functional analysis - kitlamm/seite/lecturenotes/media/fa1617.pdf · 1 introduction Ω...

Functional Analysis

Karlsruher Institut für Technologie (KIT)

Prof. Dr. Tobias Lamm

Winter term 2016/17

Contents

1 Introduction 3

2 Metric and normed spaces 7

3 Compactness (in metric spaces) 15

4 Extension of linear functionals 26

5 Uniform boundedness principle 36

6 Lp-spaces 43

7 The dual space of C0(X) 51

8 Weak convergence 67

9 Hilbert spaces 75

10 Sobolev spaces and elliptic boundary value problems 86

11 Compact and Fredholm operators 99

12 Spectral theory for compact operators 119

13 Semigroups 132

2

1 Introduction

We are interested in solutions of linear equations Lu = f , where L : X → Y is anoperator between vector spaces over R or C. In the following examples we illustratehow various results which were shown to be true on finite-dimensional vector spacesin Linear Algebra are no longer true on infinite-dimensional spaces.

Example: Let L = ∆; X = C2(Rn), Y = C0(Rn). Then

L : X → Y is linear:L(λx+ µy) = λLx+ µLy

In general we are interested in the case dimX = dimY =∞.

Example: Let X = x = (x1, x2, ...) : xi ∈ R be the sequence space and let A : X →X

A((x1, x2, ...)) = (0, x1, x2, ...)

be the shift-map. Then A is injective but not surjective.

Now, let B : X → X be defined by

B((x1, x2, ...)) = (x2, x3, ...).

Then B is surjective but not injective.

Example: Let X = C0([−π, π]) and consider the map A : X → X,

(Au)(t) = sin(t)u(t)

for all t ∈ [−π, π]. Assume that A has an eigenvalue λ ∈ R (or λ ∈ C). Hence thereexists u ∈ C0([−π, π]) with u 6= 0, so that

sin(t)u(t) = λu(t) ∀t ∈ [−π, π].

But then t : u(t) 6= 0 ⊂ t : sin t = λ which gives the contradiction u ≡ 0.

Central example: The Dirichlet principle for elliptic boundary value problems. Let

3

1 Introduction

Ω ⊂ Rn be open, connected, bounded and assume that ∂Ω is smooth.

Given f ∈ C∞(Ω) we want to find a solution u ∈ C∞0 (Ω) of the elliptic boundaryvalue problem

−∆u = f in Ωu = 0 on ∂Ω

This can be done by using a variational approach and for this one considers thefunctional

F(v) := 12

∫Ω|Dv|2 dλ−

∫Ωfv dλ,

where λ = n-dimensional Lebesgue measure.

Lemma 1.1. Let u ∈ C∞0 (Ω) satisfy

F(u) ≤ F(v) ∀v ∈ C∞0 (Ω),

then −∆u = f in Ω.

Proof. Let η ∈ C∞c (Ω), i.e. sptη ⊂⊂ Ω. We estimate for every ε ∈ R

F(u) ≤ F(u+ εη)

= 12

∫|Du|2dλ+ ε

∫〈Du,Dη〉dλ+ ε2

2

∫|Dη|2dλ−

∫fudλ− ε

∫fηdλ

= F(u) + ε

(∫〈Du,Dη〉dλ−

∫fηdλ

)+ ε2

2

∫|Dη|2dλ.

Since this has to be true for every ε, we conclude that ∀η ∈ C∞c (Ω) we have

0 =∫

(〈Du,Dη〉 − fη)dλ = −∫

(∆u+ f)ηdλ

and the fundamental theorem of the calculus of variations then implies that −∆u =f in Ω.

I Minimize F(v) in C∞0 (Ω).Formally this is similar to the problem

II Minimize F (x) = 12 |x|

2 − 〈a, x〉 in Rn, a, x ∈ Rn.

Theorem 1.2. On a finite dimensional vector space V all norms are equivalent.For ‖ · ‖1, ‖ · ‖2 on V there exists 0 < m < M <∞, such that

m‖v‖1 ≤ ‖v‖2 ≤M‖v‖1 ∀v ∈ V.

Proof. 1) Without loss of generality we can assume that V = Rn. Otherwise choose

4

1 Introduction

basis v1, ..., vn of V and define

f : Rn → V, f(x) =n∑i=1

xivi

Then we look at the new norms

‖x‖1,2 = ‖f(x)‖1,2

on Rn.

2) Next we can always assume that ‖ · ‖2 = ‖ · ‖∞ and hence it remains to show:

m‖ · ‖∞ ≤ ‖ · ‖ ≤M‖ · ‖∞.

For this we note that

‖x‖ = ‖n∑i=1

xiei‖ ≤n∑i=1|xi|‖ei‖

≤ ‖x‖∞n∑i=1‖ei‖ =: M‖x‖∞

Hence ‖ · ‖ : Rn → R is Lipschitz continuous with constant M with respect to ‖ · ‖∞:

|‖x‖ − ‖y‖| ≤ ‖x− y‖ ≤M‖x− y‖∞.

We also know that x ∈ Rn : ‖x‖∞ = 1 is compact and therefore

‖ · ‖ : x ∈ Rn : ‖x‖∞ = 1 → R

attains its infimum. Let m := inf‖x‖ : ‖x‖∞ = 1 ⇒ ∃x0 ∈ Rn such that ‖x0‖∞ =1, ‖x0‖ = m ⇒ m > 0⇒ ∀x ∈ Rn :

‖x‖ = ‖x‖∞‖x

‖x‖∞‖ ≥ m‖x‖∞.

Solution of II: Choose a minimising sequence xj ∈ Rn:

F (xj)→ infF (x) : x ∈ Rn

⇒ 12 |xj − a|

2 = 12(|xj |2 − 2〈a, xj〉+ |a|2)

= F (xj) + 12 |a|

2 ≤ const.

5

1 Introduction

⇒ (xj) is bounded. Bolzano-Weierstrass now implies that up to a subsequencexj → x0 ∈ Rn.

⇒ F (x0) = limj→∞

F (xj) = infx∈Rn

F (x)

and hence x0 is a minimiser.

Problems when copying this approach for I

1. On C0(I), I = [0, 1] not all norms are equivalent: Take

‖u‖L2(I) :=(∫

Iu2dλ

)1/2

‖u‖C0(I) := max|u|(x) : x ∈ I.

and a sequence

un(x) =

1− nx, 0 ≤ x ≤ 1/n0, otherwise.

Then ‖un‖C0 = 1 for all n ∈ N but ‖un‖L2 ≤ 1/√n→ 0 as n→∞ and so the

norms are not equivalent.

minimising F in C∞0 (Ω) will only give a bound for the quantity

‖uj‖L2(Ω) + ‖Duj‖L2(Ω) ≤ C

and this will eventually force us to study Sobolev spaces.

2. The Bolzano-Weierstrass theorem is not true in infinite dimensional vectorspaces. For this we consider

lp(R) = x = (xi)i∈N : xi ∈ R, ‖x‖p =( ∞∑i=1

(xi)p)1/p

<∞

l∞(R) = same but with ‖x‖∞ := sup |xi| <∞.

Now look at sequence xk ∈ lp(R), 1 ≤ p ≤ ∞ defined by xk = (0, ..., 0, 1, 0, ...).We have that xik = δik, ‖xk‖lp = 1 but this sequence has no converging subse-quence, since

‖xj − xk‖p = 21/p ≥ 1, j 6= k.

6

2 Metric and normed spaces

Definition 2.1. Let X be a set. A map d : X×X → [0,∞) is called a metric (and(X, d) a metric space) if for all x, y, z ∈ X there holds

1. d(x, y) = 0 ⇔ x = y.

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

3. d(x, z) ≤ d(x, y) + d(y, z).

The metric space (X, d) is called complete, if every Cauchy sequence converges inX. ((xk) is a Cauchy sequence, ⇔ d(xk, xl) < ε for all k, l large enough).

In the next theorem we show that for every metric space there exists a completion.

Theorem 2.2. For every metric space (X, d) there exists a complete metric space(X, d) and an isometric map

J : (X, d)→ (X, d)

such that J(X) is dense in X.

Uniqueness: If (X, d) is a complete metric space and J : X → X is isometric suchthat J(X) is dense in X, then there exists a unique isometry φ : X → X withJ = φ J .

(X, d)J

zzJ

$$(X, d)

φ// (X, d)

Proof. We start with the existence of (X, d). For this we define CS(X) = x =(xi) : x is Cauchy sequence in X and we introduce the equivalence relation x ∼ y

⇔ limi→∞

d(xi, yi) = 0. We now define X := CS(X)/ ∼.

Claim 1: limi→∞

d(xi, yi) exists

7

For i, j large enough we have

d(xi, xj), d(yi, yj) < ε

2

and hence

d(xi, yi) ≤ d(xi, xj) + d(xj , yj) + d(yi, yj)< ε+ d(xj , yj)

which implies that |d(xi, yi)− d(xj , yj)| < ε and hence (d(xi, yi)) is a Cauchysequence in R. Additionally we remark that for a Cauchy sequence (xi) and asubsequence (xik) we have that (xi) ∼ (xik) since d(xk, xik)→ 0.

Claim 2: The function d([x], [y]) := limi→∞

d(xi, yi) where (xi) ∈ [x], (yi) ∈ [y] is anorm on X.

We note that d is well-defined since for (xi), (xi) ∈ [x] and (yi), (yi) ∈ [y] wehave

d(xi, yi) ≤ d(xi, xi) + d(xi, yi) + d(yi, yi)

and hence |d(xi, yi)− d(xi, yi)| → 0 as i→∞.

Now for d([x], [y]) = 0 we conclude limi→∞

d(xi, yi) = 0 and therefore x ∼ y whichis equivalent to [x] = [y]. The other two properties of a metric are easy toverify.

Claim 3: The map J : X → X, J(x) = [(x, x, x, ...)] is isometric and J(X) is densein X.

By definition we have d(J(x), J(y)) = d(x, y) and hence J is isometric. For[x] ∈ X we choose (xi) ∈ [x] and then we conclude for k large enough

d([x], J(xk)) = limi→∞

d(xi, xk) < ε.

Therefore J(X) is dense in X.

Claim 4: The metric space (X, d) is complete.

We let [xk] be a Cauchy sequence in X, i.e. limi→∞

d(xik, xil) < ε for all k, l ≥ k(ε).Without loss of generality we can assume that

d(xik, xjk) <

1k∀i, j ∈ N

8

Now we look at the diagonal sequence (yk) = (xkk). For k, l ≥ k(ε) we have

d(yk, yl) ≤ d(xkk, xik) + d(xik, xil) + d(xil, xll)

<1k

+ ε+ 1l

and hence (yk) is a Cauchy sequence in X. We have to show that [xk] → [y].In order to do this we note that

d(xik, yi) ≤ d(xik, xjk) + d(xjk, x

ji ) + d(xji , xii)

<1k

+ d(xjk, xji ) + 1

i

and for j →∞ this implies

d(xik, yi) <1k

+ ε+ 1i.

Therefore we get d([xk], [y]) = limi→∞

d(xij , yi) ≤ 1k + ε which shows that [xk]→

[y].

In order to show the uniqueness statement of the theorem we let (X, d) and (X, d)be complete spaces with two isometric (and hence injective) maps J : X → X andJ : X → X. Therefore the map

φ : J(X)→ X, φ(J(x)) = J(x)

is well defined and we have

d(φ(J(x)), φ(J(y))) = d(J(x), J(y)= d(x, y)= d(J(x), J(y)).

This shows that φ : J(X)→ X is isometric and therefore φ|J(X) is uniformly continu-ous. Since J(X) is dense in X and X is complete, we can use exercise 1 from the exer-cise sheet 1 in order to conclude that there exists a unique continuous extension (stilldenoted by φ) φ : X → X which is isometric. Moreover φ(X) ⊃ φ(J(X)) = J(X) isdense in X by definition of J . And φ(X) is complete with respect to d.

Now exercise 4 from sheet 1 says that if we have a complete space (X, d) and ifY ⊂ X is also complete, then Y is closed subspace. Hence φ(X) is closed in X.Altogether this shows that φ(X) = X, and hence φ : X → X is an isometry.

Definition 2.3. Let X be an R- or C vector space. A function ‖ · ‖ : X → [0,∞) iscalled a norm, if the following three properties are satisfied:

9

1. ‖x‖ = 0 ⇔ x = 0 (definite).

2. ‖λx‖ = |λ|‖x‖ for all λ ∈ K, x ∈ X.

3. ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X.

On X we have the induced metric d(x, y) = ‖x−y‖. Is (X, d) complete, then (X, ‖·‖)is called a Banach space.

Definition 2.4. Let (X, ‖·‖X), (Y, ‖·‖Y ) be normed spaces. A linear map A : X → Y

is called bounded, if

‖A‖ = supx 6=0

‖Ax‖Y‖x‖X

= sup‖x‖X=1

‖Ax‖Y <∞.

‖A‖ is called the operator norm of A. The space

L(X,Y ) := A : X → Y |A linear, ‖A‖ <∞

is a normed space with the operator norm.

Lemma 2.5. For a linear map A : (X, ‖ · ‖)→ (Y, ‖ · ‖) the following statements areequivalent.

1. A is bounded.

2. A is continuous.

3. A is continuous in 0.

Proof. 1. ⇒ 2. For all x, y ∈ X we estimate

‖Ax−Ay‖ = ‖A(x− y)‖ ≤ ‖A‖‖x− y‖

and therefore A is even Lipschitz continuous.

3. ⇒ 1. For ε = 1 ∃δ > 0 such that

‖Ax‖ = ‖Ax− 0‖ ≤ 1 ∀‖x‖ ≤ δ

Hence we conclude for x 6= 0 that

‖Ax‖ = ‖x‖δ‖A(δx

‖x‖

)‖ ≤ ‖x‖

δ

and therefore ‖A‖ ≤ 1δ .

10

Theorem 2.6. Let (X, ‖ · ‖), (Y, ‖ · ‖) be normed spaces and assume that Y is aBanach space. Then L(X,Y ) with the operator norm is a Banach space.

Proof. Let (Ai) be a Cauchy sequence in L(X,Y ). This implies that for every x ∈ Xwe have

‖Aix−Ajx‖ ≤ ‖Ai −Aj‖‖x‖ < ε‖x‖ ∀i, j ∈ I(ε)

and hence the sequence (Aix) is a Cauchy sequence in Y for all x ∈ X. Since Y isa Banach space, this shows that the limit

Ax := limi→∞

Aix

exists for all x ∈ X. It follows right away that A is linear. Next we estimate

|‖Ai‖ − ‖Aj‖| ≤ ‖Ai −Aj‖ < ε ∀i, j ≥ I(ε)

and this shows that (‖Ai‖) is a Cauchy sequence in R. Again we conclude that thelimit Λ := lim

i→∞‖Ai‖ exists and we get

‖Ax‖ = limi→∞‖Aix‖ ≤ lim

i→∞‖Ai‖‖x‖ = Λ‖x‖

and therefore ‖A‖ ≤ Λ which implies that A ∈ L(X,Y ). Finally we observe that forevery x ∈ X

‖Aix−Ax‖ = limj→∞

‖Aix−Ajx‖ ≤ ε‖x‖ ∀i ≥ I(ε)

and hence‖Ai −A‖ < ε ∀i ≥ I(ε),

which shows that Ai → A in the operator norm.

Definition 2.7. Let X be a normed space. The Banach space X ′ := L(X,K) (K = Ror C) with the operator norm is called the dual space of X.

Theorem 2.8. Let X be a Banach space and let V ⊂ X be a closed subspace. ThenX/V with the norm

‖[x]‖ := infv∈V‖x+ v‖

is also a Banach space.

Proof. 1. We first show that ‖ · ‖ is indeed a norm on X/V .

(i) ‖[x]‖ ≥ 0 and ‖[x]‖ = 0 iff [x] = 0

11

The non-negativity is obvious and for the second claim we note that if‖[x]‖ = 0 then there exists a sequence vk ∈ V so that x + vk → 0 andhence vk → −x ∈ V since V is closed. hence [x] = 0.

(ii) For λ ∈ K we have ‖λ[x]‖ = |λ|‖[x]‖.

Since this is obvious for λ = 0 we assume that λ ∈ K\0 and we calculate

‖λ[x]‖ = ‖[λx]‖ = infv∈V‖λx+ v‖

= |λ| infv∈V

∥∥∥∥x+ v

λ

∥∥∥∥ = |λ|‖[x]‖.

(iii) Triangle inequality

For this we estimate

‖[x1] + [x2]‖ = ‖[x1 + x2]‖= inf

v1,v2∈V‖x1 + x2 + v1 + v2‖

≤ infv1∈V

‖x1 + v1‖+ infv2∈V

‖x2 + v2‖

= ‖[x1]‖+ ‖[x2]‖.

2. Next we show that X/V with this norm is complete.

In order to do this we note that

‖[y]− [x]‖ = infv∈V‖y − x+ v‖

implies that if ‖[y]− [x]‖ < ε, then there exists y ∈ [y] so that

‖y − x‖ < ε.

Now let [xi] be a Cauchy sequence in X/V . Without loss of generality weassume that

‖[xi + 1]− [xi]‖ < 2−i ∀i ∈ N .

We choose inductively xi ∈ [xi] so that

‖xi+1 − xi‖ < 2−i.

This can be done as follows: Set x1 = x1 and assume that xi has already been

12

found for some i ∈ N. Then we know that

‖[xi+1]− [xi]‖ = ‖[xi+1]− [xi]‖ < 2−i

and hence it follows from the above remark that there exists xi+1 ∈ [xi+1] sothat

‖xi+1 − xi‖ < 2−i.

By this construction (xi) is a Cauchy sequence in X and therefore x := limi→∞

xi

exists. Moreover

‖[xi]− [x]‖ = ‖[xi]− [x]‖ = ‖[xi − x]‖ ≤ ‖xi − x‖ → 0

as i→∞ and this finishes the proof.

Definition 2.9. Let X be a K-Vector space. The function

〈·, ·〉 : X ×X → K

is called a scalar product, if the following three properties are satisfied:

1. For all λ, µ ∈ K we have

〈λx+ µy, z〉 = λ〈x, z〉+ µ〈y, u〉〈x, λy + µz〉 = λ〈x, y〉+ µ〈x, z〉.

2. 〈x, y〉 = 〈y, x〉.

3. 〈x, x〉 ≥ 0 for all x ∈ X and 〈x, x〉 = 0 iff x = 0.

For every scalar product one can define the so called Euclidean norm by ‖x‖ :=√〈x, x〉.

Theorem 2.10. Let (X, 〈·, ·〉) be a scalar product space. Then we have for allx, y ∈ X:

1. |〈x, y〉| ≤ ‖x‖‖y‖.

2. ‖x+ y‖ ≤ ‖x‖+ ‖y‖.

Proof. 1. Without loss of generality we can assume that ‖x‖ = 1 = ‖y‖ and byreplacing x by xeiθ for some appropriate θ, we can also assume that 〈x, y〉 ≥ 0.

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Under these conditions we calculate

‖x‖‖y‖ − 〈x, y〉 = 12(‖x‖2 + ‖y‖2 − 2〈x, y〉)

= 12‖x− y‖

2 ≥ 0.

Statement 2 follows from the first one as follows:

‖x+ y‖2 = ‖x‖2 + 2Re(〈x, y〉) + ‖y‖2 ≤ ‖x‖2 + 2‖x‖‖y‖+ ‖y‖2 = (‖x‖+ ‖y‖)2.

Remark 1. If ‖ · ‖ is the euclidean norm of a scalar product, then

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2) ∀x, y ∈ X.

Vice versa a norm satisfying this equation defines a scalar product by

• 〈x, y〉 = 14(‖x+ y‖2 − ‖x− y‖2), K = R.

• 〈x, y〉 = 14(‖x+ y‖2 − ‖x− y‖2) + i

4(‖x+ iy‖2 − ‖x− iy‖2), K = C.

Definition 2.11. A Hilbert space is a scalar product space (X, 〈·, ·〉) which iscomplete with respect to the induced norm ‖x‖ =

√〈x, x〉.

Example: For Ω ⊂ Rn the space X = L2(Ω) with the scalar product

〈f, g〉 =∫

Ωfgdλn

and the Lebesgue measure λn is complete by the Fischer-Riesz theorem.

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3 Compactness (in metric spaces)

Definition 3.1. A set K ⊂ X of a metric space (X, d) is called compact, if thefollowing holds:

Every family Uλ, λ ∈ Λ, of open sets with K ⊂ ⋃λ∈Λ Uλ has a finite subfamily Uλi,1 ≤ i ≤ N with K ⊂ ⋃Ni=1 Uλi.

Theorem 3.2. Let (X, d) be a metric space. For K ⊂ X the following are equiva-lent:

1. K is compact.

2. K is complete and K is precompact (i.e. for all % > 0 K is covered by finitelymany balls B%(xi), 1 ≤ i ≤ N , xi ∈ K).

3. K is sequentially compact (i.e every sequence (xi) ⊂ K has a convergentsubsequence).

Proof. 3. ⇒ 2.: The completeness of K is clear since every Cauchy sequence whichhas a converging subsequence must converge. Now assume thatK is not precompact.Then we can choose x1 ∈ K and inductively xk ∈ K, k ∈ N, such that xk /∈B%(x1) ∪ · · · ∪ B%(xk−1). The sequence (xk) ⊂ K has no converging subsequence,since d(xk, xl) ≥ % for all k 6= l and this contradicts the assumption in 3.

