and wecan construct a basis {x = e1, y = e2, e3, ..., en}, and a corresponding dual basis{f1, f2, ..., fn}, where fi(ej) = δij . Then f1(x) = 1 6= f1(y).

2. If x = αy, α 6= 1, then we take {x = e1, e2, e3, ..., en}, and {f1, f2, ..., fn}. Thenf1(x) = 1 6= f1(y) = α.

Problem 2 (4 Points). Let X and Y be normed spaces and Tn : X → Y , n = 1, 2, 3, .., bebounded linear operators. Show that convergence Tn → T implies that for every ǫ > 0 thereis an N such that for all n > N and all x in any given closed ball we have ‖Tnx−Tx‖ < ǫ.

Solution. Let B be a closed ball in X. B is bounded, i.e., if x ∈ B, then ‖x‖ < K for someK. Let N be such that ‖Tn − T‖ < ǫ

Kfor all n > N . Then

‖Tnx − Tx‖ = ‖(Tn − T )x‖ ≤ ‖Tn − T‖‖x‖ < ǫ.

Problem 3 (4 Points). Show that y ⊥ xn and xn → x together imply x ⊥ y.

Solution. Suppose that y ⊥ xn and xn → x. Then

〈x, y〉 = limn→∞

〈xn, y〉 = 0.

Problem 4 (4 Points). Let M be a total set in an inner product space X. If 〈v, x〉 = 〈w, x〉for all x ∈ M , show that v = w.

Solution. We have by assumption that 〈v − w, x〉 = 0 for all x ∈ M . M is total in X, sov − w ∈ ¯spanM = X. By the continuity of the inner product we have 〈v − w, v − w〉 =‖v − w‖2 = 0.

Problem 5 (4 Points). If S and T are bounded self-adjoint operators on a Hilbert spaceH and α and β are real, show that L = αS + βT is self-adjoint.

Solution. Let L = αS + βT . Then we have

L∗ = (αS + βT )∗ = (αS)∗ + (βT )∗ = αS∗ + βT ∗ = αS∗ + βT ∗ = αS + βT = L.

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7 Final Exam

Problem 1 (4 Points). What can you say about the reflexivity of lp, 1 ≤ p ≤ ∞?

Solution. lp, For 1 < p < ∞ is reflexive using the fact that the dual space of lp is lq, where1p

+ 1q

= 1. On the other hand, l1 and l∞ are nonreflexive spaces. For l1 we can use theargument that a separable normed space with a nonseparable dual cannot be reflexive.Concerning l∞, we know that c0, the space of convergent sequences with limit equal tozero is a subspace of l∞ and that the dual of c0 is l1. It follows that the dual of l∞ is largerthan l1, and therefore l∞ is not reflexive.

Problem 2 (4 Points). Let X be a separable Banach space and M ⊂ X′

a boundedset. Show that every sequence of elements of M contains a subsequence which is weak*convergent to an element of X

′.

Solution. Any sequence (fn) in M is bounded, say‖fn‖ ≤ r. Since X is separable, itcontains a countable dense subset V , which we can arrange in a sequence (xm). Since

|fn(xm)| ≤ ‖fn‖‖xm‖ ≤ r‖xm‖,

we see that for fixed m the sequence (fn(xm)) is bounded, so that it has a subsequence A1

which converges at x1, and A1 has a subsequence A2 which converges at x2, ...etc.; hence(fnk

(x)), where fn1 ∈ A1, fn2 ∈ A2, ..., is a subsequence which converges at every elementof V . Since V is dense in X and X is complete the statement follows.

Problem 3 (4 Points). Show that an open mapping need not map closed sets onto closedsets.

Solution. The mapping T : R2 → R defined by (x1, x2) → (x1) is open, it maps the closedset {(x1, x2)|x1x2 = 1} ⊂ R2 onto the set R − {0} which is not closed in R.

Problem 4 (4 Points). Let X and Y be normed spaces and X compact. If T : X → Y isa bijective closed linear operator, show that T−1 is bounded.

Solution. T−1 is closed, being the inverse of a closed operator. Hence T−1 : Y → X, whereT−1 is closed and X is compact, is bounded.

Problem 5 (4 Points). Let f be an integrable function on the measure space (X,B, µ).Show that given ǫ > 0, there is a δ > 0 such that for each measurable set E with µE < δwe have

|∫

E

f | < ǫ.

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Solution. Let f be an integrable function on the measure space (X,B, µ). Let fn(x) = f(x)if |f(x)| ≤ n, fn(x) = n if f(x) ≥ n, fn(x) = −n if f(x) ≤ in. Then |f − fn| → 0 a.e. and|f − fn| ≤ |f | for all n. By the dominated convergence theorem we have

∫

|f − fn| → o.

Let ǫ > 0 be given. We can pick N such that

∫

|f − fn| <ǫ

2for all N.

Let δ < ǫ2N

. If E is measurable with µE < δ, then

|∫

E

f | = |∫

E

(fN + (f − fN))| ≤∫

E

|fN | +∫

E

|f − fN | ≤ δN +

∫

|f − fN | ≤ ǫ.

97