2. ⇒ 1. We assume that K ⊂ ⋃λ∈Λ Uλ but K is not covered by finitely many

of the sets Uλ. Then we construct balls Bk = B2−k(xk), xk ∈ K, k ∈ N so thatBk ∩ Bk−1 6= ∅ and Bk ∩ K is not covered by finitely many of the sets Uλ. Theconstruction is done as follows: First we let B0 := X and then we assume thatB1, ..., Bk−1 have already been found. We cover Bk−1 ∩K by finitely many balls

B2−k(yi), yi ∈ K so thatB2−k(yi) ∩Bk−1 6= ∅.

This is possible since K is assumed to be precompact. Now at least one of the ballsB2−k(yi) is not covered by finitely many of the sets Uλ and we let Bk be one of theseballs. The sequence of centres (xk) is a Cauchy sequence in K since for l > k large

15

enough

d(xk, xl) ≤ d(xk, xk+1) + · · ·+ d(xl−1, xl) ≤l∑

i=k2−i < ε.

Since K is assumed to be complete the limit x = lim xk exists. Since K ⊂ ⋃λ∈Λ Uλ

there has to exist a λ0 ∈ Λ so that x ∈ Uλ0 and therefore Bk ⊂ Uλ0 for k largeenough, which is a contradiction to the fact that Bk is not covered by finitely manyof the sets Uλ.

1. ⇒ 3.: We assume that there exists a sequence (xk) ⊂ K which has no accu-mulation point in K. Hence, for every x ∈ K there exists a radius %x > 0 so thatxk ∈ B%x(x) for at most finitely many k ∈ N. Now K ⊂

⋃x∈K Bρx(x) and since

K is compact this implies that K ⊂ ⋃Ni=1Bρxi (xi) for some N ∈ N and xi ∈ K,

1 ≤ i ≤ N . Hence xk ∈ K for at most finitely many k which is a contradiction.

Lemma 3.3. Let (X, ‖ · ‖) be a normed space and let V ( X be a closed subspace.For all θ < 1 there exists xθ ∈ X such that ‖xθ‖ = 1 and dist(xθ, V ) ≥ θ.

Proof. By our assumptions there exists an element y ∈ X\V with dist(y, V ) > 0.For every θ < 1 there exists an element vθ ∈ V with ‖y − vθ‖ ≤ 1

θdist(y, V ). Wedefine xθ := y−vθ

‖y−vθ‖ and we note that for every v ∈ V

‖xθ − v‖ = ‖y − vθ‖−1‖y − (vθ − ‖y − vθ‖v)‖ ≥ ‖y − vθ‖−1dist(y, V ) ≥ θ,

where we used that vθ − ‖y − vθ‖v ∈ V .

Theorem 3.4. Let (X, ‖ · ‖) be a normed space. Then B1(0) ⊂ X is compact if andonly if dimX <∞.

Proof. ⇐ This follows immediately from Theorem 1.2, the Bolzano-Weierstrass the-orem and Theorem 2.2.

⇒We assume that dimX =∞. Then we can inductively choose a sequence xk ∈ X,k ∈ N, with ‖xk‖ = 1 and dist(xk,Spanx1, . . . , xk−1) ≥ 1

2 . For this constructionwe use Lemma 3.3 and the fact that finite-dimensional subspaces of a normed spaceare always closed.

The sequence (xk) has no converging subsequence since by construction ‖xk−xj‖ ≥ 12

for all k 6= j and hence it follows from Theorem 3.2 that B1(0) is not compact.

Definition 3.5. Let X,Y be metric spaces. We define the spaces of bounded respec-

16

tively continuous functions between X and Y by

B(X,Y ) := f : X → Y : f(X) is boundedC0(X,Y ) := f : X → Y : f is continuous.

Theorem 3.6. Let X,Y be metric spaces and assume that Y is complete. Then wehave that

1. dB(f, g) := supd(f(x), g(x)) : x ∈ X is a complete metric on B(X,Y ).

2. (C0∩B(X,Y )) is a closed subset of (B(X,Y ), dB) and hence complete as well.

Proof. 1. It is clear that dB is a metric. Now let fk ∈ B(X,Y ) be a Cauchy sequence.It follows that

d(fk(x), fl(x)) < ε

for all x ∈ X and k, l ≥ k(ε). In particular the sequence (fk(x)) is a Cauchy sequencein Y for all x ∈ X. Since Y is complete the limit

f(x) := limh→∞

fh(x)

exists and if we let l →∞ in the above estimate, we conclude for all x ∈ X and allk ≥ k(ε)

d(fk(x), f(x) ≤ ε.

This implies that

dB(fk, f)→ 0

as k → ∞. Now it remains to show that f ∈ B(X,Y ) but this follows from theestimate

d(f(x), f(x0)) ≤ d(f(x), fk(x)) + d(fk(x), fk(x0)) + d(fk(x0), f(x0)) <∞,

where x0 is a fixed point in X and k is chosen large enough.

2. Now we let fk ∈ (C0 ∩ B)(X,Y ) be a Cauchy sequence. As in part 1. we showthat fk converges to some f ∈ B(X,Y ) with respect to the dB-norm. It remains toshow that f ∈ C0(X,Y ). For this we let ε > 0 and we choose k(ε) so that

d(fk(x), f(x) ≤ ε

17

for all k ≥ k(ε) and all x ∈ X. Since fk is continuous, there exists δ > 0 such that

d(fk(x), fk(x0)) < ε

for all x, x0 ∈ X with d(x, x0) < δ. Arguing as above we get

d(f(x), f(x0)) ≤ d(f(x), fk(x)) + d(fk(x), fk(x0)) + d(fk(x0), f(x0)) < 3ε,

for all k ≥ k(ε) and all x, x0 ∈ X with d(x, x0) < δ.

Definition 3.7. Let (X, d) and (Y, d) be two metric spaces. The oscillation off : X → Y is the function

ωf : (0,∞)→ [0,∞], ωf (δ) = supd(x1,x2)<δ

d(f(x1), f(x2)).

If ω(δ)→ 0 as δ → 0, then f is called uniformly continuous.

A family F of maps f : X → Y is called equicontinuous if

supf∈F

ωf (δ)→ 0 as δ → 0.

Example: Let 0 < α ≤ 1. The α-Hölder constant of f : (X, d)→ (Y, d) is definedby

[f ]α := supx 6=y

d(f(x), f(y))d(x, y)α

and f is called α-Hölder continuous if [f ]α <∞. We estimate for x 6= y

d(f(x), f(y)) ≤ d(f(x), f(y))d(x, y)α d(x, y)α

and henceωf (δ) ≤ [f ]αδα.

Therefore every α-Hölder continuous function is uniformly continuous and the familyof functions F = f : X → Y : [f ]α ≤ Λ is equicontinuous.

Theorem 3.8 (Theorem of Arzela-Ascoli). Let X,Y be metric spaces and assumethat X is compact and Y is complete. For a family F ⊂ C0(X,Y ) the followingthree statements are equivalent:

1. F is relatively compact in (C0(X,Y ), dB(·, ·)) (i.e. F is compact).

2. Every sequence fk ∈ F has a subsequence fkj which converges to f ∈ C0(X,Y )(with respect to the metric dB).

18

3. F is equicontinuous and for all f ∈ F the set f(x) : x ∈ X is relativelycompact in Y .

Proof. 3. ⇒ 2.: Since X is compact there exists a countably dense subset D ⊂ X.For example we can cover X by finitely many balls with radius 1/ν for every ν ∈ Nand then D = centres of all these balls is countable and dense in X. By the as-sumptions, after choosing a subsequence, for every x ∈ D the limit f(x) : lim

k→∞fk(x)

exists and with a diagonal sequence argument we obtain the existence of the limitf(x) : lim

k→∞fk(x) for every x ∈ D.

Now we let x1, x2 ∈ D, such that d(x1, x2) ≤ δ and we obtain again by the assump-tion in 3.

d(f(x1), f(x2)) = limk→∞

d(fk(x1), fk(x2))

≤ lim supk→∞

ωfk(δ),

≤ supf∈F

ωf (δ)→ 0

as δ → 0. Therefore f |D is uniformly continuous. By exercise 4 on the exercisesheet 1 there exists a unique continuous extension (still denoted by f) f : X → Y .It remains to show that fk converges uniformly to f .

For this we let ε > 0 and we choose δ > 0 such that ωf (δ) < ε for all f ∈ F . Nextwe note that Bδ(x) : x ∈ D is an open cover of X and since X is compact wechoose a finite subcover Bδ(x) : x ∈ Dδ where Dδ ⊂ D is a finite set. Moreover,we choose k0 ∈ N so that for all k ≥ k0

maxx′∈Dδ

d(fk(x′), f(x′)) ≤ ε.

Now we let x ∈ X, k ≥ k0 and we choose x′ ∈ Dδ so that d(x, x′) ≤ δ. This yields

d(fk(x), f(x)) ≤ d(fk(x), fk(x′)) + d(fk(x′), f(x′)) + d(f(x′), f(x))≤ ωfk(δ) + ε+ ωf (δ) ≤ 3ε.

Taking the supremum over all x ∈ X we conclude that dB(fk, f)→ 0 as k →∞.

2. ⇒ 1.: By assumption 2. the set F is sequentially compact and hence it followsfrom Theorem 3.2 that F is compact.

1. ⇒ 3.: We show first that F is equicontinuous. By Theorem 3.2 the set F is

19

precompact and hence for every ε > 0 there exist f1, . . . , fN ⊂ F so that

F ⊂ ∪Ni=1Bε(fi).

This shows that for every f ∈ F there exists i ∈ 1, . . . , N so that dB(f, fi) ≤ ε

and therefore we conclude for x, y ∈ X with d(x, y) ≤ δ

d(f(x), f(y)) ≤ d(f(x), fi(x)) + d(fi(x), fi(y)) + d(fi(y), f(y))≤ 2ε+ max

1≤i≤Nωfi(δ).

Taking the supremum over all x, y ∈ X with d(x, y) ≤ δ and all f ∈ F yields

supf∈F

ωf (δ) ≤ 2ε+ max1≤i≤N

ωfi(δ)

Since fi ∈ F is continuous and hence uniformly continuous since X is compact, itfollows that

limδ→0

max1≤i≤N

ωfi(δ) = 0

and hence we conclude that

lim supδ→0

(supf∈F

ωf (δ)) ≤ 2ε

which implies that F is equicontinuous.

Finally we look at the sequence zk = fk(x) with fk ∈ F . After choosing a subse-quence we have that fk → f ∈ C0(X,Y ) uniformly, in particular zk = fk(x)→ f(x)which finishes the proof.

Next we let (X, d(·, ·)) be a metric space and let : X → R. Then we define theC0,α-norm of u by

‖u‖C0,α(X) := ‖u‖C0(X) + [u]α,X = supx∈X|u(x)|+ sup

x 6=y∈X

|u(x)− u(y)|d(x− y)α .

Theorem 3.9. The space C0,α(X) = u ∈ C0(X) : ‖u‖C0,α(X) <∞ with the C0,α-norm is a Banach space.

Proof. Let uk ∈ C0,α(X) be a Cauchy sequence. Then uk ∈ (C0 ∩B)(X,R) and ukis a Cauchy sequence with respect to ‖ · ‖C0(X). Hence it follows from Theorem 2.6that uk converges to u ∈ C0(X) with respect to ‖ · ‖C0(X). It remains to show thatu ∈ C0,α(X) and that ‖uk − u‖C0,α(X) → 0.

20

In order to show this we let x, y ∈ X, x 6= y and we estimate

|u(x)− u(y)d(x, y)α = lim

k→∞

|uk(x)− uk(y)|d(x, y)α ≤ lim

k→∞‖uk‖C0,α(X) <∞

which shows that u ∈ C0,α(X). Finally,

|(u− uk)(x)− (u− uk)(y)|d(x, y)α = lim

l→∞

|(ul − uk)(x)− (ul − uk)(y)|d(x, y)α

≤ lim supl→∞

‖ul − uk‖C0,α(X) → 0

as k →∞ and thus uk → u in C0,α(X).

Theorem 3.10. Let X be a compact metric space and let uk ∈ C0,α(X), 0 < α ≤ 1,with ‖uk‖C0,α(X) ≤ Λ <∞ for all k ∈ N. Then there exists u ∈ C0,α(X), so that upto a subsequence uk → u in C0,β(X) for all 0 ≤ β < α.

Proof. It follows from example before Theorem 3.8 that for all δ > 0 and all k ∈ N

ωuk(δ) ≤ [uk]α,Xδα ≤ Λδα.

Hence the family of functions uk is equicontinuous. Additionally uk(x) is uni-formly bounded for every x ∈ X and therefore uk(x) is relatively compact for allx ∈ X. By Theorem 3.8 there exists u ∈ C0(X) so that up to a subsequence uk → u

in C0(X). Now u ∈ C0,α(X) since

|u(x)− u(y)|d(x, y)α ≤ lim

l→∞

|ul(x)− ul(y)d(x, y)α ≤ Λ.

In order to show that uk → u in C0,β(X) we fix δ > 0 and we look at two cases:

1. d(x, y) ≤ δ

|(u− uk)(x)− (u− uk)(y)|d(x, y)β = d(x, y)α−β lim

l→∞

|(ul − uk)(x)− (uk − ul)(y)d(x, y)α

≤ 2Λδα−β,

2. d(x, y) > δ

(u− uk)(x)− (u− uk)(y)d(x, y)β ≤ 2δ−β‖u− uk‖C0(X).

21

For both cases it follows that

lim supk→∞

[u− uk]β,X ≤ 2Λδα−β → 0

as δ → 0.

In the following we let Ω ⊂ Rn be open and bounded, k ∈ N0, 0 < α ≤ 1. Foru ∈ Ck(Ω) we define the norms

‖u‖Ck(Ω) :=∑

0≤|γ|≤k‖Dγu‖C0(Ω)

‖u‖Ck,α(Ω) := ‖u‖Ck(Ω) +∑|γ|=k

[Dγu]α,Ω

and the sets

Ck(Ω) := u ∈ Ck(Ω): Dγu is continously extendable to Ω ∀|γ| ≤ kCk,α(Ω) := u ∈ Ck(Ω) : ‖u‖Ck,α(Ω) <∞.

Theorem 3.11. Let Ω ⊂ Rn be open and bounded. Then

1. (Ck(Ω), ‖u‖Ck(Ω)) is a Banach space.

2. (Ck,α(Ω), ‖u‖Ck,α(Ω)) is a Banach space.

Proof. 1. The case k = 0 follows from Theorem 3.6. For the case k = 1 we argueas follows: Let uj ∈ C1(Ω) be a Cauchy sequence with respect to ‖u‖C1(Ω). Thenit follows that uj → u in C0(Ω) and ∂iuj → vi in C0(Ω) for all 1 ≤ i ≤ n. Hencewe have to show that ∂iu = vi since then ui → u in C1(Ω). In order to see this wechoose x0 ∈ Ω and we let s ∈ R be small. Then

uj(x0 + sei) = uj(x0) +∫ s

0∂iuj(x0 + tei)dt

and the uniform convergence of uj resp. ∂iuj implies that

u(x0 + sei) = u(x0) +∫ s

0vi(x0 + tei)dt.

Hence it follows from the fundamental theorem of calculus that u ∈ C1(Ω) with∂iu(x0) = vi(x0) for all 1 ≤ i ≤ n. The general case now follows by induction usingthe same argument.

2. Let uj ∈ Ck,α(Ω) be a Cauchy sequence with respect to the norm ‖ · ‖Ck,α(Ω). Itfollows from 1. that uj → u in Ck(Ω). For |γ| = k we have Dγuj → vγ in C0,α(Ω)

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by Theorem 3.9 and hence we conclude that vγ = Dγu which directly implies thatuj → u ∈ Ck,α(Ω).

Definition 3.12. We say that Ω ⊂ Rn satisfies a chord-arc condition with constantκ ∈ [1,∞), if for all x, y ∈ Ω there exists a path γ ∈ C1([0, 1],Ω) with γ(0) = x,γ(1) = y and L(γ) ≤ κ|x− y|.

Theorem 3.13. Let Ω ⊂ Rn be open, bounded and assume that it satisfies a chord-arc condition with constant κ. Moreover, let k, l ∈ N0, α, β ∈ [0, 1], with k+α > l+β.Is uj ∈ Ck,α(Ω) a sequence with ‖uj‖Ck,α(Ω) ≤ Λ for all j ∈ N, then there exists asubsequence uj → u in C l,β(Ω).

Proof. Step 1: k = l = 0:

From the assumptions it follows that α > β and the result is then a consequence ofTheorem 3.10.

Step 2: We have ‖u‖C0,1(Ω) ≤ κ‖u‖C1(Ω).

In order to show this, let x, y ∈ Ω, γ ∈ C1([0, 1],Ω), γ(0) = x, γ(1) = y, L(γ) ≤κ|x− y|. Then we estimate

|u(x)− u(y)| =∣∣∣∣∫ 1

0

d

dtu(γ(t))dt

∣∣∣∣=∣∣∣∣∫ 1

0Du(γ(t))γ′(t)dt

∣∣∣∣≤ ‖Du‖C0(Ω)L(γ)≤ κ‖Du‖C0(Ω)|x− y|

and hence it follows that

‖u‖C0,1(Ω) = ‖u‖C0(Ω) + [u]1,Ω ≤ ‖u‖C0(Ω) + κ‖Du‖C0(Ω) ≤ κ‖u‖C1(Ω).

Step 3: k = l ≥ 1:

In this case we have again 1 ≥ α > β ≥ 0 and Step 1, 2 imply inductively thatuj → u in Ck(Ω). Step 1 yields that Dγuj → vγ in C0,β(Ω) for all |γ| = k. As in theproof of Theorem 3.11 one then obtains Dγu = vγ for all |γ| = k and hence uj → u

in Ck,β(Ω).

Step 4: k > l ≥ 0:

By step 2 we know that ‖uj‖Cl,1(Ω) ≤ C‖uj‖Ck,α(Ω). Now we consider two subcases

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1. α > 0: In this case it follows from step 3 that uj → u in Ck(Ω) and thenstep 2 implies that uj → u in C l,1(Ω). Applying step 3 again yields uj → u inC l,β(Ω).

2. α = 0, β = 1: Here we conclude that k ≥ l + 2 and step 2 implies

‖uj‖Ck−1,1(Ω) ≤ C‖uj‖Ck(Ω) ≤ CΛ.

Now step 3 implies uj → u in Ck−1(Ω) and hence also in C l,β(Ω).

Next we apply these results in order to study the kernel of a linear differentialoperator of second order. We let Ω ⊂ Rn be a bounded C∞-domain and we introducethe elliptic boundary value problem (BVP)

(∗) ∑n

i,j=1 aij(x)∂2i,ju(x) +∑n

i=1 bi(x)∂iu(x) + c(x)u(x) = f(x) ∀x ∈ Ωu = 0 on ∂Ω

We assume that there exist constants 0 < λ ≤ Λ <∞ such that

1. max1≤i,j≤n ‖aij‖C0,α(Ω), max1≤i≤n ‖bi‖C0,α(Ω), ‖c‖C0,α(Ω) ≤ Λ.

2. ∑ni,j=1 aij(x)ξiξj ≥ λ|ξ|2 for all x ∈ Ω, ξ ∈ Rn.

We define the operator L : C20 (Ω) → C0(Ω), where C2

0 (Ω) = u ∈ C2(Ω): u =0 on ∂Ω by

Lu :=n∑

i,j=1aij∂

2iju+

n∑i=1

bi∂iu+ cu.

It follows from the above assumptions that

‖Lu‖C0(Ω) ≤ Λ

n∑i,j=1‖∂2

iju‖C0(Ω) +n∑i=1‖∂iu‖C0(Ω) + ‖u‖C0(Ω)

= Λ‖u‖C2(Ω)

and therefore‖L‖ ≤ Λ

which shows that L ∈ L(C20 (Ω), C0(Ω)).

In order to proceed we need the so called Schauder estimates (see the book ofGilbarg-Trudinger for a proof).

24


Theorem 3.14 (Schauder estimates). Let Ω and L be as above and let u ∈ C20 (Ω)

solve Lu = f . If f ∈ C0,α(Ω), for α > 0, then u ∈ C2,α(Ω) with

‖u‖C2,α(Ω) ≤ C(‖f‖C0,α(Ω) + ‖u‖C0(Ω))

where C = C(n, α, λ,Λ,Ω).

As an application we show that the kernel of L is always finite-dimensional.

Theorem 3.15. The kernel of L : C20 (Ω)→ C0(Ω) is finite dimensional.

Proof. On ker(L) we choose the ‖u‖C0(Ω)-norm. Let uk ⊂ ker(L) be a sequence with‖uk‖C0(Ω) ≤ 1 for all k ∈ N. Since uk is a solution of Luk = 0 we can use Theorem3.14 to conclude

‖uk‖C2,α(Ω) ≤ C‖uk‖C0(Ω) ≤ C.

Hence it follows from Theorem 3.13 that there exists a subsequence ukj so thatukj → u ∈ C2(Ω) with respect to the C2-norm. We conclude that u ∈ ker(L) with

‖u‖C0(Ω) ≤ 1 and therefore BC0(Ω)1 (0) ∩ kerL is sequentially compact which, by

Theorem 3.4, shows that kerL has to be finite dimensional.

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4 Extension of linear functionals

In this chapter we show how to extend linear functionals in infinite-dimensionalspaces.

Theorem 4.1 (Hahn-Banach). Let X be a R-vector space and let p : X → R be asublinear function, i.e.

p(x+ y) ≤ p(x) + p(y)p(λx) = λp(x), ∀ x, y ∈ X, λ ≥ 0.

Moreover, let V ⊂ X be a linear subspace and let φ : V → R be linear with φ(v) ≤p(v) for all v ∈ V . Then there exists a linear map Φ: X → R with Φ|V = φ andΦ(x) ≤ p(x) for all x ∈ X.

We show this theorem in two steps.

Proof. Step 1: Let x /∈ V ⇒ there exists an extension Φ of φ on V ⊕Rx with Φ ≤ p.

In order to show this, we make the ansatz Φ(v + αx) := φ(v) + αs for all α ∈ Rwith s = Φ(x). Hence Φ is linear and Φ|V = φ. Moreover Φ ≤ p if and only if thereexists a number s ∈ R so that φ(v) + αs ≤ p(v + αx) for all v ∈ V and α ∈ R. Butthis is equivalent to

s ≤ p(v+αx)−φ(v)α , ∀α ≥ 0, v ∈ V

s ≥ φ(v′)−p(v′−α′x)α′ , ∀α′ > 0, v′ ∈ V.

and by setting w = v/α this is in turn equivalent tos ≤ p(w + x)− φ(w), ∀w ∈ Vs ≥ φ(w′)− p(w′ − x), ∀w′ ∈ V

We can choose s as we wish if and only if φ(w′) − p(w′ − x) ≤ p(w + x) − φ(w) ⇔φ(w + w′) ≤ p(w + x) + p(w′ − x) for all w,w′ ∈ V . But since p is assumed to besublinear it follows that

p(w + x) + p(w′ − x) ≥ p(w + w′) ≥ φ(w + w′)

26

for all w,w′ ∈ V and hence we conclude that the desired value of s exists.

Step 2: In order to show the result in full generality we need to use Zorn’s Lemma.For this we recall a few definitions. A set E with an ordering ≤ is called partiallyordered if the following properties are satisfied for all a, b, c ∈ E

• a ≤ a,

• a ≤ b and b ≤ a implies a = b and

• a ≤ b, b ≤ c implies a ≤ c.

A subset M ⊂ E is called totally ordered if for all a, b ∈ M we have either a ≤ b

or b ≤ a. An element b ∈ E is called upper bound of M if a ≤ b for all a ∈ Mand an element s ∈ E is called maximal element if s ≤ a for some a ∈ E impliesa = s.

Finally, we call (E,≤) inductively ordered if every totally ordered subset M ⊂ Ehas an upper bound. With all these definitions we are now able to formulate the

Lemma of Zorn: Let (E,≤) be inductively ordered, then E has a maximal element.

In order to apply this result in our situation we let E be the set of pairs (W,ψ),where W is a linear subspace of X with V ⊂ W and ψ : W → R is linear withψ|V = φ and ψ(w) ≤ p(w) for all w ∈W . Moreover we define the ordering ≤ by

(W1, ψ1) ≤ (W2, ψ2)⇔W1 ⊂W2 and ψ2|W1 = ψ1.

It is easy to check that with these definitions (E,≤) is partially ordered.

Now we claim that (E,≤) is inductively ordered. For this we let M = (Wi, ψi)i∈J ⊂E be totally ordered (Wi ⊂Wj for all j ≥ i) and we defineW = ⋃

i∈JWi, ψ : W → Rwith ψ|Wi = ψi. We note that

• W is a linear subspace with V ⊂W . In order to see this let w1, w2 ∈W , thusthere exist i1, i2 ∈ J so that w1 ∈ Wi1 , w2 ∈ Wi2 . Without loss of generalitywe assume i1 ≤ i2 and therefore w1, w2 ∈ Wi2 . Thus λw1 + µw2 ∈ Wi2 ⊂ W

for all λ, µ ∈ R.

• The function ψ is well-defined since for w ∈ Wi1 ∩Wi2 with i1 < i2 we havew ∈Wi1 and hence ψ|Wi2

(w) = ψ|Wi1(w).

• Using the same argument it also follows that ψ is linear.

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• It follows from the definition that ψ ≤ p on W .

We conclude that (W,ψ) is an upper bound forM and therefore (E,≤) is inductivelyordered. By the Lemma of Zorn there exists a maximal element (W,Φ) of E. IfW 6= X, then we use step 1 in order to extend Φ to W ⊕ Rx where x ∈ X\W .This gives a contradiction to maximality of (W,Φ) and hence W = X and Φ is theextension we were looking for.

Theorem 4.2. Let X be a normed K-vector space (K = R or C) and let V ⊂ X bea linear subspace with the induced norm. For every φ ∈ V ′ there exists a Φ ∈ X ′

with Φ|V = φ and ‖Φ‖X′ = ‖φ‖V ′.

Proof. We split the proof into two parts:

1. K = R.

Here we define the sublinear function p(x) := ‖φ‖V ′‖x‖ for all x ∈ X. For allv ∈ V we conclude φ(v) ≤ ‖φ‖V ′‖v‖ = p(v) and hence we can apply Theorem4.1 in order to conclude that there exists a linear function Φ: X → R withΦ|V = φ and Φ(x) ≤ p(x) = ‖φ‖V ′‖x‖ for all x ∈ X. This shows that‖Phi‖X′ ≤ ‖φ‖V ′ and the reverse inequality is obvious since Φ is an extensionof φ.

2. K = C.

In this case V is a complex subspace and φ is C-linear. Hence for φ1 := Re(φ),φ2 := Im(φ) we get

iφ(x) = φ(ix), ∀x ∈ V⇔ (−φ2 + iφ1)(x) = φ1(ix) + iφ2(ix), ∀x ∈ V

⇔ φ2(x) = −φ1(ix), ∀x ∈ V⇔ φ(x) = φ1(x)− iφ1(ix) ∀x ∈ V.

Now we consider X,V as R-vector spaces. Since φ is C-linear it follows thatφ1 is R-linear and it follows from the definition of the operator norm that‖φ1‖V ′ ≤ ‖φ‖V ′ . By applying case 1 we get the existence Φ1 ∈ X ′R withΦ1|V = φ1 and ‖Φ‖X′ = ‖φ1‖V ′ ≤ ‖φ‖V ′ . Now we define

Φ(x) := Φ1(x)− iΦ1(ix)

and it follows from the above argument that Φ is C-linear with Φ|V = φ. Itremains to show that ‖Φ‖X′ = ‖φ‖V ′ . In order to show this, we let x ∈ X and

28


we express Φ in polar coordinates Φ(x) = reiθ. We conclude

|Φ(x)‖ = r = Re(e−iθΦ(x))= Re(Φ(e−iθx))= Φ1(e−iθx)≤ ‖φ‖V ′‖x‖

and therefore ‖Φ‖X′ ≤ ‖φ‖V ′ . The reverse inequality is again trivial.

Theorem 4.3. Let (X, ‖ · ‖) be a normed vector space and let V ⊂ X be a subspace.For every x0 ∈ X with dist(x0, V ) > 0 there exists a φ ∈ X ′ with φ|V ≡ 0, ‖φ‖ = 1and φ(x0) = dist(x0, V ).

Proof. We start by defining a functional φ on V ⊕Kx0 by

φ(v + αx0) = αdist(x0, V )

for all v ∈ V and all α ∈ K. Hence φ|V = 0 and φ(x0) = dist(x0, V ). We claim that‖φ‖(V⊕Kx0)′ = 1. In order to show this we note that without loss of generality wecan assume α 6= 0 and then we estimate

|φ(v + αx0)| ≤ |α|dist(x0, v)

≤ |α|∥∥∥∥x0 −

(− 1αv

)∥∥∥∥= ‖v + αx0‖.

Hence we get‖φ‖(V⊕Kx0)′ ≤ 1

and therefore φ ∈ (V ⊕Kx0)′.

In order to show the reverse inequality we note that since dist(x0, V ) > 0 for everyε > 0 there exists vε ∈ V so that ‖x0−vε‖ ≤ (1+ε)dist(x0, V ) and hence φ(x0−vε) =dist(x0, V ) ≥ ‖x0−vε‖

1+ε which yields

‖φ‖(V⊕Kx0)′ ≥1

1 + ε∀ε > 0.

Therefore the claim is proved and we finish the proof by using Theorem 4.2 in orderto extend φ to all of X.

Lemma 4.4. Let X be a normed space.

29

1. For all x0 ∈ X there exists φ ∈ X ′ so that ‖φ‖ = 1 and φ(x0) = ‖x0‖.

2. If φ(x0) = 0 for all φ ∈ X ′ then x0 = 0.

Proof. Claim 1 follows from Theorem 4.3 applied to V = 0 since then dist(x0, V ) =‖x0‖ ⇒ and claim 2 follows from Claim 1.

Theorem 4.5. Let (X, ‖ · ‖) be a normed space. The canonical map J : X → X ′′ :=(X ′)′

(Jx)(φ) = φ(x), ∀x ∈ X, φ ∈ X ′

is an isometric embedding.

Proof. We have to show that ‖Jx‖X′′ = ‖x‖X and we split this into two parts:

(i) Let φ ∈ X ′ with ‖φ‖X′ ≤ 1. Then we have

|(Jx)(φ)| = |φ(x)‖ ≤ ‖φ‖X′‖x‖X ≤ ‖x‖x

which implies ‖Jx‖X′′ ≤ ‖x‖X .

(ii) Let x 6= 0 and choose φ as in Lemma 4.4, 1. Then ‖φ‖X′ = 1 and

Jx(φ) = φ(x) = ‖x‖X

which implies‖Jx‖X′′ = sup

‖φ‖X′≤1|Jx(φ)| ≥ ‖x‖X .

Remark: A Banach space X is called reflexive if the canonical map J : X → X ′′ isadditionally surjective. Examples of reflexive spaces are all finite dimensional spaces,all Hilbert spaces, lp(R) for 1 < p < ∞ and Lp(µ) again for 1 < p < ∞. On theother hand, the spaces l1(R), l∞(R), L1(µ), L∞(µ) and C0(Ω) are not reflexive.

Theorem 4.6. Let (X, ‖ · ‖) be a normed space. If the dual space X ′ is separablethen X is also separable (recall that a metric space is called separable if there existsa countably dense subset).

Proof. By the assumption of the theorem the set φ ∈ X ′ : ‖φ‖X′ = 1 has acountably dense subset φkk∈N. Now we choose xk ∈ X so that ‖xk‖X = 1 andφk(xk) ≥ 1/2 for all k ∈ N.

30

We claim that span xk is dense in X, i.e. span xk = X. In order to see thiswe argue by contradiction and we assume that there exists some x0 ∈ X withdist(x0, V ) > 0. By Theorem 4.3 there exists φ ∈ X ′ with ‖φ‖ = 1, φ(x0) =dist(x0, V ) and φ|V = 0. Since φk is dense in φ ∈ X ′ : ‖φ‖X′ = 1 we get

0 = φ(xk) = φk(xk)− (φk − φ)(xk) ≥12 −

14 > 0

if k is chosen appropriately.

Now we come to a geometric version of the Hahn-Banach theorem.

Definition 4.7. Let (X, ‖·‖) be a normed space. Two sets A,B ⊂ X are separatedby φ ∈ X ′ if

supx∈A

φ(x) ≤ infy∈B

φ(y).

Lemma 4.8. Let (X, ‖ · ‖) be a normed space and let K ⊂ X be open and convexwith 0 ∈ K. Then the function p(x) := inft > 0: x/t ∈ K is sublinear and

K = x ∈ X : p(x) < 1.

Proof. Since K is open we can choose ρ > 0 so that B2ρ(0) ⊂ K and thereforeρ x‖x‖ ∈ K for all x 6= 0 and hence p(x) ≤ ‖x‖ρ <∞ for all x 6= 0.

Now we claim that x/t ∈ K is equivalent to the fact that p(x) < t. In order to seethis we note that x/t ∈ K implies that x/s ∈ K for some s < t since K is open.But now it follows from the definition of p that p(x) ≤ s < t. On the other hand, ifp(x) < t then there exists s ∈ (p(x), t). As K is convex, K is also starshaped withrespect to 0 ∈ K and since x/s ∈ K we conclude that also x/t = s

txs ∈ K.

It remains to show that p is sublinear. For this we let λ > 0 and we calculate

p(λx) = inft > 0: (λx)/t ∈ K= λ inft/λ > 0: x/(t/λ) ∈ K= λp(x).

Next, we let λ > p(x) and µ > p(y) which implies x/λ, y/µ ∈ K and therefore

λ

λ+ µ

x

λ+ µ

λ+ µ

y

µ= x+ y

λ+ µ∈ K.

Hencep(x+ y

λ+ µ) < 1 ⇔ p(x+ y) < λ+ µ.

31

Finally, we let λ p(x) and µ p(y) and we get

p(x+ y) ≤ p(x) + p(y).

Lemma 4.9. Let (X, ‖ · ‖) be a normed space and let K ⊂ X be open and convex.Then for x0 /∈ K there exists φ ∈ X ′ so that φ(x) < φ(x0) for all x ∈ K.

Proof. Without loss of generality we assume that 0 ∈ K and we let p be as in Lemma4.8. We define ϕ(tx0) := t on Rx0 and we claim that ϕ ≤ p on Rx0. For t > 0 wehave ϕ(tx0) = t and tx0

t = x0 /∈ K which yields p(tx0) ≥ t = ϕ(tx0). For t ≤ 0 wehave ϕ(tx0) = t ≤ 0 ≤ p(tx0) and hence the claim is proved.

By Theorem 3.1 there exists φ : X → R linear with φ(x) ≤ p(x) for all x ∈ X andφ|Rx0 = ϕ. Since by Lemma 3.8 we know that p(x) < 1 for all x ∈ K we concludethat φ(x) < 1 = ϕ(x0) = φ(x0) for all x ∈ K. Since 0 ∈ K and K is open, thereexists ρ > 0 so that B2ρ(0) ⊂ K and hence for all x 6= 0 we have

φ(x) ≤ p(x) = ‖x‖ρp

(ρx

‖x‖

)<‖x‖ρ

which implies‖φ‖ ≤ 1

ρ

and hence φ ∈ X ′.

Theorem 4.10 (Hahn-Banach for convex sets). Let (X, ‖ · ‖) be a normed spaceand let A,B ⊂ X be convex> Moreover, we assume that A is open and A ∩ B = ∅.Then there exists φ ∈ X ′ which separates A and B.

Proof. We define the set

K := x− y : x ∈ A, y ∈ B=⋃Y ∈Bx− y : x ∈ A

and we note that K is open and convex:

λ(x1 − y1) + µ(x2 − y2) = λx1 + µx2 − (λy1 + µy2) ∈ K

for all x1, x2 ∈ A, y1, y2 ∈ B, λ, µ ∈ [0, 1] with λ+ µ = 1. Since A ∩B = ∅ we havethat 0 /∈ K and by Lemma 4.9 we get the existence of a φ ∈ X ′ with

φ(z) < φ(0) = 0

32

for all z ∈ K, which implies that

φ(x) < φ(y)

for all x ∈ A, y ∈ B. Taking the supremum over all x ∈ A and the infimum over ally ∈ B finishes the proof.

Next we show that under certain assumptions two convex sets can be strictly sepa-rated.

Theorem 4.11. Let (X, ‖ · ‖) be a normed vector space and let A,B ⊂ X be convexwith A ∩ B = ∅. Moreover, we assume that A is closed and B is compact. Then Aand B can be strictly separated, i.e. there exists φ ∈ X ′ with

supx∈A

φ(x) < infy∈B

φ(y).

Proof. For ρ > 0 we define the sets

Aρ := A+Bρ(0) = x+ z : x ∈ A, z ∈ Bρ(0),Bρ := B +Bρ(0) = y + z : y ∈ B, z ∈ Bρ(0)

and we note that Aρ and Bρ are open and convex. Moreover, Aρ ∩ Bρ = ∅, if ρ issmall enough, since the assumptions of the theorem imply that dist(A,B) > 0. Nowwe can apply Theorem 3.10 to Aρ and Bρ and hence there exists φ ∈ X ′ with

φ(x) + ρφ(z) = φ(x+ ρz) ≤ φ(y + ρz′) ≤ φ(y) + ρφ(z′)

for all x ∈ A, z, z′ ∈ B1(0) and y ∈ B. Next we take the supremum over all z ∈ B1(0)and the infimum over all z′ ∈ B1(0) to get

φ(x) + ρ‖φ‖ ≤ φ(y)− ρ‖φ‖

for all x ∈ A, y ∈ B. Hence

supx∈A

φ(x) + 2ρ‖φ‖ ≤ infy∈B

φ(y)

and since 2ρ‖φ‖ > 0 this shows the result.

Lemma 4.12. Let (X, ‖ · ‖) be a normed vector space and let K ⊂ X be closed andconvex with 0 /∈ K. Then there exists φ ∈ X ′ with ‖φ‖ = 1 and

supx∈K

φ(x) ≤ −dist(0,K).

33

Proof. We set R := dist(0,K) > 0 and we note that by Theorem 4.10 there existsφ ∈ X ′ separating the sets K and BR(0). We normalise φ so that ‖φ‖ = 1 and weestimate

supx∈K

φ(x) ≤ infy∈BR(0)

φ(y) = R infy∈B1(0)

φ(y) = −R‖φ‖ = −R.

Remark: We can not drop the assumption that the set A is open in Theorem 3.10.Namely, if we let A be a dense subspace of a Banach space X and if B = x0,x0 /∈ A then the assumption that there exists a φ ∈ X ′ with φ(x) < φ(x0) for allx ∈ A gives a contradiction as follows: Since A is dense in X and φ is continuouswe have φ(x) ≤ φ(x0) for all x ∈ X and therefore φ ≡ 0.

Lemma 4.13. Let Ω ⊂ Rn be open and bounded satisfying a chord-arc conditionwith constant κ. For u, v ∈ Ck,α(Ω) we have uv ∈ Ck,α(Ω) and

‖uv‖Ck,α(Ω) ≤ C(k,Ω)‖u‖Ck,α(Ω)‖v‖Ck,α(Ω).

Proof. Sheet 4, exercise 4.

Next we look again at the elliptic differential operator

L : C2,α0 (Ω)→ C0,α(Ω),

Lu =n∑

i,j=1ai,j∂

2i,ju+

n∑i=1

bi∂iu+ cu

with the same assumptions as in chapter 3. It follows from Lemma 4.13 that

‖Lu‖C0,α(Ω) ≤ C‖u‖C2,α(Ω)

for all u ∈ C2,α(Ω).

Theorem 4.14. Let Ω and L be as above, then the image of the operator L : C2,α0 (Ω)→

C0,α(Ω) is closed in C0,α(Ω).

Proof. It follows from Theorem 3.15 that the kernel of L is finite-dimensional. Byexercise 2 on sheet 4 there exists X ⊂ C2,α

0 (Ω) closed so that

C2,α0 (Ω) = ker(L)⊕X

We claim that there exists a µ > 0 so that ‖Lu‖C0,α(Ω) ≥ µ‖u‖C2,α(Ω) for all u ∈ X.

34

Assuming that the claim is wrong we get the existence of a sequence uk ∈ X with

‖Luk‖C0,α(Ω) ≤1k‖uk‖C2,α(Ω).

By replacing uk with uk‖uk‖C2,α(Ω)

we can assume that ‖uk‖C2,α(Ω) = 1. By Theorem3.13 there exists a subsequence so that uk → u in C0(Ω) and Theorem 3.14 implies

‖uk − ul‖C2,α(Ω) ≤ C(‖Luk − Lul‖C0,α(Ω) + ‖uk − ul‖C0) < ε

for all k, l large enough. Hence (uk) is a Cauchy sequence in C2,α0 (Ω) and by Theorem

3.11 this implies that uk → u in C2,α(Ω) and u|∂Ω = 0. Now

‖Lu− Luk‖C0,α(Ω) ≤ ‖u− uk‖C2,α(Ω) → 0

as k → ∞ which gives Lu = 0 since ‖Luk‖C0,α(Ω) ≤ 1k → 0 as k → ∞. Hence

u ∈ ker(L) but since X is closed we also have u ∈ X which gives u ≡ 0 whichcontradicts the fact that 1 = ‖uk‖C2,α(Ω) → ‖u‖C2,α(Ω). So the claim is true.

Now we let fk ∈ Image(L) be a sequence with fk → f in C0,α(Ω). By definitionthere exists a uk ∈ X such that Luk = fk and the claim implies that

‖uk − ul‖C2,α(Ω) ≤1µ‖Luk − Lul‖C0,α(Ω) = 1

µ‖fk − fl‖C0,α(Ω) → 0

as k, l → ∞ and hence (uk) is a Cauchy sequence. By Theorem 3.11 we concludethat uk → u in C2,α(Ω) with u|∂Ω = 0 and hence Luk → Lu in C0,α(Ω). We concludethat Lu = f and f ∈ Image(L).

35

5 Uniform boundedness principle

Definition 5.1. Let X,Y be Banach spaces. A map T ∈ L(X,Y ) is called invert-ible if T is bijective and T−1 ∈ L(Y,X).

Theorem 5.2 (Neumann series). Let X and Y be two Banach spaces. Then wehave:

1. If T ∈ L(X,Y ) is invertible and if S ∈ L(X,Y ) satisfies ‖S − T‖ < ‖T−1‖−1,then S is also invertible.

2. The set of invertible maps in L(X,Y ) is open.

Proof. Statement 2 follows directly from statement 1. In order to show statement1, we let Q ∈ L(X,X), Q = T−1(T − S) and we calculate

S = T − (T − S) = T (idX −T−1(T − S)) =: T (idX −Q).

Since ‖Q‖ ≤ ‖T−1(T−S)‖ ≤ ‖T−1‖‖T−S‖ < 1 it remains to show that the operatoridX −Q is invertible. For this we define AN := ∑N

n=0Qn, N ∈ N, and we estimate∥∥∥∥∥

N∑n=0

Qn∥∥∥∥∥ ≤

N∑n=0‖Q‖n ≤

∞∑n=0‖Q‖n = 1

1− ‖Q‖ <∞

and hence the operator A := ∑∞n=0Q

n ∈ L(X,X) is well-defined. We get

AN (idX −Q) =N∑n=0

(Qn −Qn+1) = idX −QN+1 −→ idX

as N →∞. Similarly one shows that (idX −Q)A = idX .

Theorem 5.3 (Continuity principle). Let X,Y be Banach spaces and let L(·) :[0, 1] → L(X,Y ) be a continuous family of operators. Assume that there exists aconstant c > 0, such that

‖x‖ ≤ c‖L(t)x‖,

for all t ∈ [0, 1] and all x ∈ X. Then L(1) is invertible if L(0) is invertible.

36

Proof. Assume that L(t) is invertible for some t ∈ [0, 1]. Then we can use theassumption of the Theorem with x = L(t)−1y in order to get

‖L−1(t)‖ ≤ c.

Theorem 4.2 then implies that L(s) is invertible for all s ∈ [0, 1] for which

‖L(s)− L(t)‖ < 1c.

Since L(·) is uniformly continuous on [0, 1] we can choose τ > 0 so that ‖L(s) −L(t)‖ < 1/c for all s, t ∈ [0, 1] with |s − t| < 2τ . Next we decompose [0, 1] =[0, τ ]∪ [τ, 2τ ]∪ · · ·∪ [Nτ, 1] and since L(0) is invertible it follows that L(τ), . . ., L(1)is invertible.

Definition 5.4. Let (X, d) be a metric space. A subset S ⊂ X is called nowheredense if (S) = ∅. Moreover, S is of second category if S is not a countableunion of nowhere dense subsets.

Theorem 5.5 (Baire category theorem). Let X 6= ∅ be a complete metric space.Then X is of second category, i.e. if X = ⋃∞

n=1An with closed sets An, then thereexists at least one n0 ∈ N so that

An0 6= ∅.

Proof. We argue by contradiction and we assume that X = ⋃∞n=1An where all An

are closed andAn = ∅. Now we construct closed balls Bn := Brn(xn), n ∈ N so that

• rn ∈ (0, 1/n].

• Bn ∩An = ∅.

• Bn+1 ⊂ Bn.

This is done inductively as follows: First the set X\A1 6= ∅ is open and hence theexistence of B1 follows. Now we assume that B1, ..., Bn−1 have been constructedand we note that Bn−1\An 6= ∅ is again open which implies the existence of Bn.

For m ≥ n we have d(xn, xm) ≤ 2/n→ 0 as n→∞ and hence (xn)n∈N is a Cauchysequence in X which implies that lim xn = x ∈ X exists. Now by constructionx ∈

⋂∞n=1Bn and therefore x /∈ An for all n ∈ N. But this contradicts the assumption

that X = ⋃∞n=1An.

Lemma 5.6. Let (X, d) be a complete metric space with X 6= ∅ and let (Y, ‖ · ‖)be a normed space. Assume that the family F ⊂ C0(X,Y ) is pointwise uniformlybounded, i.e.

S(x) = supf∈F‖f(x)‖ <∞ ∀x ∈ X,

37

then there exists x0 ∈ X and ρ > 0 so that

supd(x,x0)≤ρ

S(x) <∞.

Proof. We define the closed sets

An :=⋂f∈Fx ∈ X : ‖f(x)‖ ≤ n

and we note that x ∈ An if and only if S(x) ≤ n. Hence X = ⋃∞n=1An and by

Theorem 5.5 there exists Bρ(x0) ⊂ An for some n ∈ N. This implies S(x) ≤ n forall x ∈ Bρ(x0)and therefore

supx∈Bρ(x0)

S(x) ≤ n.

Theorem 5.7 (Banach-Steinhaus, uniform boundedness principle). Let X be a Ba-nach space and let Y be a normed space. If the family F ⊂ L(X,Y ) is pointwiseuniformly bounded, then

supT∈F‖T‖ <∞

and hence F is uniformly bounded.

Proof. It follows from Lemma 5.6 that there exist x0 ∈ X, ρ > 0 and c <∞ so that

‖Tx‖ ≤ c, ∀x ∈ Bρ(x0), ∀T ∈ F .

Now let x ∈ X be arbitrary, then we have

Tx = ‖x‖ρ

(T

(x0 + ρ

x

‖x‖

)− T (x0)

)and hence

‖Tx‖ ≤ ‖x‖ρ

2c.

This shows that‖T‖ ≤ 2c

ρ

for all T ∈ F .

Lemma 5.8. Let X be a Banach space and let φk ∈ X ′ with φk(x) → φ(x) for allx ∈ X. Then

supk‖φk‖ <∞.

38

Proof. We note that it follows from the convergence assumption that supk |φk(x)| <∞ for all x ∈ X. Hence Theorem 5.7 implies

supk‖φk‖ <∞.

Theorem 5.9 (Open mapping theorem). Let X and Y be Banach spaces and letT ∈ L(X,Y ) be surjective. Then T is open, i.e. T (Ω) is open for all Ω ⊂ X open.

Remark: If T is open then there exists ρ > 0 so that T (B1(0)) ⊃ Bρ(0) and henceT (BR(0)) ⊃ BRρ(0) for all R > 0 which implies that T surjective.

Proof. We split the proof into three steps.

Step 1: We have that T (B1(0)) ⊃ Bδ(0) for some δ > 0.

We have that Y = ⋃k∈N T (Bk(0)) since T is surjective. Hence we can apply

Theorem 5.5 in order to get that T (Bk(0)) ⊃ Bε(y0) for some y0 ∈ Y , ε > 0and k ∈ N. Thus for all η ∈ Y with ‖η‖ < ε there exists a sequence xj ∈ Bk(0)so that

Txj → y0 + η.

In particular, there exists a sequence ξj ∈ Bk(0) with Tξj → y0 and therefore

T

(xj − ξj

2k

)→ 1

2k (y0 + η − y0) = η

2k .

Since ‖xj−ξj2k ‖ < 1 this shows that

T (B1(0)) ⊃ Bδ(0),

with δ := ε2k .

Step 2: We have that Bδ(0) ⊂ T (B2(0)) ⊂ T (B3(0)).

We let y ∈ Bδ(0) and we note that from the Step 1 we get the existence of apoint x0 ∈ B1(0) so that

‖y − Tx0‖ < δ/2 ⇔ 2(y − Tx0) ∈ Bδ(0).

Inductively there exist points xk ∈ B1(0) with

yk+1 := 2(yk − Txk) ∈ Bδ(0), y0 := y.

39

Now for N ∈ N we have

T

(N∑k=0

2−kxk)

=N∑k=0

2−k(yk −

yk+12

)= y − 2−N−1yN+1

N→∞−→ y

and since ‖xk‖ ≤ 1 it follows that∑∞k=0 2−k‖xk‖ ≤ 2. Therefore(∑n

k=0 2−kxk)

is a Cauchy sequence in X and the limit x := ∑∞k=0 2−kxk ∈ X exists and

satisfies ‖x‖ ≤ 2 and Tx = y.

Step 3: Let Ω ⊂ X be open and let x0 ∈ Ω and y0 := Tx0 ∈ Y . There exists s > 0so that B3s(x0) ⊂ Ω and hence it follows from Step 2 that

Bδs(y0) = y0 +Bδs(0) ⊂ Tx0 + T (B3s(0)) = T (B3s(x0)) ⊂ T (Ω).

Theorem 5.10 (Inverse mapping theorem). Let X,Y be Banach spaces and assumethat T ∈ L(X,Y ) is bijective. Then T−1 ∈ L(Y,X).

Proof. The open mapping theorem, Theorem 5.9, implies that there exists ρ > 0with Bρ(0) ⊂ T (B1(0)). Hence T−1(Bρ(0)) ⊂ B1(0) or T−1(B1(0)) ⊂ B1/ρ(0) whichimplies ‖T−1‖ ≤ 1/ρ.

Example: Let ‖ · ‖1, ‖ · ‖2 be two complete norms on the vector space X. Assumethat there exists a constant c < ∞ with ‖x‖2 ≤ c‖x‖1 for all x ∈ X. Then therealso exists a constant c′ > 0 with

c′‖x‖1 ≤ ‖x‖2 ≤ c‖x‖1

for all x ∈ X, since by the assumption the map id : (X, ‖ · ‖1) → (X, ‖ · ‖2) iscontinuous and Theorem 4.10 then implies that the inverse map id : (X, ‖ · ‖2) →(X, ‖ · ‖1) is also continuous.

Theorem 5.11 (Closed graph theorem). Let X,Y be Banach spaces and let T : X →Y be linear. The following two statements are equivalent:

1. The graph of T , G(T ) := (x, Tx) : x ∈ X ⊂ X × Y , is a closed subspace ofX × Y .

2. The map T is continuous.

40


Proof. We note that X×Y with the norm ‖(x, y)‖ = max‖x‖X , ‖y‖Y is a Banachspace (see the exercises) and the maps

PX : X × Y → X, PX(x, y) = x,

PY : X × Y → Y, PY (x, y) = y

are continuous since ‖PX‖, ‖PY ‖ ≤ 1.

2. ⇒ 1.: We let (xn, Txn) → (x0, y0) ∈ X × Y and hence y0 = limn→∞

Txn = Tx0

since T is continuous. Therefore (x0, y0) ∈ G(T ) and G(T ) is closed.

1. ⇒ 2. Since G(T ) is closed (G(T ), ‖ · ‖X×Y ) is a Banach space (see exercise sheet1). Now the map PX |G(T ) : G(T ) → X is continuous and bijective and hence itfollows from Theorem 5.10 that (PX |G(T ))−1 : X → G(T ) is also continuous. ThenT is also continuous since T = PY (PX |G(T ))−1 : X → Y .

Example: (Theorem of Hellinger-Toeplitz) Let H be a Hilbert space with a scalarproduct 〈·, ·〉 and let A : H → H be linear with

〈Ax, y〉 = 〈x,Ay〉 ∀x, y ∈ H.

Then A is continuous.

We will show this using the closed graph theorem and hence we let (xn, Axn) ∈ G(A)with (xn, Axn)→ (x0, y0). Now we have for all z ∈ H

〈Axn, z〉

= // 〈xn, Az〉

〈y0, z〉〈x0, Az〉 = 〈Ax0, z〉

which implies that Ax0 = y0 and therefore (x0, y0) ∈ G(A).

Lemma 5.12. Let Y be a closed subspace of the Banach space X. Let Z be analgebraic complement of Y , i.e. X = Y ⊕ Z and let PY : X → Y be the linearprojection onto the first summand. Then PY is continuous if and only if Z closed.

Proof. "⇒": This is obvious since Z = ker(PY ) is closed.

"⇐": Define the map T : (Y × Z, ‖(y, z)‖Y×Z = max‖y‖Y , ‖z‖Z) → (X, ‖ · ‖X)between the two banach spaces by T (y, z) = y + z and note that ‖T‖ ≤ 2 and T isbijective by the assumption of the Lemma. It follows from Theorem 5.10 that T−1

is also continuous. Next we let π1 : Y × Z → Y , π1(y, z) = y and we note that π1 is

41


continuous since ‖π‖ ≤ 1. Hence it follows that PY = π1 T−1 is the compositionof two continuous maps and therefore it is continuous as well.

42

6 Lp-spaces

Let µ be a measure on X and let f : X → R be a µ-measurable function. Then theLp-norm of f , 1 ≤ p ≤ ∞, is defined by

‖f‖Lp(µ) =

(∫|f |pdµ)1/p , 1 ≤ p <∞

ess sup |f | = infs > 0: |f(x)| ≤ s, µ-a.e., x ∈ X, p =∞.

The space f : X → R : ‖f‖Lp < ∞ is a vector space since the map t 7→ tp isconvex and therefore |f + g|p = 2p|(f + g)/2|p ≤ 2p−1(|f |p + |g|p). We also definethe set

N = f : X → R : f(x) = 0, µ-a. e., x ∈ X.

Definition 6.1. The Lp-space is defined by

Lp(µ) = f : X → R : f measurable, ‖f‖Lp(µ) <∞/N

and ‖ · ‖Lp is a norm on Lp(µ). Moreover the normed space (Lp(µ), ‖ · ‖Lp(µ)) iscomplete by the theorem of Fischer-Riesz.

Theorem 6.2. Let µ = Ln on Rn. Then Lp(Rn) is separable for 1 ≤ p <∞.

Proof. It is enough to show this result for the space C0c (Rn) since C0

c (Rn) is densein Lp(Ln). Let Bν(0) be the ν-Ball in Rn and for every ν ∈ N we cover the compactset Bν(0) by finitely many balls B1/ν(xν,j), 1 ≤ j ≤ jν . For 1 ≤ j ≤ jν we definethe functions

χν,j(x) :=

1− νd(x, xν,j), if d(x, xν,j) < 1/ν0, if d(x, xν,j) ≥ 1/ν

∈ C0c (Rn). and

χν,0(x) := dist(x,Bν(0)).

Moreover, we let ην,j(x) := χν,j(x)∑jνk=0 χν,k(x)

for j = 0, ..., jν . It follows that

• ην,j ∈ C0c (Rn), for all j = 1, ..., jν and

• ∑jνj=0 ην,j(x) = 1 for all x ∈ Rn.

43

6 Lp-spaces

Now we let u ∈ C0c (Rn) and we assume that u ≡ 0 on Rn \Bν(0). We let αj := u(xν,j)

for 1 ≤ j ≤ jν and we note that for all x ∈ Rn∣∣∣∣∣∣u(x)−jν∑j=1

αjην,j(x)

∣∣∣∣∣∣ =

∣∣∣∣∣∣jν∑j=1

(u(x)− αj)ην,j(x)

∣∣∣∣∣∣≤ sup

d(x,y)≤1/ν|u(x)− u(y)|

jν∑j=1

ην,j(x)

≤ ωu(1ν

)→ 0

as ν → ∞. Note that the first equality holds since for x /∈ Bν(0) we have u(x) = 0and for x ∈ Bν(0) it follows that ην,0 ≡ 0. Hence we have shown that span ην,j : ν ∈N, 1 ≤ j ≤ jν is dense in C0

c (Rn). Now we just look linear combinations of thesefunctions with rational coefficients and this is a countably dense subset of C0

c (Rn).

We remark that the same argument works for a Radon measure µ on a σ-finte metricspace X since in this case C0

c (X) is also dense in Lp(µ).

Next we want to characterise the dual spaces of the Lp-spaces. For this we definefor all 1 ≤ q ≤ ∞ with 1

p + 1q = 1 the map

J : Lq(µ)→ Lp(µ)′, (Jv)(u) := 〈Jv, u〉 =∫Xuvdµ

and we note that the Hölder inequality yields |(Jv)(u)| ≤ ‖u‖Lp‖v‖Lq . Hence ‖Jv‖ =sup‖u‖Lp≤1 |(Jv)(u)| ≤ ‖v‖Lq and therefore

‖J‖ = sup‖v‖Lq≤1

‖Jv‖ ≤ 1,

which implies that J is continuous.

Lemma 6.3. Let p, q ∈ [1,∞] with 1p + 1

q = 1 and assume that the measure µ is σ-finite if q =∞. Then the map J : Lq(µ)→ Lp(µ) is an isometry, i.e. ‖Jv‖ = ‖v‖Lqfor all v ∈ Lq(µ).

Proof. Without loss of generality we assume that ‖v‖Lq = 1 and we split the proofinto two parts.

1. q <∞

44

6 Lp-spaces

Here we define the map

Dq : v ∈ Lq(µ) : ‖v‖Lq = 1 → u ∈ Lp(µ) : ‖u‖Lp = 1, Dq(v) = |v|q−2v

and we calculate

‖Dq(v)‖Lp =(∫|v|(q−1)pdµ

)1/p=(∫|v|qdµ

)1/p= 1

which shows that Dq is well-defined. Now

(Jv)(Dq(v)) =∫Dq(v)vdµ =

∫|v|qdµ = 1 = ‖Dq(v)‖Lp

and therefore‖Jv‖ ≥ 1 = ‖v‖Lq

which, together with the argument before the lemma, finishes the proof forq <∞.

2. q =∞

In this case p = 1 and ‖v‖L∞ = 1. By the assumption of the lemma there existssets E1 ⊂ E2 ⊂ . . . so that Ej is µ-measurable for all j ∈ N with µ(Ej) < ∞and X = ⋃∞

j=1Ej . We let

Ej,δ := x ∈ Ej : |v(x)| ≥ 1− δ

and we note that ‖v‖L∞ = 1 implies that µ(Ej,δ) > 0 if j is large enough.Next we let u = (signv)χEj,δ ∈ L1(µ) and we calculate

(Jv)(u) =∫uvdµ =

∫Ej,δ

|v|dµ ≥ (1− δ)µ(Ej,δ) = (1− δ)‖u‖L1

which implies‖Jv‖ ≥ (1− δ)

for all δ > 0 and therefore

‖Jv‖ ≥ 1 = ‖v‖L∞ .

Theorem 6.4 (Riesz). Let 1 < q ≤ ∞, 1 ≤ p < ∞ with 1p + 1

q = 1 and let µ be aσ-finite measure on X in the case p = 1 ⇔ q =∞. Then the map

J : Lq(µ)→ Lp(µ)′, (Jv)(u) =∫Xuvdµ

45

6 Lp-spaces

is a surjective isometry.

Proof. It follows from Lemma 6.3 that we only have to show that J is surjective.

We first consider the case p > 1:

For this we let φ ∈ Lp(µ)′ and we assume without loss of generality that ‖φ‖ = 1.Our goal is to find a function v ∈ Lq(µ) with φ = Jv. We let

S := u ∈ Lp(µ) : ‖u‖Lp = 1

and we note that supu∈S φ(u) = 1. Now if we assume there exists a function v0 ∈Lq(µ) with Jv0 = φ, then it follows from Lemma 6.3 that ‖v0‖Lq = ‖Jv0‖ = ‖φ‖ = 1.Therefore ‖Dq(v0)‖Lp = 1 and if we define Dq(v0) = |v0|q−2v0 =: u0 it follows thatu0 ∈ S with

φ(u0) = φ(Dq(v0)) = Jv0(Dq(v0)) =∫X|v0|qdµ = 1

which means that φ attains its maximum on S in u0. Therefore our idea is nowto show the existence of a maximum u0 ∈ S of φ|S . In order to do this we letuk ∈ S be a minimising sequence with φ(uk) → 1 = supφ(u) : u ∈ S. Next weuse the uniform convexity of the Lp-spaces for 1 0 there exists δ > 0 such that for all‖u‖Lp = ‖v‖Lp = 1 with

∥∥u+v2∥∥Lp≥ 1− δ it holds that

‖u− v‖Lp < ε.

Now for every δ > 0 and k, l large enough we have that

1− δ ≤ 12(φ(uk) + φ(ul)) = φ

(uk + ul

2

)≤∥∥∥∥uk + ul

2

∥∥∥∥Lp

and therefore‖uk − ul‖Lp < ε

which shows that (uk) is a Cauchy sequence in the complete space Lp(µ) and henceuk → u0 ∈ S in Lp(µ) and φ(u0) = 1.

Next we show that v0 := Dp(u0) ∈ Lq(µ) satisfies Jv0 = φ. For this we defineG : Lp(µ) → R, G(u) :=

∫X |u|pdµ and we let |t| < 1 and u ∈ Lp(µ) be arbitrary.

Then∂t|u0 + tu|p = |u0 + tu|p−2(u0 + tu)up

46

6 Lp-spaces

and this function is in L1(µ) since

|∂t|u0 + tu|p| ≤ (|u0 + |u|)p−1|u|p ≤ 2pp(|u0|p + |u|p).

Hence we can apply the theorems on parameter differentiation of integrals and weget

d

dt‖u0 + tu‖−1

Lp |t=0 = d

dt|t=0G(u0 + tu)1/p

= −1pG(u0)−1/p−1 d

dtG(u0 + tu)|t=0

= −1pp

∫X|u0|p−2u0udµ

= −(J(Dp(u0)))(u).

Since u0 ∈ S is a maximum of φ|S this yields

0 = d

dtφ

(u0 + tu

‖u0 + tu‖Lp

)|t=0

= d

dt|t=0(‖u0 + tu‖−1

Lp (φ(u0) + tφ(u)))

= φ(u)− φ(u0)(J(Dp(u0)))(u0)= φ(u)− J(Dp(u0))

from which the desired result follows.

Next we consider the case p =∞:

Here we let φ ∈ L1(µ)′ and for A ⊂ X measurable with µ(A) < ∞ and p ≥ 1 wedefine

φp : Lp(µ)→ R, φp(u) = φ(χAu)

and we note that the Hölder inequality implies

|φp(u)| ≤ ‖φ‖‖χAu‖L1 ≤ ‖φ‖µ(A)1/q‖u‖Lp

for 1p + 1

q = 1 and hence φp ∈ Lp(µ)′. Therefore we can use the result we just showedfor p > 1 and there exists vp ∈ Lq(µ) such that for all u ∈ Lp(µ)

(Jvp)(u) =∫uvpdµ = φp(u) = φ(χAu).

47

6 Lp-spaces

Next we note that

(J(χX\Avp))(u) =∫uχX\Avpdµ

= (Jvp)(χX\Au)= φ(χAχX\Au) = 0

and therefore J(χX\Avp) = 0 which by Lemma 6.3 implies vp = 0 almost everywhereon X\A.

Next we let 1 q′ and since vp = 0 a.e. onX\A we get vp ∈ Lq

′(µ) and hence

(Jvp′)(u) =∫uvp′dµ = φp′(µ) = φ(χAχAu)

= φp(χAu)

=∫uχAvpdµ

=∫uvpdµ = (Jvp)(u)

where we used that χAu ∈ Lp(µ). Therefore J(vp′ − vp) = 0 and we conclude againfrom Lemma 6.3 that vp′ = vp almost everywhere. Thus we define the functionv := vp for some p > 1 and we note that for all q <∞

‖v‖Lq = ‖Jv‖ = ‖φp‖ ≤ ‖φ‖µ(A)1/q.

Letting q →∞ we conclude with the help of exercise 2, sheet 6 that

‖v‖L∞ ≤ ‖φ‖.

Therefore v ∈ L∞(µ) with∫uvdµ = φ(χAu) for all u ∈ Lp(µ), p > 1 and by density

als for all u ∈ L1(µ).

Now we let X = ⋃k∈NAk, µ(Ak) < ∞, Ak measurable, A1 ⊂ A2 ⊂ . . . and we let

vk be constructed as above with A replaced by Ak. Hence ‖vk‖L∞ ≤ ‖φ‖ and fork < k′, u ∈ L1(µ) we conclude∫

u(χAkvk′)dµ =∫

(χAku)vk′dµ

= φ(χA′kuχAk)

= φ(χAku) =∫uvkdµ

which shows that vk′ |Ak = vk almost everywhere. There the function v ∈ L∞(µ),‖v‖L∞ ≤ ‖φ‖ with v|Ak = vk is well-defined and by the Lebesgue convergence

48

6 Lp-spaces

theorem we get

(Jv)(u) =∫uvdµ = lim

k→∞

∫uχAkvdµ = lim

k→∞φ(χAku) = φ(u).

Lemma 6.5 (Uniform convexity of Lp, 1 0 such that for all u, v ∈ Lp(µ) we have

12(‖u‖pLp + ‖v‖pLp)−

∥∥∥∥u+ v

2

∥∥∥∥pLp≥C‖u− v‖pLp , p ≥ 2C(‖u‖p−2

Lp + ‖v‖p−2Lp )‖u− v‖2Lp , 1 < p ≤ 2

.

Proof. The function f : R → R, f(x) = |x|p satisfies f ′(x) = p|x|p−2x for all x ∈ Rand

f ′′(x) = p(p− 1)|x|p−2, for all x 6= 0 if p < 2.

Hence f is convex and it follows that for all x0, x1 ∈ R

12(|x0|p + |x1|p)−

∣∣∣∣x0 + x12

∣∣∣∣p > 0.

Now we claim that we can improve this estimate to

12(|x0|p + |x1|p)−

∥∥∥∥x0 + x12

∥∥∥∥ ≥c(p)|x0 − x1|p, p ≥ 2c(p)(|x0|+ |x1|)p−2|x0 − x1|2, 1 < p ≤ 2

.

(6.1)

In order to show this we note that without loss of generality we can assume thatx1 = 1 and x0 = x ∈ [−1, 1]. Next we denote the left hand side of (6.1) by

σ(x) := 12(1 + |x|p)−

∣∣∣∣1 + x

2

∣∣∣∣pand the right hand side by

τ(x) :=c(p)|1− x|p, p ≥ 2c(p)(1 + |x|)p−2(1− x)2, 1 0

49

6 Lp-spaces

and

τ(1) = τ ′(1) = 0

τ ′′(1) =

0, p > 2c(p)2p−1, 1 0 so that σ′′(1)−µτ ′′(1) > 0 and a δ = δ(p) > 0 suchthat σ(x)− µτ(x) > 0 for all 1− δ ≤ x ≤ 1. For all −1 ≤ x ≤ 1− δ we estimate

σ(x) ≥ σ(1− δ) ≥ σ(1− δ)c(p)2p τ(x)

since τ(x) ≤ c(p)2p for all x ∈ [−1, 1]. Therefore σ(x) ≥ c(p)τ(x) for c(p) :=min

µ(p), σ(1−δ)

c(p)2pand hence the claim is proved.

The claim directly yields the statement of the lemma for p ≥ 2 by integration. For1 ≤ p < 2 we use the claim and the Hölder inequality with the dual exponents 2

2−pand 2

p in order to get

‖u− v‖2Lp =(∫

(|u|+ |v|)p2 (2−p)(|u|+ |v|)

p2 (p−2)|u− v|pdµ

)2/p

≤(∫

(|u|+ |v|)p)(2−p)/p (∫

(|u|+ |v|)p−2|u− v|2dµ)

≤ c(p)(‖u‖Lp + ‖v‖Lp)2−p(1

2‖u‖pLp + ‖v‖pLp −

∥∥∥∥u+ v

2

∥∥∥∥pLp

).

50

7 The dual space of C0(X)

In this chapter we want to characterise the dual space of C0(X) for a compact metricspace X.

Definition 7.1. An outer measure µ on a metric space X is called Borel regular,if

1. every Borel set B is µ-measurable.

2. for all S ⊂ X there exists a Borel set B ⊃ S with µ(B) = µ(S).

The measure µ ist called a Radon measure, if µ(K) <∞ for all K ⊂ X compact.

Lemma 7.2. Let µ be a Borel regular measure on the metric space X and let E ⊂ X.Then the measure (µxE)(S) := µ(E ∩ S) for all S ⊂ X is again Borel regular, ifone of the following two conditions is satisfied.

1. E is a Borel set.

2. E is σ-finite, i.e. E = ⋃j∈NEj for Ej µ-measurable and µ(Ej) <∞ for all j.

Proof. We first show that every Borel set B is µxE-measurable. This follows fromthe fact that for all S ⊂ X we have

(µxE)(S) = µ(E ∩ S)= µ((E ∩ S) ∩B) + µ((E ∩ S)\B)= µ(E ∩ (S ∩B)) + µ(E ∩ (S\B))= (µxE)(S ∩B) + (µxE)(S\B).

We split the rest of the proof into three parts.

1. E is a Borel set.

For S ⊂ X we choose a Borel set B1 ⊃ E ∩ S with µ(B1) = µ(E ∩ S). Then

51

B = B1 ∪ (X\E) is also a Borel set with B ⊃ S and

(µxE)(S) ≤ (µxE)(B) = µ(E ∩B)= µ(B1 ∩ E) ≤ µ(B1)= µ(E ∩ S) = (µxE)(S).

Therefore we conclude that (µxE)(S) = (µxE)(B).

2. E is µ-measurable with µ(E) <∞.

Choose a Borel set B ⊃ E with µ(B) = µ(E). For every S ⊂ X we have

(µxB)(S) = µ(B ∩ S) ≤ µ(E ∩ S) + µ(B\E)= µ(E ∩ S) + µ(B)− µ(E) = (µxE)(S)

and hence µxB = µxE and we are back in case 1.

3. E = ⋃∞j=1Ej with Ej µ-measurable and µ(Ej) <∞, Ej ⊂ Ej+1 for all j.

For S ⊂ X we use step 2 in order to conclude that there exist Borel setsBj ⊃ S with (µxEj)(Bj) = (µxEj)(S). The set B := ⋂∞

j=1Bj is then again aBorel set with B ⊃ S and

(µxE)(B) = µ

∞⋃j=1

(Ej ∩B)

= limj→∞

µ(Ej ∩B)

≤ lim supj→∞

µ(Ej ∩Bj) ≤ lim supj→∞

µ(Ej ∩ S)

≤ µ(E ∩ S) = (µxE)(S).

Theorem 7.3 (Caratheodory criterion). Let µ be an outer measure on the metricspace X with the property that if A and B are subsets of X with dist(A,B) > 0 then

µ(A ∪B) = µ(A) + µ(B).

Then all Borel sets are µ-measurable.

Proof. We show that under the assumption every closed set is µ-measurable andsince the Borel σ-algebra is generated by the closed sets this implies the full state-

52

ment. Hence we let C ⊂ X be closed and we have to show that for every S ⊂ X

µ(S) ≥ µ(S ∩ C) + µ(S\C).

We assume without loss of generality that µ(S) <∞ and we define

Ck :=x ∈ X : dist(x,C) ≤ 1

k

for k ∈ N. Since dist(S ∩ C, S\Ck) ≥ 1

k > 0 theassumption of the theorem implies

µ(S ∩ C) + µ(S\Ck) = µ((S ∩ C) ∪ (S\Ck)) ≤ µ(S).

Hence it remains to show that

µ(S\C) = limk→∞

µ(S\Ck).

For this we define for every k ∈ N

Sk :=x ∈ S : 1

k + 1 < dist(x,C) ≤ 1k

and we have S\C = (S\Ck) ∪

⋃∞j=k Sj since C is closed. We estimate

µ(S\C) ≤ µ(S\Ck) +∞∑j=k

µ(Sj)

and thus it remains to show that that the series ∑∞j=1 µ(Sj) converges. Since forj ≥ i+ 2

dist(Si, Sj) ≥1

i+ 1 −1j> 0,

we get by induction and by using the assumption that for all N ∈ N

N∑i=1

µ(S2i) = µ

(N⋃i=1

S2i

)≤ µ(S) <∞,

N∑i=1

µ(S2i−1) = µ

(N⋃i=1

S2i−1

)≤ µ(S) <∞.

Therefore

µ(S\C) ≤ limk→∞

µ(S\Ck) + limk→∞

∞∑j=k

µ(Sj) = limk→∞

µ(S\Ck) ≤ µ(S\C)

which proves the theorem.

Theorem 7.4. Let µ be a Borel regular measure on the metric space (X, d) andassume that X = ⋃∞

j=1Xj with µ(Xj) < ∞ and Xj open for all j ∈ N. Then wehave

53

1. µ(A) = infA⊂U open

µ(U) for all A ⊂ X.

2. µ(A) = supC⊂A closed

µ(C) for all A ⊂ X µ-measurable.

Proof. We show the result again in three steps.

1. Proof of 1. for µ(X) <∞.

We claim thatA := A ⊂ X : A Borel, 1. holds

is the Borel σ-Algebra. Then it follows from Definition 6.1 that 1. is truefor all A ⊂ X. We show that A is closed under countable intersections andunions. For this we let Aj ∈ A and hence there exist Uj ⊃ Aj open with

µ(Uj\Aj) = µ(Uj)− µ(Aj) < 2−jε,

where we used that Aj is µ-measurable. It follows that

µ

∞⋂j=1

Uj

− µ ∞⋂j=1

Aj

= µ

∞⋂j=1

Uj\∞⋂j=1

Aj

≤ µ ∞⋃j=1

(Uj\Aj)

< ε.

Since µ(X) < ∞ we have limk→∞

µ(⋂∞

j=1 Uj)

= µ(⋂∞

j=1 Uj)and thus for every

ε > 0 there exists a k ∈ N so that

µ

k⋂j=1

Uj

< µ

∞⋂j=1

Aj

+ ε.

and hence ∞⋂j=1

Aj ∈ A.

Additionally, we have

µ

∞⋂j=1

Aj

= µ

∞⋃j=1

Uj

− µ ∞⋃j=1

Uj\∞⋃j=1

Aj

≥ µ

∞⋃j=1

Uj

− µ ∞⋃j=1

(Uj\Aj)

≥ µ ∞⋃j=1

Uj

− εand thus ∞⋃

j=1Aj ∈ A.

54

Next we define the system A′ := A ∈ A : X\A ∈ A and we note that

A′ ⊂ A ⊂ Borel σ-algebra.

The system A′ is by definition closed under taking complements and it is alsoclosed under taking countable unions since

X\∞⋃j=1

Aj =∞⋂j=1

(X\Aj).

Moreover, A′ contains all open sets since for C ⊂ X closed we have

C =∞⋂j=1x ∈ X : dist(x,Cj) > 1/j ∈ A

and hence A′ and A are the Borel σ-algebra.

2. Proof of 1. for X = ⋃∞j=1Xj with Xj open and µ(Xj) <∞.

It follows from Lemma 7.2 that µxXj is a Borel regular measure. Step 1 thenimplies that for all ε > 0 there exists Uj ⊃ A open with

(µxXj)(Uj) = (µxXj)(A) + 2−jε

for all A ⊂ X. We can assume that A is a Borel set since this does notchange the measure as µ and µxXj are both Borel regular. Hence A is µxXj-measurable by Lemma 7.2 and

(µxXj)(Uj\A) = (µxXj)(Uj)− (µxXj)(A) ≤ 2−jε.

This implies

µ

∞⋃j=1

(Uj ∩Xj)

= µ

∞⋃j=1

(Uj ∩Xj) ∩A

+ µ

∞⋃j=1

(Uj ∩Xj)\A

= µ(A) + µ

∞⋃j=1

Xj ∩ (Uj\A)

≤ µ(A) + ε

and hence statement 1. is shown to be true

3. Proof of 2.

We start again with the case µ(X) <∞ and we note that statement 1. impliesthat for all ε > 0 there exists U ⊃ X\A open with µ(U) < µ(X\A) + ε. We

55

note that C := X\U ⊂ A is closed and

µ(C) = µ(X\U) = µ(X)− µ(U) > µ(X)− µ(X\A)− ε = µ(A)− ε,

where we used that A is µ-measurable in the last equality sign.

Now if X = ⋃∞j=1Xj with Xj open and µ(Xj) <∞, there exists Cj ⊂ A closed

so that

(µxXj)(Cj) ≥ (µxXj)(A)− ε.

Without loss of generality we can assume that C1 ⊂ C2 ⊂ . . .. Therefore

µ(A) = limj→∞

µ(Xj ∩A) ≤ limj→∞

µ(Cj) + ε.

Altogether this finishes the proof of statement 2.

For the rest of this chapter we assume that the metric space (X, d) is σ-compact,i.e. for all x ∈ X and ρ > 0 we have that Bρ(x) is compact.

Lemma 7.5. Let (X, d) be a σ-compact metric space and let µ be Radon measureon X. Then we have

1. µ(A) = infA⊂U open

µ(U) for all A ⊂ X.

2. µ(A) = supK⊂A compact µ(K) for all A ⊂ X which are µ-measurable.

Proof. Since µ is assumed to be a Radon measure we have that µ(Bρ(x)) < ∞ forall x ∈ X and all ρ > 0. Moreover, X = ⋃∞

j=1Bj(0) and hence statement 1. followsfrom statement 1. of Theorem 7.4. Next, for every closed set C ⊂ X and everyj ∈ N the set C ∩Bj(0) is compact and

limj→∞

µ(C ∩Bj(0)) = µ

∞⋃j=1

(C ∩Bj(0))

= µ(C),

thus statement 2. follows from statement 2. in Theorem 7.4.

Now we come to the representation problem. For this we let µ be a Radon measureon (X, d) and we let η : X → Rk be µ-measurable with |η(x)| = 1 for µ-a.e. x ∈ X.

56


We define the linear form

φ : C0c (X,Rk)→ R, φ(f) =

∫X〈f, η〉dµ

and we note that

|φ(f)| ≤ C(K)‖f‖C0(X) for sptf ⊂ K (7.1)

with C(K) = µ(K). Hence, for every compact K the linear form φ is continuouson the subspace of functions f ∈ C0

c (X) whose support is contained in K. If X iscompact, then φ is continuous on all of C0(X). We call a linear form φ satisfying (7.1)a linear functional on C0

c (X) and our goal is to show that every linear functional onC0c (X) has an integral representation as above. The measure µ will be the following

one.

Definition 7.6. Let φ ∈ C0c (X,Rk) → R be a linear functional. Define the associ-

ated variational measure |φ| : P(X)→ [0,∞] in two steps:

1. |φ|(U) = supφ(f) : |f | ≤ 1, sptf ⊂ U for U ⊂ X open,

2. |φ|(E) = inf|φ|(U) : U ⊃ E, U open for E ⊂ X arbitrary.

Note that the two steps are consistent since one has in step 1 that |φ|(U) ≤ |φ|(V )for U ⊂ V .

Theorem 7.7. The variational measure |φ| is a Radon measure.

Proof. We split the proof into four steps.

1. |φ| is an outer measure.

For u = ∅ only the function f ≡ 0 is admissible in the definition of |φ andhence |φ|(∅) = 0. Next we let Uj , j ∈ N, be open and we let f ∈ C0

c (X) with|f | ≤ 1 and sptf ⊂ ⋃∞j=1 Uj . By Theorem 3.2 there exists N ∈ N so that

sptf ⊂N⋃j=1

Uj .

Next we choose a partition of unity χj ∈ C0c (X, [0, 1]) with sptχj ⊂ Uj and∑N

j=1 χj = 1 on sptf (see the remark after this theorem for the existence ofsuch a partition of unity). For fj := χjf ∈ C0

c (X) we conclude sptfj ⊂ Uj ,

57

|fj | ≤ 1 and f = ∑Nj=1 fj . Hence we get

φ(f) =N∑j=1

φ(fj) ≤N∑j=1|φ|(Uj) ≤

∞∑j=1|φ|(Uj).

Taking the supremum over all such functions f , we have

|φ|

∞⋃j=1

Uj

≤ ∞∑j=1|φ|(Uj).

Now we let E ⊂ ⋃∞j=1Ej with E,Ej arbitrary. For ε > 0 we choose open setsUj ⊃ Ej with |φ|(Uj) < |φ|(Ej) + 2−jε, which is possible by Definition 7.6. Itfollows that E ⊂ ⋃∞j=1 Uj and

|φ|(E) ≤ |φ|

∞⋃j=1

Uj

≤ ∞∑j=1|φ|(Uj) <

∞∑j=1|φ|(Ej) + ε.

The subadditivity of |φ| follows from letting ε 0.

2. Borel sets are |φ|-measurable.

Let A,B ⊂ X with dist(A,B) > 0. By the Caratheodory criterion, Theorem7.3, we have to show that

|φ|(W ) ≥ |φ|(A) + |φ|(B), ∀W ⊃ (A ∪B) open.

For δ > 0 sufficiently small the sets U := Bδ(A)∩W and V := Bδ(B)∩W aredisjoint. Let f, g ∈ C0

c (X,Rk) with sptf ⊂ U , sptg ⊂ V and |f | ≤ 1, |g| ≤ 1.Then spt(f + g) ⊂ (sptf ∪ sptg) ⊂W and |f + g| ≤ 1 on X. This implies

φ(f) + φ(g) = φ(f + g) ≤ |φ|(W ).

Taking the supremum over all such functions f and g implies

|φ|(U) + |φ|(V ) ≤ |φ(W )|.

3. |φ is Borel regular

For E ⊂ X with |φ|(E) <∞ we choose Uj ⊃ E open with |φ|(Uj) ≤ |φ|(E)+ 1j

and without loss of generality we assume U1 ⊃ U2 ⊃ . . .. Then the set B :=⋂∞j=1 Uj ⊃ E is a Borel set and

|φ|(B) = limj→∞

|φ|(Uj) ≤ |φ|(E).

58

4. |φ| is a Radon measure

Let K ⊂ X be compact. Since X is assumed to be σ-compact there existsU ⊃ K open with U compact and it follows from (7.1) that

|φ|(K) ≤ |φ|(U) ≤ C(U) <∞.

Remark on the existence of the partition of unity:The partition of unity used in the above proof can be constructed as follows: LetK ⊂ X be compact and assume that K ⊂ ⋃λ∈Λ Uλ where Uλ is open for all λ ∈ Λ.For x ∈ K choose r(x) > 0 and λ(x) ∈ Λ so that

B2r(x)(x) ⊂ Uλ(x).

By Theorem 3.2 it follows that there exist points x1, . . . , xN with K ⊂ ⋃Nj=1Brj (xj)with rj := r(xj). Next we choose a non-negative function xj ∈ C0

c (X) with

χj =

1, on Brj (xj)0, on X\B2rj (xj)

and we define χ := ∑Nj=1 χj . Then we have χ ≥ 1 on K, hence χ > 1/2 on an open

neighbourhood of K. Now we choose η ∈ C0c (X),with sptη ⊂ U and η|K = 1 and

we finally defineχj := ηχj

χ∈ C0

c (X).

By construction we have sptχj ⊂ B2rj (xj) ⊂ Uλ(xj) and ∑Nj=1 χj = 1 on K.

Before we come to the representation theorem we need the following result ofLusin.

Theorem 7.8 (Lusin). Let µ be a Radon measure on the metric space (X, d) andlet A ⊂ X with µ(A) <∞ and let ε > 0. If the function g : X → R is µ-measurable,then there exists another function g ∈ C0(X) so that

µ(x ∈ A : g(x) 6= g(x)) < ε and ‖g‖C0(X) ≤ supx∈A|g(x)|.

Proof. For j ∈ N, k ∈ Z we consider the sets

Aj,k = x ∈ A : kj≤ f(x) ≤ k + 1

j.

59

Since we know from Lemma 7.2 that µxA is a Radon measure, we get that forevery ε > 0 there exist compact sets Kj,k ⊂ Aj,k with µ(Aj,k\Kj,k) < 2−j−|k|ε/3.Therefore

limN→∞

µ

A\ N⋃k=−N

Kj,k

= µ

A\ ∞⋃k=−∞

Kj,k

<∞∑

k=−∞2−j−|k|ε/3 < 2−jε.

For Nj sufficiently large we then also get µ(Kj) < 2−jε, where Kj = ⋃Njk=−Nj Kj,k.

Letting K = ⋂∞j=1Kj we conclude

µ(A\K) ≤ µ

∞⋃j=1

A\Kj

≤ ∞∑j=1

µ(A\Kj) < ε.

Next we consider the functions fj : A → R, fj(x) = kj for x ∈ Aj,k. Since the sets

Kj,k ⊂ Aj,k are compact they have positive distance from each other and thus fjis locally constant and hence continuous on Kj ⊃ K. Moreover, for all x ∈ Aj,k

we have |f(x) − fj(x)| ≤ 1j → 0 as j → ∞. Therefore f |K is the uniform limit

of a sequence of continuous functions and hence continuous itself. The result nowfollows from the Tietze extension theorem, which says that, on every metric spaceX a continuous function on a closed set C ⊂ X, f : C → R, can be extended to afunction f ∈ C0(X) with ‖f‖C0(X) = supx∈C |f(x)|.

Theorem 7.9 (Representation Theorem of Riesz-Markow-Radon). Let (X, d) bea σ-compact metric space. Then for every linear functional φ on C0

c (X,Rk) thereexists a Radon measure µ and a µ-measurable function η : X → Rk with |η| = 1µ-a.e., so that

φ(f) =∫X〈f, η〉dµ ∀f ∈ C0

c (X,Rk).

The pair µ, η is uniquely determined and we have µ = |φ|.

Proof. We start with the proof of the uniqueness statement. For this we assumethat there exists another Radon measure λ and a λ-measurable function ξ : X → Rk

with |ξ = 1 λ-a.e. and so that∫X〈f, η〉dµ =

∫X〈f, ξ〉dλ

for all f ∈ C0c (X). Let U ⊂ X be open with sptf ⊂ U and |f | ≤ 1. Then

we have φ(f) ≤ λ(U) and hence, by taking the supremum over such functions f ,µ(U) = |φ|(U) ≤ λ(U) for all U ⊂ X open. Statement 1 of Lemma 7.5 then impliesthat µ ≤ λ. Next we let K ⊂ X be compact. For U ⊃ K open with U compact and

60

ε > 0 we can use Lusin’s theorem to obtain a function ξ ∈ C0(X,Rk) such that

λ(E) < ε where E = x ∈ U : ξ(x) 6= ξ(x) and‖ξ‖C0(X) ≤ sup

x∈U|ξ(x)‖ ≤ 1.

Next we choose a function χ ∈ C0c (X) with 0 ≤ χ ≤ 1, sptχ ⊂ U and χ ≡ 1 on K.

We obtain

µ(U) ≥ φ(χξ)

=∫U〈χξ, ξ〉dλ

=∫Uχdλ+

∫Uχ(〈ξ, ξ〉 − 1)dλ

≥ λ(K)− 2λ(E)≥ λ(K)− 2ε.

Letting ε → 0 and U K we conclude µ(K) ≥ λ(K) for all K compact andstatement 2 of Lemma 7.5 then implies µ(B) ≥ λ(B) for all B ⊂ X Borel and thusµ = λ on Borel sets.

For E ⊂ X arbitrary there exist Borel sets B,B′ ⊃ E with

µ(B) = µ(E), λ(B′) = λ(E).

Without loss of generality we assume B = B′ since otherwise we look at B∩B′, andthis shows that λ = µ.

Finally, we fix v ∈ Rk and for f ∈ C0c (X) we have fv ∈ C0

c (X,Rk). Hence∫X〈fv, η〉dµ =

∫X〈fv, ξ〉dµ, ∀f ∈ C0

c (X).

By exercise sheet 7 we know that C0c (X) is dense in L1(µ) and therefeore∫

X〈fv, η〉dµ =

∫X〈fv, ξ〉dµ, ∀f ∈ L1(µ),

which by Lemma 6.3 implies 〈v, ξ〉 = 〈v, η〉 µ-almost everywhere. Inserting v =e1, ..., en we finally get ξ = η µ-almost everywhere and this finishes the proof of theuniqueness part of the statement.

In order to show the existence part we let µ = |φ|. For v ∈ Rk with |v| = 1 we define

φv : C0c (X)→ R, φv(f) = φ(fv).

61


Our goal is to show that φv can be extended as a linear functional to L1(µ). In orderto estimate φv we consider the functional

ϕ : C0c (X,R+

0 )→ R+0 , ϕ(f) = supφ(g) : g ∈ C0

c (X,Rk), |g| ≤ f.

We claim that for U ⊂ X open we have

µ(U) = supϕ(χ) : χ ∈ C0c (X,R+

0 ), sptχ ⊂ U, χ ≤ 1. (7.2)

This is true since for g ∈ C0c (X,Rk) with sptg ⊂ U and |g| ≤ 1 we have

φ(g) ≤ ϕ(|g|) ≤ supϕ(χ) : χ ∈ C0c (X,Rk), sptχ ⊂ U, χ ≤ 1.

and by taking the supremum over all such functions g it follows that

µ(U) ≤ supϕ(χ) : χ ∈ C0c (X,R+

0 ), sptχ ⊂ U, χ ≤ 1

On the other hand, we let χ ∈ C0c (X,R+

0 ) with χ ≤ 1 and sptχ ⊂ U and obtain

ϕ(χ) = supφ(g) : g ∈ C0c (X,Rk), |g| ≤ χ ≤ µ(U)

and these two inequalities imply the claim.

We split the rest of the proof into four steps.

Step 1: ϕ is a semilinear functional on C0c (X), i.e.

(i) ϕ(αf) = αϕ(f) for all f ∈ C0c (X,R+

0 ), α ≥ 0.

(ii) ϕ(f1 + f2) = ϕ(f1) + ϕ(f2), for all f1, f2 ∈ C0c (X,R+

0 ).

In order to show this we note that (i) follows directly from the definition of ϕ. For(ii) we choose functions g1, g2 ∈ C0

c (X,Rk) with |gi| ≤ fi and φ(g1) ≥ ϕ(fi) − ε Itfollows from an appropriate choice of the sign

ϕ(f1) + ϕ(f2)− 2ε ≤ |φ(g1)|+ |φ(g2)| = |φ(g1 ± g2)| ≤ ϕ(f1 + f2),

and with ε 0 we get ϕ(f1) + ϕ(f2) ≤ ϕ)f1 + f2). For the other inequality we letg ∈ C0

c (X,Rk) with |g| ≤ f1 + f2 and we define

gi :=

fif1+f2

g, f1 + f2 > 00, f1 + f2 = 0

62

It follows that |gi| ≤ fi and in particular gi ∈ C0c (X,Rk). Since g = g1 + g2 we get

|φ(g)| ≤ |φ(g1)|+ φ(g2)| ≤ ϕ(f1) + ϕ(f2)

and thus

ϕ(f1 + f2) ≤ ϕ(f1) + ϕ(f2).

Step 2: We have ϕ(f) =∫X fdµ for all f ∈ C0

c (X,R+0 ).

For ε > 0 we choose points 0 = t0 < . . . < tN <∞ with |ti − ti−1| < ε, tN > max fand

µ(f−1ti) = 0, ∀1 ≤ i ≤ N.

Note that this is possible since the set of all t > 0 so that µ(f−1t) > 0 is countablesince µ(sptf) <∞. Since f is continuous we have that these sets Ui := f−1((ti−1, ti))are open.

Now we choose a function χi ∈ C0c (X,R+

0 ) with sptχi ⊂ Ui and χi ≤ 1 for all1 ≤ i ≤ N . It follows that ∑N

i=1 ti−1χi ≤ f and since ϕ is monotone by definition,we obtain from step 1

N∑i=1

ti−1ϕ(χi) = ϕ

(N∑i=1

ti−1χi

)≤ ϕ(f).

Taking the supremum over the functions χi and using (7.2) we get∑Ni=1 ti−1µ(Ui) ≤

ϕ(f) and

∫Xfdµ ≤

N∑i=1

tiµ(Ui) ≤N∑i=1

(ti−1 + ε)µ(Ui) ≤ ϕ(f) + εµ(sptf),

which as ε 0 implies∫X fdµ ≤ ϕ(f).

In order to show the reverse inequality we choose Vi ⊃ U i ∪ f−1ti open for all1 ≤ i ≤ N with

µ(Vi) ≤ µ(U i) + ε/N.

There exist χi ∈ C0c (Vi) with sptχi ⊂ Vi, 0 ≤ χi ≤ 1 and χi ≡ 1 on U i ∪ f−1ti.

63


This gives f ≤∑Ni=1 tiχi and then

ϕ(f) ≤N∑i=1

tiϕ(χi)

≤N∑i=1

(ti−1 + ε)µ(Vi)

≤N∑i=2

(ti−1 + ε)(µ(Ui) + ε

N

)+ ε

(µ(U1) + ε

N

)≤∫Xfdµ+ 2εµ(sptf) + ε‖f‖C0 + Cε

where we used again (7.2). This shows the reverse inequality after letting ε 0.

Step 3: There exists a µ-measurable function η : X → Rk with Φ(g) =∫X〈g, η〉dµ

for all g ∈ C0c (X,Rk).

For f ∈ C0c (X) we have φv(f) = φv(f+)− φv(f−) and hence by step 2

|φv(f)| ≤ |φv(f+)|+ |φv(f−)| ≤ ϕ(f+) + ϕ(f−) =∫X|f |dµ.

It follows from the exercises that C0c (X) is dense in (L1(µ), ‖ · ‖L1) and hence there

exists a unique continuous extension φv : L1(µ) → R with |φv(f)| ≤ ‖f‖L1(µ), i.e.φv ∈ L1(µ)′. By Theorem 6.4 there exists a function ηv ∈ L∞(µ) so that for allf ∈ L1(µ) we have

φv(f) =∫fηvdµ.

If we let v = ei, 1 ≤ i ≤ n, and if we define

η : X → Rk, η =k∑i=1

ηeiei,

then we obtain for all g ∈ C0c (X,Rk)

φ(g) =k∑i=1

φei(gi) =k∑i=1

∫Xgiηeidµ

=∫X〈g, η〉.

Step 4: |η| = 1 µ-a.e.

64

Let U ⊂ X be open with µ(U) <∞. Then

µ(U) = supφ(g) : g ∈ C0c (X,Rk), sptg ⊂ U, |g| ≤ 1 ≤

∫U|η|dµ.

We define the µ-measurable function η : X → R by

η(x) :=

η(x)|η(x)| , η(x) 6= 00, η(x) = 0,

and we note that Lusin’s theorem implies that there exists K ⊂ U compact withµ(U\K) < ε and a function ˜η ∈ C0

c (X,Rk) with ˜η|K = η. We choose χ ∈ C0c (U)

with χ ≡ 1 on K, 0 ≤ χ ≤ 1 and we define g := χ˜η and we note that |g| ≤ 1 andsptg ⊂ U . Therefore

µ(U) ≥ φ(g) =∫U〈g, η〉dµ =

∫K〈g, η〉dµ+

∫U\K〈g, η〉dµ

=∫K|η|dµ+


=∫U|η|dµ−

∫U\K|η|dµ+


→∫U|η|dµ

as ε 0. Henceµ(U) =

∫U|η|dµ

for all U ⊂ X open.

Next we let A ⊂ X be a Borel set. Then it follows from Lemma 7.5 that there existsan open set U ⊃ A with µ(U\A) < ε. Now∣∣∣∣µ(A)−

∫A|η|dµ

∣∣∣∣ ≤ µ(U)− µ(A) +∣∣∣∣µ(U)−

∫U|η|dµ

∣∣∣∣+ ∫U|η|dµ−

∫A|η|dµ ≤ cε

and therefore µ(A) =∫A |η|dµ for all Borel sets A ⊂ X.

Now we assume that there exists a set E ⊂ X with 0 < µ(E) < ∞ and so that|η| > 1 on E. By Lusin’s theorem there exist a compact set K ⊂ E and a functionη ∈ C0(X,Rk) with

µ(E\K) < ε and η|K = η

which implies |η| > 1 on K. But then we conclude

0 <∫K

(|η| − 1)dµ =∫K|η|dµ− µ(K) = 0,

65

which gives a contradiction. The same argument shows that there can’t exists a setE ⊂ X with 0 < µ(E) <∞ and |η| < 1 on E. Since

X =∞⋃i=1

Bi(0) with µ(Bi(0)) <∞

it follows that |η| = 1 µ-a.e.

66

8 Weak convergence

Definition 8.1. Let X be a Banach space.

1. A sequence xk converges weakly to x in X (xk x) if ϕ(xk) → ϕ(x) for allϕ ∈ X ′.

2. A sequence ϕk converges weakly ∗ to ϕ in X ′ (ϕk∗ ϕ) if ϕk(x)→ ϕ(x) for

all x ∈ X.

Example 1. 1. fk f in Lp(µ) for 1 ≤ p < ∞ ⇔∫fkgdµ →

∫fgdµ for all

g ∈ Lq(µ) with 1p + 1

q = 1.

2. fk∗ f in Lp(µ) = Lq(µ)′ for 1 < p ≤ ∞ ⇔

∫fkgdµ →

∫fgdµ for all

g ∈ Lq(µ) with 1p + 1

q = 1.

Therefore weak and weak ∗ convergence agree if 1 < p <∞.

Theorem 8.2. Let X and Y be Banach spaces. Then the following statements hold

1. Weak and weak ∗ limits are unique.

2. Norm convergence implies weak convergence.

3. Weakly and weakly ∗ converging sequences are bounded.

4. If xk x then ‖x‖ ≤ lim infk→∞ ‖xk‖ and if ϕk∗ ϕ then ‖ϕ‖ ≤ lim infk→∞ ‖ϕk‖.

5. If xk → x and ϕk∗ ϕ then ϕk(xk)→ ϕ(x). The same is true if xk x and

ϕk → ϕ.

6. For T ∈ L(X,Y ) and xk x in X it follows that Txk Tx in Y .

Proof. 1. The statement is clear for weak ∗ convergence since this is just pointwiseconvergence. Now let xk x and xk y. For all ϕ ∈ X ′ this implies

ϕ(x− y) = ϕ(x)− ϕ(y) = limϕ(xk)− limϕ(xk) = 0.

67

8 Weak convergence

It follows from Lemma 4.4 that x = y.

2. Since |ϕ(xk) − ϕ(x)| ≤ ‖ϕ‖‖xk − x‖ → 0 for all ϕ ∈ X ′, we get that normconvergence implies weak convergence. Similarly

|ϕ(x)− ϕk(x)‖ ≤ ‖ϕ− ϕk‖‖x‖ → 0

for all x ∈ X.

3. The statement for the weak * convergence follows directly from Lemma 5.8.Now we let xk x in X and we let J : X → X ′′ be the canonical embedding.Then

(Jxk)(ϕ) = ϕ(xk)→ ϕ(x) = (Jx)(ϕ)

for all ϕ ∈ X ′ and therefore Jxk∗ Jx in X ′′ = (X ′)′. Thus Jxk is uniformly

bounded in X ′′ and since we have shown in Theorem 4.5 that J is an isometry,it follows that xk is also uniformly bounded in X.

4. Let ϕk∗ ϕ in X ′. By choosing a subsequence we can assume that lim

k→∞‖ϕk‖

exists and is finite. For ‖x‖ ≤ 1 we obtain

|ϕ(x)| = limk→∞

|ϕk(x)| ≤ limk→∞

‖ϕ‖ = lim infk→∞

‖ϕk‖.

For xk x in X we conclude as in the statement above that Jxk∗ Jx in X ′′

and hence ‖Jx‖ ≤ lim inf ‖Jxk‖. Again by Theorem 4.5 this shows that

‖x‖ ≤ lim inf ‖xk‖.

5. Assume that xk x in X and ϕk → ϕ in X ′. Then we estimate

|ϕk(xk)− ϕ(x)| ≤ |ϕk(xk)− ϕ(xk) + |ϕ(xk)− ϕ(x)|≤ ‖ϕk − ϕ‖‖xk‖+ |ϕ(xk)− ϕ(x)| → 0

since ‖xk‖ is uniformly bounded by statement 3. The other case follows simi-larly.

6. Let ψ ∈ Y ′, T ∈ L(X,Y ) and assume that xk x in X. Then

ψ(Txk) = (ψ T )(xk)→ (ψ T )(x) = ψ(Tx)

as ψ T ∈ X ′. Since this is true for all ψ ∈ Y ′, we get Txk Tx in Y .

68

8 Weak convergence

Theorem 8.3. Let X be a separable Banach space. Then every uniformly boundedsequence ϕk ∈ X ′ has a subsequence which converges weakly ∗ to a ϕ ∈ X ′, i.e. theset ϕ ∈ X ′ : ‖ϕ‖ ≤ R is weakly ∗ sequentially compact for every R <∞.

Proof. Let xn : n ∈ N be dense in X. Let ‖ϕk‖ ≤ c < ∞ for all k ∈ N and letxn : n ∈ N be dense in X. Then we have for all k, n ∈ N

|ϕk(xn)| ≤ ‖ϕk‖‖xn‖ ≤ c‖xn‖

and a diagonal sequence argument as in the proof of the Arzela-Ascoli theorem thenimplies that there exists a subsequence (wlog ϕk itself) so that limϕk(xn) = ϕ(xn)for all n ∈ N. Thus the map ϕ(x) = lim

k→∞ϕk(x) is well-defined on Y = span x1, ....

We have that ϕ ∈ Y ′ since for all x ∈ Y we have

|ϕ(x)| = limk→∞

|ϕk(x)| ≤ c‖x‖.

It follows from the exercise sheet 1 that there exists a unique continuous extensionϕ ∈ X ′ with ‖ϕ‖ ≤ c. It remains to show that ϕk(x)→ ϕ(x) for all x ∈ X, but thisfollows from the estimate

|ϕ(x)− ϕk(x)| ≤ |ϕ(x)− ϕ(xk)|+ |ϕ(xn)− ϕk(xn)|+ |ϕk(xn)− ϕk(x)| ≤ ε,

where we first choose n ∈ N so that ‖x − xn‖ < ε/(3c) and then we choose k0 ∈ Nsuch that |ϕ(xn)− ϕk(xn)| ≤ ε/3 for all k ≥ k0.

Next we recall that a Banach space X is called reflexive if the canonical mapJX : X → X ′′, (JXx)(ϕ) = ϕ(x), for all ϕ ∈ X ′, is surjective. It will follow from theresults below and in the next chapter that all Hilbert spaces and all Lp-spaces with1 < p <∞ are reflexive.

Lemma 8.4. Let X be a Banach space.

1. If X is reflexive, then every closed subspace Y ⊂ X is also reflexive.

2. X is reflexive if and only if X ′ is reflexive.

Proof. 1. Let λ ∈ Y ′′ and define Λ: X ′ → R by Λ(ϕ) = λ(ϕ|Y ) for all ϕ ∈ X ′.Since

|Λ(ϕ)| = |λ(ϕ|Y )| ≤ ‖λ‖‖ϕ|Y ‖ ≤ ‖λ‖‖ϕ‖

it follows that Λ ∈ X ′′. By the assumption of the lemma there exists x ∈ Xso that Λ = JXx. Next we assume x /∈ Y , i.e. dist(x, y) > 0. By Theorem 4.3

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8 Weak convergence

there exists ϕ ∈ X ′ with ϕ|Y = 0 and ϕ(x) = dist(x, Y ) > 0 and we get thecontradiction

0 = λ(ϕ|Y ) = Λ(ϕ) = (JXx)(ϕ) = ϕ(x) > 0.

Hence x ∈ Y and for an arbitrary ϕ ∈ Y ′ we use Theorem 4.2 in order toobtain ϕ ∈ X ′ with ϕ|Y = ϕ. We get

(JY x)(ϕ) = ϕ(x) = ϕ(x) = (JXx)(ϕ) = Λ(ϕ) = λ(ϕ|Y ) = λ(ϕ)

and therefore JY x = λ.

2. We start with the implication "⇒". For this we let φ ∈ X ′′′ and we note thatφ JX ∈ X ′. For Λ ∈ X ′′ we calculate

(JX′(φ JX))(Λ) =Λ(φ JX)=(JX(J−1

X Λ))(φ JX)=(φ JX)(J−1

X Λ) = φ(Λ),

where we used that JX is invertible by Theorem 4.5 since X is assumed to bereflexive. Therefore we conclude JX′(φ JX) = φ.

Next we show the implication "⇐". From what we just proved it follows thatif X ′ reflexive then also X ′′ = (X ′)′ is reflexive and we claim that JX(X) is aclosed subspace of X ′′. For this we let JXxk be a Cauchy sequence in JX(X)and since JX is an isometry this implies that xk is a Cauchy sequence in theBanach space X. Hence xk → x ∈ X and JXxk → JXx. Thus it followsfrom the first part of this lemma that JX(X) is reflexive. It follows that X isreflexive since we have the general

Fact: Let X and Y be Banach spaces and let T ∈ L(X,Y ) be invertible. Then

X is reflexive ⇔ Y is reflexive.

In order to show this we look at the following diagram

XT //

JX

Y

JY

X ′′T ′′ // Y ′′

For Λ ∈ X ′′ and ψ ∈ Y ′ we define

(T ′′Λ)(ψ) = Λ(ψ T ) ∈ X ′.

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8 Weak convergence

Moreover, for Λ ∈ Y ′′ and ϕ ∈ X ′, we define the map T ′′ : Y ′′ → x′′ by

(T ′′Λ)ϕ = Λ(ϕ T−1).

We calculate

(T ′′ T ′′)(Λ)(ψ) = T ′′(T ′′(Λ))(ψ) = (T ′′Λ)(ψ T ) = Λ(ψ)

and thus T ′′ T ′′ = idY ′′ . Similarly, one shows that T ′′ T ′′ = idX′′ and henceT ′′ ∈ L(X ′′, Y ′′) is invertible.

Now we assume that X is reflexive, i.e. JX is surjective, and we claim thatJY = T ′′ JX T−1, which implies that also JY is surjective and Y is reflexive.In order to see this we let y ∈ Y , ψ ∈ Y ′ and we calculate

(T ′′JXT−1y)(ψ) =(JXT−1y)(ψ T )=(ψ T )(T−1y)=ψ(y) = (JY y)(ψ).

Theorem 8.5. Let X be a reflexive Banach space. Then every bounded sequencein X has a weakly converging subsequence in X, i.e. the set x ∈ X : ‖x‖ ≤ R isweakly sequentially compact for all R <∞.

Proof. Let (xk) be a sequence in X with ‖xk‖ ≤ c for all k ∈ N and define the spaceY := span xk : k ∈ N. By definition Y is separable and closed. Hence it followsfrom Lemma 8.4 that Y is reflexive and moreover Y ′′ = JY Y is separable, which,by Theorem 4.6, implies that Y ′ is separable. Now (JY xk)k∈N ⊂ Y ′′ = (Y ′)′ with‖JY xk‖ ≤ c and by Theorem 8.3 there exists Λ ∈ Y ′′, so that for all ϕ ∈ Y ′

ϕ(xk) = (JY xk)(ϕ)→ Λ(ϕ)

up to a subsequence. Define x := J−1Y (Λ) ∈ Y and observe that for all ϕ ∈ Y ′

ϕ(xk)→ Λ(ϕ) = (JY x)(ϕ) = ϕ(x).

It remains to show that this convergence property holds for all ϕ ∈ X ′:

ϕ(xk) = (ϕ|Y )(xk)→ (ϕ|Y )(x) = ϕ(x).

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8 Weak convergence

Lemma 8.6. The space Lp(µ) is reflexive for 1 < p <∞.

Proof. Let φ ∈ Lp(µ)′′ and thus φ : Lp(µ)′ → R is linear and bounded. Let

Jq : Lq(µ)→ Lp(µ)′

be the duality map of Theorem 6.4 and hence φJq ∈ Lq(µ)′. By Theorem 6.4 thereexists f ∈ Lp(µ) such that for all g ∈ Lq(µ)

(φ Jq)(g) =∫Xfgdµ.

Let JLp : Lp(µ) → Lp(µ)′′ be the canonical embedding and note that for every g ∈Lq(µ)

(JLpf)(Jqg) = (Jqg)(f) =∫Xfgdµ = (φ Jq)(g) = φ(Jqg).

As Jq is surjective, Theorem 6.4 gives us JLpf = φ.

Application: Assume that ‖uk‖Lp ≤ C for all k ∈ N.

Case 1: If 1 < p <∞ then Lp(µ) reflexive and Theorem 8.5 implies that there existsa subsequence uk u in Lp(µ) ⇔

∫X ukvdµ→

∫X uvdµ for all v ∈ Lq(µ) with

p−1 + q−1 = 1.

Case 2: If p =∞ then L∞ = (L1)′, L1 is separable and it follows from Theorem 8.3that there exists a subsequence uk

∗ u in L∞(µ) ⇔

∫ukvdµ→

∫uvdµ for all

v ∈ L1(µ).

Theorem 8.7. Let X be a Banach space and let K ⊂ X be convex and closed. Thenfor every sequence xk ∈ K with xk x ∈ X we have x ∈ K, i.e. K is weakly closed.

Proof. Without loss of generality we can assume that x = 0. If 0 /∈ K, thendist(0,K) =: ρ > 0 and by Lemma 4.12 there exists ϕ ∈ X ′ with ‖ϕ‖ = 1 andϕ(y) ≤ −ρ for all y ∈ K and we get the contradiction

0 = ϕ(0) = limk→∞

ϕ(xk) ≤ −ρ < 0.

Lemma 8.8. If X is a reflexive Banach space and K ⊂ X is closed and convex,

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8 Weak convergence

then for every x0 ∈ X there exists x ∈ K with

‖x− x0‖ = dist(x0,K).

Proof. Without loss of generality we can assume that x0 = 0. By choosing a mini-mizing sequence xk ∈ K, i.e. ‖xk‖ → dist(0,K), we can use Theorem 8.5 in orderto conclude that xk x up to a subsequence and by Theorem 8.7 it follows thatx ∈ K. Finally, by Theorem 8.2 we get

dist(0,K) ≤ ‖x‖ ≤ lim inf ‖xk‖ = dist(0,K).

Definition 8.9. A normed space (X, ‖ · ‖) is called uniformly convex if for allε > 0 there exists δ > 0 so that the following implication is true(

‖x‖ = 1 = ‖y‖,∥∥∥∥x+ y

2

∥∥∥∥ ≥ 1− δ)⇒ ‖x− y‖ < ε,∀x, y ∈ X.

We have seen in Lemma 6.5 that all Lp-spaces with 1 < p < ∞ are uniformlyconvex. Moreover, all Hilbert spaces H are uniformly convex, since for all x, y ∈ Hwith ‖x‖2 = 1 = ‖y‖2 and

∥∥∥x+y2

∥∥∥ ≥ 1− δ it follows that

4 = ‖x+ y‖2 + ‖x− y‖2 ≥ 4(1− δ)2 + ‖x− y‖2

and hence ‖x− y‖ ≤ 2√

2δ =: ε/2.

Lemma 8.10. Let (X, ‖ · ‖) be uniformly convex and let xn, yn be sequences in X

with lim sup ‖xn‖ ≤ 1, lim sup ‖yn‖ ≤ 1 and

lim∥∥∥∥xn + yn

2

∥∥∥∥ = 1.

Then ‖xn − yn‖ → 0.

Proof. We note that ρn := ‖xn‖ → 1, σn := ‖yn‖ → 1 and we define the newsequences ξn := xn

ρn, ηn := yn

σn. It follows that

12‖ξn + ηn‖ =1

2

∥∥∥∥xnρn + ynρn− ynρn

+ ynσn

∥∥∥∥≥1

2

( 1ρn‖xn + yn‖ −

∣∣∣∣ 1σn− 1ρn

∣∣∣∣ ‖yn‖)→ 1.

Hence we can use the uniform convexity of X to get ‖ξn − ηn‖ → 0 which shows

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8 Weak convergence

that

‖xn − yn‖ =‖ρnξn − ρnηn + ρnηn − σnηn‖=|ρn|‖ξn − ηn‖+ ‖ηn‖|ρn − σn‖ → 0.

Theorem 8.11. Let (X, ‖ · ‖) be uniformly convex. Then the following two state-ments are equivalent.

(i) xn → x with respect to the norm ‖ · ‖.

(ii) xn x and lim supn→∞ ‖xn‖ ≤ ‖x‖.

Proof. It is obvious that statement (i) implies statement (ii).

In order to show that (ii) implies (i) we assume without loss of generality that x 6= 0.Then we get that ρn := max‖xn‖, ‖x‖ > 0 and ρn → ‖x‖ by Theorem 8.2 and theassumption. Now

ξn := xnρn

x

‖x‖=: ξ

since ϕ(xn/ρn) = 1ρnϕ(xn) → 1

‖x‖ϕ(x) = ϕ(ξ) for all ϕ ∈ X ′. Theorem 8.2 thusyields

1 = ‖ξ‖ = ‖ξn‖ ≤ lim inf∥∥∥∥1

2(ξ + ξn)∥∥∥∥

and by Lemma 8.10 we conclude ‖ξn − ξ‖ → 0 or equivalently

‖xn − x‖ → 0.

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9 Hilbert spaces

Recall that (X, 〈·, ·〉) is a Hilbert space if the norm ‖x‖ =√〈x, x〉 is complete.

Theorem 9.1 (Projection Theorem). Let X be a Hilbert space and let K ⊂ X beclosed and convex. Then for every x ∈ X there exists a unique x0 ∈ K with

1. ‖x− x0‖ = dist(x,K) = inf‖x− z‖ : z ∈ K.

2. Moreover, Re〈x−x0, z−x0〉 ≤ 0 for all z ∈ K and x0 is the unique point withthis property.

We write x0 = PK(x) with PK : X → K and we call PK the nearest pointprojection.

Proof. In order to show statement 1 we choose a minimising sequence xk ∈ K andhence

‖x− xk‖ → d := dist(x,K).

Next we use the parallelogram identity

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2)

in order to get

‖xk − xl‖2 = 2(‖xk − x‖2 + ‖xl − x‖2)− 4∥∥∥∥xk + xl

2 − x∥∥∥∥2→ 0

as k, l→∞, where we used that xk+xl2 and therefore

∥∥∥xk+xl2 − x

∥∥∥2≥ d2. Hence (xk)

is a Cauchy sequence and thus xk → x0 with x0 ∈ K and ‖x − x0‖ = d since K isclosed.

Now let x0, x1 ∈ K be two such points as in statement 1. It follows again from theparallelogram identity that

‖x0 − x1‖2 = 2(‖x0 − x‖2 + ‖x1 − x‖2)− 4∥∥∥∥x0 − x1

2 − x∥∥∥∥2≤ 2(d2 + d2)− 4d2 = 0

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9 Hilbert spaces

and thus x0 = x1.

In order to show statement 2 we let z ∈ K and 0 ≤ ε ≤ 1. It follows from statement1 that

d2 ≤‖x− ((1− ε)x0 + εz)‖2 = ‖x− x0 − ε(z − x0)‖2

=‖x− x0‖2 + ε2‖z − x0‖2 − 2εRe〈x− x0, z − x0〉=d2 + ε2‖z − x0‖2 − 2εRe〈x− x0, z − x0〉

HenceRe〈x− x0, z − x0〉 ≤

ε

2‖z − x0‖2

and for ε 0 we getRe〈x− x0, z − x0〉 ≤ 0.

Finally, we let x0 ∈ K be another point so that Re〈x− x0, y− x0〉 ≤ 0 for all y ∈ K.Inserting y = x0 and z = x0 in the above inequality for x0 get

Re〈x− x0, x0 − x0〉 ≤ 0 and Re〈x− x0, x0 − x0〉 ≤ 0

Adding these two inequalities yields

‖x0 − x0‖2 = Re‖x0 − x0‖2 = Re〈x0 − x0, x0 − x0〉 ≤ 0

and thereforex0 = x0.

Lemma 9.2. Let X be a Hilbert space and let K ⊂ X be closed and convex. Then thenearest point projection PK : X → K is Lipschitz continuous with constant L = 1.

Proof. Let x, y ∈ X with x 6= y, and recall from Theorem 9.1 that for all z1 ∈ Kresp. z2 ∈ K

Re 〈x− PK(x), z1 − PK(x)〉 ≤ 0 Re 〈y − PK(y), z2 − PK(Y )〉 ≤ 0.

Now we insert z2 := PK(x) and z1 := PK(y) and we add the two inequalities inorder to get

Re〈y − PK(y) + PK(x)− x, PK(x)− PK(y)〉 ≤ 0.

Therefore

‖PK(x)− PK(y)‖2 ≤ Re〈x− y, PK(x)− PK(y)〉 ≤ ‖x− y‖‖PK(x)− PK(y)‖,

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9 Hilbert spaces

which implies‖PK(x)− PK(y)‖ ≤ ‖x− y‖.

Definition 9.3. Let X be a Hilbert space and let M ⊂ X. Then the closed subspace

M⊥ := x ∈ X : 〈x, y〉 = 0 ∀y ∈M

is the orthogonal complement of M .

Lemma 9.4. Let X be a Hilbert space and let Y ⊂ X be a closed subspace. ThenPY : X → Y is linear and ker(Py) = Y ⊥. Moreover, we have that X = Y ⊕ Y ⊥.

Proof. Recall from Theorem 9.1 that PY (x) = x0 if and only if Re〈x−x0, z−x0〉 ≤ 0for all z ∈ Y . Now we replace z ∈ Y by x0 + λy ∈ Y for all y ∈ Y and λ ∈ K Thisyields

Re〈x− x0, λy〉 ≤ 0

for all λ ∈ K, y ∈ Y . Choosing λ = 〈x − x0, y〉 then gives that PY (x) = x0 if andonly if

〈x− x0, y〉 = 0

for all y ∈ Y .

Next, we let PY (x) = x0, PY (z) = z0, λ, µ ∈ K and we note that λx0 + µz0 ∈ Y .Moreover,

〈(λx+ µz)− (λx0 + µz0), y〉 = 0

for all y ∈ Y and hence

PY (λx+ µz) = λx0 + µz0 = λPY (x) + µPY (z)

which implies that PY is linear and continuous with norm 1 (see Lemma 9.2).

We also not that PY (x) = 0 if and only if 〈x, y〉 = 0 for all y ∈ Y and thus if andonly if x ∈ Y ⊥.

Finally, we let x ∈ X be arbitrary and we note that

PY (x− PY (x)) =PY (x)− P 2Y (x) = 0

since PY |Y = id and therefore x = PY (x)+(x−PY (x)) with PY (x) ∈ Y , (x−PY (x)) ∈Y ⊥ and hence X = Y ⊕ Y ⊥.

Theorem 9.5 (Riesz representation for Hilbert spaces). Let X be a Hilbert space.

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9 Hilbert spaces

Then the map

R : X → X ′, (Ry)(x) = 〈x, y〉

is a surjective, conjugate linear isometry.

Proof. 1. R is surjective.

Let ϕ ∈ X ′ with ϕ 6= 0. Then kerϕ = ϕ−1(0) is a closed subspace and Lemma9.4 implies that there exists v ∈ (ker(ϕ))⊥ with ‖v‖ = 1 and ϕ(v) 6= 0. Forx ∈ X we have

x = x− ϕ(x)ϕ(v)v + ϕ(x)

ϕ(v)v ∈ (kerϕ)⊕ (kerϕ)⊥

and when we take the scalar product with v we obtain

〈x, v〉 = ϕ(x)ϕ(v)

and thusϕ(x) = 〈x, ϕ(v)v〉,

i.e. ϕ = R(ϕ(v)v).

2. R(y) ∈ X ′.

For this we not that for all α, β ∈ K and x1, x2, y ∈ X we have

〈αx1 + βx2, y〉 = α〈x1, y〉+ β〈x2, y〉

and therefore

(Ry)(αx1 + βx2) = α(Ry)(x1) + β(Ry)(x2).

Moreover, we estimate with the help of the Cauchy-Schwarz inequality

|(Ry)(x)| = |〈x, y〉| ≤ ‖x‖‖y‖

and hence ‖Ry‖ ≤ ‖y‖.

3. R is conjugate linear.

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9 Hilbert spaces

We again let α, β ∈ K and x, y1, y2 ∈ X. Then

(R(αy1 + βy2))(x) =〈x, αy1 + βy2〉=α〈x, y1〉+ β〈x, y2〉=α(Ry1)(x) + β(Ry2)(x).

4. R is an isometry.

For y ∈ X we choose x = y/‖y‖ and conclude

(Ry)(x) =⟨y

‖y‖, y

⟩= ‖y‖.

Together with step 2 of the proof this shows that

‖Ry‖ = ‖y‖.

Theorem 9.6. Hilbert spaces are reflexive.

Proof. Let X be a Hilbert space and let x′′ ∈ X ′′. Define ϕ : X → K by

ϕ(y) = x′′(Ry),

where R is as in Theorem 9.5. Then ϕ is linear and

|ϕ(y)| ≤‖x′′‖‖Ry‖ ≤ ‖x′′‖‖y‖

and therefore ‖ϕ‖ ≤ ‖x′′‖ which shows that ϕ ∈ X ′. Next we define x := R−1ϕ ∈ Xand it follows that

x′′(Ry) =ϕ(y) = 〈y, x〉 = 〈x, y〉 = (Ry)(x) = (Jx)(Ry).

Thus x′′ = Jx and the canonical embedding J is surjective.

Definition 9.7. Let X,Y be Banach spaces and let T ∈ L(X,Y ). The (Banach)adjoint is defined by

T ′ : Y ′ → X ′, T ′ψ := ψ T.

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9 Hilbert spaces

We have that ‖T ′‖ ≤ ‖T‖ since for all ψ ∈ Y ′ with ‖ψ‖ ≤ 1 we estimate

‖T ′ψ‖ = sup‖x‖≤1

|T ′ψ(x)| = sup‖x‖≤1

|ψ(Tx)|

≤‖ψ‖ sup‖x‖≤1

|Tx| ≤ ‖T‖.

Theorem 9.8. Let X,Y be Hilbert spaces over K. For T ∈ L(X,Y ) there exists aunique map T ∗ ∈ L(Y,X) with

〈Tx, y〉 = 〈x, T ∗y〉 ∀x ∈ X, y ∈ Y.

for all x ∈ X and y ∈ Y .

Proof. 1. Uniqueness:

Assume that there exist two maps S1, S2 ∈ L(Y,X) with

〈Tx, y〉 = 〈x, S1y〉 and 〈Tx, y〉 = 〈x, S2y〉

for all x ∈ X and y ∈ Y . Subtracting these equations yields

0 = 〈x, (S1 − S2)y〉

for all x ∈ X and y ∈ Y and hence

S1y = S2y

for all y ∈ Y , which shows that S1 ≡ S2.

2. Existence:

We look at the diagram

XT //

RX

Y

RY

X ′T ′// Y ′

where RX : X → X ′ and RY : Y → Y ′ are the maps from Theorem 9.5 and weclaim that the map T ∗ = R−1

X T ′RY ∈ L(Y,X) satisfies the desired property.

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9 Hilbert spaces

For this we calculate for all x ∈ X and y ∈ Y

〈x, T ∗y〉 =〈x,R−1X T ′RY y〉

=(T ′RY y)(x)=(RY y)(Tx)=〈Tx, y〉.

It follows from Theorem 9.5 that and the remark after Definition 9.7 that ‖T ∗‖ ≤‖T‖. Indeed, we have equality in this estimate (see the exercises).

Lemma 9.9. Let B : X ×X → K be a bounded sesquilinear form on a Hilbert spaceX, i.e. for all x, y, z ∈ X and α, β ∈ K we have

‖B‖ := sup‖x‖,‖y‖≤1

|B(x, y)| <∞,

B(αx+ βy, z) = αB(x, z) + βB(y, z),B(z, αx+ βy) = αB(z, x) + βB(z, y).

Then there exists a unique map T = TB ∈ L(X,X) with ‖T‖ = ‖B‖ and so that forall x, y ∈ X

B(x, y) = 〈Tx, y〉 ∀x, y ∈ X.

Proof. The uniqueness statement follows in the same way as we have shown unique-ness in the proof of Theorem 9.8.

In order to show the existence of T we let x ∈ X be fixed and we define

ϕx(y) := B(x, y).

It follows from the properties of B that ϕx is linear, ‖ϕx‖ ≤ ‖B‖‖x‖ <∞ and henceϕx ∈ X ′. By Theorem 9.5 there exists a map T : X → X, T := R−1(ϕx) with

B(x, y) = 〈y, Tx〉

or equivalentlyB(x, y) = 〈Tx, y〉.

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9 Hilbert spaces

For all α, β ∈ K and x, y, z ∈ X we have

〈T (αx+ βy), z〉 =B(αx+ βy, z)=αB(x, z) + βB(y, z)=α〈Tx, z〉+ β〈Ty, z〉=〈αTx+ βTy, z〉

which shows that T is linear. Choosing y = Tx for all x ∈ X with ‖x‖ ≤ 1, we get

‖Tx‖2 = 〈Tx, Tx〉 = B(x, Tx) ≤ ‖B‖‖Tx‖

and therefore‖T‖ ≤ ‖B‖.

The estimate ‖B‖ ≤ ‖T‖ follows from

|B(x, y)| = |〈Tx, y〉| ≤ ‖Tx‖‖y‖ ≤ ‖T‖‖x‖‖y‖.

Theorem 9.10 (Lax-Milgram). Let B : X ×X → R be a bounded bilinear form onthe R-Hilbert space X and let B be coercive, i.e.

B(x, x) ≥ λ‖x‖2

for all x ∈ X and for some λ > 0. Then the map RB : X → X ′, (RBy)(x) = B(x, y)is invertible with

‖R−1B ‖ ≤

1λ.

Additionally, if B is symmetric, then R−1B ϕ = y0 is the unique minimum of the

quadratic functional

Q : X → R, Q(y) = 12B(y, y)− ϕ(y).

Proof. It follows from Theorem 9.8 and Lemma 9.9 that there exists a map S ∈L(X,X) so that B(x, y) = 〈x, Sy〉 for all x, y ∈ X.

1. S is injective and im(S) is closed.

For x 6= 0 it follows from the coercivity of B that

0 < λ‖x‖2 ≤ B(x, x) = 〈x, Sx〉 ≤ ‖x‖‖Sx‖

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9 Hilbert spaces

which implies

‖Sx‖ ≥ λ‖x‖ (9.1)

and hence S is injective.

Next, we let yk = Sxk ∈ im(S) be a Cauchy sequence. It follows from (9.1)that ‖xk − xl‖ ≤ λ−1‖yk − yl‖ → 0 as k, l→∞ and hence xk is also a Cauchysequence. Therefore xk → x and Sxk → Sx which shows that im(S) is closed.

2. S is surjective.

Let y ∈ (im(S))⊥. Then we conclude

λ‖y‖2 ≤ B(y, y) = 〈y, Sy〉 = 0.

Hence (im(S))⊥ = 0 and it follows from Lemma 9.4 that X = im(S) ⊕(im(S))⊥ = im(S).

3. RB is invertible and ‖R−1B ‖ ≤ λ−1.

For this we show that RB = RS, where R : X → X ′ is the map from Theorem8.5, and then RB is the composition of two invertible maps. For every x, y ∈ Xwe have

(RSy)(x) = 〈x, Sy〉 = B(x, y) = (RBy)(x).

Moreover, we have that

‖R−1B ‖ =‖S−1R−1‖ = ‖S−1‖ ≤ 1

λ

where we have used (9.1) and the fact that R is an isometry. Note that as Sis surjective, for every y ∈ X there exists x ∈ X so that Sx = y and hence oneapplies (9.1) with x = S−1y in order to get ‖S−1y‖ ≤ 1

λ‖y‖.

4. The minimising property.

Now we assume that B is symmetric and we let y0 = R−1B ϕ for ϕ ∈ X ′.

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9 Hilbert spaces

Moreover, we let y = y0 + η ∈ X. Then we calculate

Q(y) =12B(y0 + η, y0 + η)− ϕ(y0 + η)

=12B(y0, y0) +B(y0, η) + 1

2B(η, η)− ϕ(y0)− ϕ(η)

=Q(y0) + 12B(η, η) +B(η, y0)− ϕ(η)

≥Q(y0) + λ

2 ‖η‖2,

where we used that ϕ(η) = (RBy0)(η) = B(η, y0). Hence we get that y0 is theunique minimiser of the functional Q.

Definition 9.11. A Hilbert space X is called Hilbert sum of closed subspaces Ej,j ∈ J if and only if

1. Ei ⊥ Ej for i 6= j.

2. ⊕j∈J Ej is dense in x. Here we only look at finite linear combinations of thespaces Ej, j ∈ J .

Theorem 9.12. Let Ej , j ∈ N0 be a system of closed, pairwise orthogonal sub-spaces of the Hilbert space X and let Pj be the corresponding nearest point projectionsonto Ej. Then the following four statements are equivalent.

1. Ej is maximal: If x ⊥ Ej for all j then x = 0.

2. X is the Hilbert sum of the sets Ej.

3. x = ∑∞j=0 Pjx for all x ∈ X (Fourier expansion).

4. ‖x‖2 = ∑∞j=0 ‖Pjx‖2 (Parseval identity).

Proof. 1. ⇒ 2.: Let V := ⊕j∈N0 Ej . By the assumption we have V ⊥ = 0 and

hence it follows from Lemma 9.4 that X = V ⊕ V ⊥ = V .

2. ⇒ 3. Let y ∈⊕Nj=0Ej for some N ∈ N . Then

‖x− y‖2 =

∥∥∥∥∥∥x−N∑j=0

Pjx+N∑j=0

Pjx− y

∥∥∥∥∥∥2

=

∥∥∥∥∥∥x−N∑j=0

Pjx

∥∥∥∥∥∥2

+

∥∥∥∥∥∥N∑j=0

Pjx− y

∥∥∥∥∥∥2

,

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9 Hilbert spaces

where we used that x−∑Nj=0 Pjx ∈

(⊕Nj=0Ej

)⊥and∑N

j=0 Pjx−y ∈ ⊕Nj=0Ej . Hence∥∥∥∥∥∥x−N∑j=0

Pjx

∥∥∥∥∥∥ = dist

x, N⊕j=0

Ej

→ 0

as N →∞ because of the assumption.

3. ⇒ 4. We calculate

‖x‖2 =〈x, x〉 = limN→∞

⟨N∑j=0

Pjx,N∑j=0

Pjx

⟩

= limN→∞

N∑i,j=0〈Pix, Pjx〉

= limN→∞

N∑i=0‖Pix‖2.

4. ⇒ 1. If x ⊥ Ej for all j ∈ N0 then Lemma 9.4 implies that Pjx = 0 for all j ∈ N0

and hence

‖x‖2 =∞∑j=0‖Pjx‖2 = 0.

85

10 Sobolev spaces and elliptic boundaryvalue problems

Definition 10.1. Let Ω ⊂ Rn be open and let u ∈ L1loc(Ω). A function g ∈ L1

loc(Ω)is called weak derivative of u with respect to xi, i = 1, ..., n, if∫

Ωu∂iη dx = −

∫Ωgη dx,

for all η ∈ C∞c (Ω). We use the notation ∂iu = g weakly.

Remarks:

1. The weak derivative is unique: If g1, g2 ∈ L1loc(Ω) are two weak derivatives,

then g := (g1 − g2) satisfies ∫Ωgηdx = 0

for all η ∈ C∞c (Ω). By the Fundamental Lemma of the Calculus of Variationsit then follows that g = 0 almost everywhere.

2. If ∂iu = g, ∂iv = g and α, β ∈ R, then ∂1(αu+ βv) = αg + βv.

3. For α = (α1, ..., αn) ∈ Nn0 the α-weak derivative Dαu = g is defined by∫ΩuDαη dx = (−1)|α|

∫Ωgη dx

for all η ∈ C∞c (Ω).

4. If Dαu = v and Dβv = g weakly, where α, β ∈ Nn0 then Dα+βu = g weaklysince ∫

ΩuDα+βη dx =

∫ΩuDα(Dβ) dx

=(−1)|α|∫

ΩvDβη dx = (−1)|α|+|β|

∫Ωgη dx

for all η ∈ C∞c (Ω).

86

10 Sobolev spaces and elliptic boundary value problems

Example: Let u(x) = |x|α. For which α ∈ R exists a weak derivative in L1loc(Rn)?

For x 6= 0 the function u is clasically differentiable and ∂iu(x) = α|x|α−2xi. Hencethe only candidate for the weak derivative is gi(x) = α|x|α−2xi. Now gi ∈ L1

loc(Rn)if and only if α− 1 > −n ⇔ α > 1− n. In order to show that gi is indeed the weakderivative for α > 1− n we let η ∈ C∞c (Ω) be arbitrary and we calculate∫u(x)∂iη(x) dx = lim

ρ→0

∫Rn \Bρ(0)

u(x)∂iη(x) dx

= limρ→0

(∫Rn \Bρ(0)

∂i(u(x)η(x)) dx−∫Rn \Bρ(0)

gi(x)η(x) dx)

= limρ→0

(−∫∂Bρ(0)

u(x)η(x)xiρdµ∂Bρ(0)(x)−

∫Rn \Bρ(0)

gi(x)η(x) dx)

=−∫Rngiη dx

where we used that∫∂Bρ(0)

|u(x)η(x)|xiρ| dµ∂Bρ(0)(x) ≤ C%n−1+α → 0

as %→ 0.

Definition 10.2. Let Ω ⊂ Rn be open and let 1 ≤ p ≤ ∞. Then we define theSobolev space

W 1,p(Ω) := u ∈ Lp(Ω): ∂iu ∈ Lp(Ω), for all 1 ≤ i ≤ n.

Moreover, we define the W 1,p-norm by

‖u‖W 1,p := ‖u‖Lp +n∑i=1‖∂iu‖Lp .

Similarly, for k ∈ N and 1 ≤ p ≤ ∞, we let u ∈ W k,p(Ω) ⇔ Dαu ∈ Lp(Ω) for all|α| ≤ k.

Theorem 10.3. The normed space (W 1,p(Ω), ‖ · ‖W 1,p) is a Banach space.

Proof. Let (uk) be a Cauchy sequence in W 1,p(Ω). Then it follows that both (uk)and (∂iuk), for all 1 ≤ i ≤ n, are Cauchy sequences in Lp(Ω). Since Lp(Ω) iscomplete (Fischer-Riesz) we conclude that that there exist functions u, vi ∈ Lp(Ω),1 ≤ i ≤ n with uk → u, ∂iuk → vi in Lp(Ω). It remains to show that vi = ∂iu

weakly. Let η ∈ C∞c (Ω). We calculate with the help of the dominated convergence

87

theorem ∫u∂iη dx = lim

k→∞

∫uk∂iη dx

=− limk→∞

∫(∂iuk)η dx

=−∫viηdx.

Thus uk → u in W 1,p(Ω).

Remark: It was shown in the exercises that

• W 1,p(Ω), 1 ≤ p <∞, is separable.

• W 1,p(Ω), 1 0 and define

uρ(x) := (ηρ ∗ u)(x) =∫Rnηρ(x− y)u(y)dy

for u ∈ L1loc(Rn). Then we have

1. uρ ∈ C∞(Rn) for all ρ > 0.

2. ‖uρ‖Lp ≤ ‖u‖Lp for all 1 ≤ p ≤ ∞.

3. If u ∈ Lp(Rn) with 1 ≤ p <∞ then ‖uρ − u‖Lp → 0 as ρ→ 0.

4. If u ∈ L∞(Rn) then uρ ∗ u in L∞(Rn) and there exists a subsequence ρi → 0so that uρi → u pointwise almost everywhere.

Example.

η(x) =

c exp(

1|x|2−1

), |x| < 1

0, |x| ≥ 1.

Proof. Statement 1. follows from differentiation results of parameter integrals.

88

For statement 2 we estimate for all 1 ≤ p ≤ ∞∫|uρ(x)|p dx =

∫ ∣∣∣∣∫ ηρ(x− y)u(y)dy∣∣∣∣p dx

Hölder≤

∫ (∫ηρ(x− y)|u(y)|p dy

)(∫ηρ(x− y) dy

)p−1dx.

Fubini=∫ (|u(y)|p

∫ηρ(x− y) dx

)dy = ‖u‖pLp .

In order to show statement 3. we define the function τz(x) = x+ z for z ∈ Rn. Wefirst show that for all 1 ≤ p <∞ we have

‖u τz − u‖Lp → 0

as z → 0. Since C0c (Rn) is dense in Lp(Rn) we choose v ∈ C0

c (Rn) so that ‖v−u‖Lp <ε3 . Then

‖u τz − u‖Lp ≤ ‖(u− v) τz‖Lp + ‖v τz − v‖Lp + ‖v − u‖Lp ≤ ε

for |z| small, since v τz → v uniformly as |z| → 0. In order to show the generalstatement we observe that∫

|uρ(x)− u(x)|p dx =∫ ∣∣∣∣∫ ηρ(x− y)(u(y)− u(x)) dy

∣∣∣∣p dxHölder≤

∫ ∫ηρ(x− y)|u(y)− u(x)|p dy dx

=∫ ∫

η(z)|u(x− ρz)− u(x)|p dz dx→ 0

as ρ→ 0.

For statement 4. we let u ∈ L∞(Rn) and this implies u ∈ Lploc(Rn) for all p ∈ [1,∞).Statement 3. then implies that there exists a subsequence uρi → u in Lploc(Rn) andhence also pointwise almost everywhere. Finally, we let v ∈ L1(Rn), η(x) := η(−x)and we conclude that∫

uρ(x)v(x) dy =∫ ∫

ηρ(x− y)u(y)v(x) dy dx

=∫u(y)

(∫ηρ(x− y)v(x) dx

)dy

=∫u(y)

(∫ηρ(y − x)v(x) dx

)dy

=∫u(y)ηρ ∗ v(y) dy

→∫u(y)v(y) dy

as ρ→ 0 by statement 3.

89

Lemma 10.5. Let Ω ⊂ Rn be open, let u ∈ Lploc(Ω) with p ∈ [1,∞) and let η beas in Theorem 10.4. Then the function uρ is well defined on the set Ωρ := x ∈Ω: dist(x, ∂Ω) > ρ and

‖uρ − u‖Lp(Ω′) → 0,

for all Ω′ ⊂⊂ Ω as ρ→ 0.

Proof. We note that

uρ(x) =∫Rnηρ(x− y)u(y)dy

=∫Bρ(0)

ηρ(y)u(x− y)dy

and hence we see that this function is well-defined for all x ∈ Ωρ. Without loss ofgenerality we let Ω′ = Ωσ for some σ > 0. Let

u(x) =u(x), if x ∈ Ωσ/20, otherwise

.

Then u ∈ Lp(Rn) and for all x ∈ Ωσ with ρ < σ2 we have

uρ(x) = (ηρ ∗ u)(x) =∫Bρ(0)

ηρ(y)u(x− y) dy

=∫B1(0)

η(z)u(x− ρz) dz =∫B1(0)

η(z)u(x− ρz) dz

=uρ(x).

Thus we conclude with the help of Theorem 10.4 that

‖uρ − u‖Lp(Ω′) ≤ ‖uρ − u‖Lp(Rn) → 0

as ρ→ 0.

Lemma 10.6 (Fundamental Lemma of the Calculus of Variations). Let Ω ⊂ Rn beopen and let u ∈ L1

loc(Ω) with∫

Ω uϕdx ≥ 0 for all ϕ ∈ C∞c (Ω) with ϕ ≥ 0. Thenu(x) ≥ 0 for almost every x ∈ Ω.

Proof. Let 0 < ρ < σ so that uρ is well defined on Ωσ, and let ϕ ∈ C∞c (Ωσ) with

90

ϕ ≥ 0. Then we have∫uρϕdx =

∫ ∫ηρ(x− y)u(y)ϕ(x) dy dx

=∫u(y)

(∫ηρ(x− y)ϕ(x) dx

)dy

=∫u(y)(ηρ ∗ ϕ)(y) dy ≥ 0

since ηρ ∗ ϕ ∈ C∞c (Ω,R+0 ). Since uρ is continuous a standard argument yields that

uρ ≥ 0 on Ωσ. Since uρi → u pointwise almost everywhere it follows that u ≥ 0almost everywhere.

Lemma 10.7. Let Ω ⊂ Rn be open, let u ∈ W 1,p(Ω) and let η be as in Theorem10.4. Then

1. ∂iuρ = (∂iu)ρ on Ωρ.

2. For 1 ≤ p <∞ and Ω′ ⊂⊂ Ω we have uρ → u in W 1,p(Ω′).

Proof. Statement 1. implies statement 2. by Lemma 10.5. In order to show state-ment 1. we calculate for all x ∈ Ωρ

∂iuρ = ∂i(ηρ ∗ u)(x) =∫ (

∂

∂xiηρ(x− y)

)u(y) dy

=−∫

∂

∂yiηρ(x− y)u(y) dy

=∫ηρ(x− y)∂iu(y) dy

=(ηρ ∗ ∂iu)(x) = (∂iu)ρ(x),

where we used the definition of the weak derivative in the third line.

Remark: If u ∈W 1,p(Rn) then uρ → u in W 1,p(Rn) by Theorem 10.4.

Lemma 10.8. Let Ω ⊂ Rn be open and let u, v ∈ W 1,p ∩ L∞(Ω). Then uv ∈W 1,p ∩ L∞(Ω) with

∂i(uv) = (∂iu)v + u(∂iv).

Proof. Let ϕ ∈ C∞c (Ω) and note that it follows from Lemma 10.5, Lemma 10.7 and

91

the dominated convergence theorem that∫Ωuv∂iϕ = lim

ρ→0

∫u(ηρ ∗ v)∂iϕ

= limρ→0

∫u∂i [(ηρ ∗ v)ϕ]− uϕ∂i(ηρ ∗ v)

Lemma 9.7= − limρ→0

∫[∂iu(ηρ ∗ v) + uϕηρ ∗ ∂iv]

=−∫

(u∂ivϕ+ ∂iuvϕ).

Remark: The same proof implies that if u ∈ W 1,p(Rn) and v ∈ C∞c (Rn) then uv ∈W 1,p(Rn) and ∂i(uv) = ∂iuv + u∂iv.

Theorem 10.9 (Meyers-Serrin). We have that

1. C∞c (Rn) is dense in W 1,p(Rn) for all 1 ≤ p <∞.

2. W 1,p ∩ C∞(Ω) is dense in W 1,p(Ω) for all 1 ≤ p <∞ and all Ω ⊂ Rn open.

Proof. 1. We know from Lemma 10.7 that W 1,p ∩ C∞(Rn) is dense in W 1,p(Rn).Hence we let u ∈W 1,p ∩ C∞(Rn) and we choose ϕ ∈ C∞c (Rn, [0, 1]) with

ϕ(x) =

1, |x| ≤ 10, |x| ≥ 2

Set ϕR(x) := ϕ(x/R) and uR(x) = ϕR(x)u(x). Then uR ∈ C∞c (Rn) and Lemma10.8 yields

∂iuR(x) = ϕR(x)∂iu(x) + 1R∂iϕ

(x

R

)u(x)

and hence

‖uR − u‖W 1,p =‖uR − u‖Lp +n∑i=1‖∂i(uR − u)‖Lp

≤

∫Rn \BR(0)

|u|p

1/p

+n∑i=1

∫Rn \BR(0)

|∂iu|p

1/p

+n∑i=1

1R‖∂iϕ

( ·R

)u‖Lp

→0

as R→∞.

2. Let Uk = Ω1/k ∩ Bk(0) for k ∈ N and let U0 := ∅. Moreover, we define Vk :=

92

Uk+1\Uk−1 for all k ∈ N and we choose a partition of unity

ϕk ∈ C∞c (Vk), ϕk ≥ 0,∞∑k=1

ϕk ≡ 1 on Ω.

For ε > 0 there exists δk > 0 so that for uk = ηδk ∗ (ϕku) we have

‖uk − ϕku‖W 1,p(Ω) ≤ 2−kε

by Lemma 10.7. For v := ∑∞k=1 uk ∈ C∞(Ω) ∩W 1,p(Ω) we conclude

‖v − u‖W 1,p ≤∞∑k=1‖uk − ϕku‖W 1,p < ε.

Remarks:

1. Let H1,p(Ω) be the completion of the space

X = u ∈ C∞(Ω): ‖u‖W 1,p(Ω) <∞

with respect to the W 1,p-norm. Every element u ∈ H1,p(Ω) is represented bya Cauchy sequence uk ∈ X and we obtain an isometric embedding H1,p(Ω)→W 1,p(Ω), u 7→ lim

k→0uk. By Theorem 10.9 this map is surjective and therefore

H1,p(Ω) ∼= W 1,p(Ω) isometrically.

2. For Ω ⊂ Rn open, bounded we have that C∞c (Ω) is not dense in W 1,p(Ω). Forthis we let u ∈ C∞c (Ω) and we note that

0 =∫

Ω∂i(uxi) =

∫Ω

(xi∂ixi + u.

If C∞c (Ω) would be dense in W 1,p(Ω), then this equality would be true for allu ∈W 1,p(Ω) but for u ≡ 1 we obtain the contradiction 0 = Ln(Ω).

3. In general we have that C∞(Ω) is not dense in W 1,p(Ω).

4. W 1,∞ ∩ C∞(Ω) is not dense in W 1,∞(Ω) since this would imply that ∂iu ∈C0(Ω) for all u ∈W 1,∞(Ω) as it is the uniform limit of a sequence of continuousfunctions. But |x| ∈W 1,∞(B1(0)) with

∂i|x| =xi|x|

/∈ C0(B1(0)).

Definition 10.10. For Ω ⊂ Rn open and 1 ≤ p < ∞ we denote by W 1,p0 (Ω) the

93

closure of C∞c (Ω) in W 1,p(Ω).

Theorem 10.11. Let Ω ⊂ Rn be open and let u ∈ W 1,1loc (Ω), f ∈ C1(R) with

‖f ′‖C0 =: L <∞. Then we have

D(f u) =(f ′ u)Du ∈ L1loc(Ω).

Proof. Let ϕ ∈ C∞c (Ωσ) for some σ > 0 and let 0 < ρ < σ. Then it follows that

∂i(f uρ) = (f ′ uρ)∂iuρ on Ωσ

and thus−∫f uρ∂iϕ =

∫(f ′ uρ)∂iuρϕ.

Next we let ρ 0 and we note that uρ → u, Duρ → Du in L1(Ωσ) by Lemma10.7. Choosing a subsequence implies that uρ → u also converges pointwise almosteverywhere in Ωσ. Hence we conclude that for ρ 0∣∣∣∣∫ (f uρ − f u)∂iϕ

∣∣∣∣ ≤L‖∂iϕ‖L∞ ∫Ωσ|uρ − u| → 0,∣∣∣∣∫ (f ′ uρ)(∂iuρ)ϕ− (f ′ u)(∂iu)ϕ

∣∣∣∣ ≤ ∫ |f ′ uρ||∂iuρ − ∂iu||ϕ|+∫|f ′ uρ − f ′ u||∂iu||ϕ| → 0.

Together, this shows that

−∫f u∂iϕ =

∫(f ′ u)∂iuϕ.

In the following we are interested in solving elliptic boundary value problems. Moreprecisely, for a ∈ L∞(Ω,Mn(R)), Ω ∈ Rn open and bounded we study the operatorL given by

Lv := −div(aDv) = −n∑

α,β=1∂α(aαβ∂βv).

The Dirichlet boundary value problem for L is given byLv = f in Ω,v = 0 on ∂Ω,

.

Definition 10.12. The bilinear form associated to L on W 1,20 (Ω) is given by

B(u, v) =∫

Ω〈Du, aDv〉 =

n∑α,β=1

∫aαβ∂αu∂βv.

94


It follows that B is bounded since

|B(u, v)| ≤ ‖a‖L∞‖Du‖L2‖Dv‖L2 ≤ ‖a‖L∞‖u‖W 1,2‖v‖W 1,2

and therefore ‖B‖ ≤ ‖a‖L∞ .

From now on we consider L as an operator

L : W 1,20 (Ω)→ (W 1,2

0 (Ω))′,

(Lv)(u) =n∑

α,β=1

∫aαβ∂αu∂βv = B(u, v).

It follows that L is continuous with ‖L‖ ≤ ‖a‖L∞ and we have that L = RB (compareTheorem 9.10). It follows from Theorem 9.10 that L is surjective if the bilinear formis coercive and this is what we want to verify in the following.

Theorem 10.13 (Poincare inequality). Let Ω ⊂ Rn be open and bounded. For1 ≤ p ≤ ∞ and all u ∈W 1,2

0 (Ω) we have

‖u‖Lp ≤ d‖Du‖Lp ,

where d = diam(Ω).

Proof. Without loss of generality we assume that u ∈ C∞c (Ω) and Ω ⊂ x : 0 ≤ xn ≤d. Set u(x) = 0 for all x /∈ Ω and let x = (ξ, xn) ∈ Rn−1×R = Rn, 0 ≤ xn ≤ d.Then we estimate

|u(ξ, xn)|p =∣∣∣∣∣∫ xn

0(∂nu)(ξ, s) ds

∣∣∣∣∣p

≤dp−1∫ d

0|∂nu(ξ, s)|p ds,

where we used that u(·, 0) = 0 and Hölders inequality. By Fubini’s theorem we nowget ∫

Ω|u(x)|p dx =

∫Rn−1

∫ d

0|u(ξ, xn)|p dxn dξ

≤dp−1∫Rn−1

∫ d

0

∫ d

0|∂nu(ξ, s)|p ds dxn dξ

=dp∫Rn−1

∫ d

0|∂nu(ξ, s)|pdsdξ

=dp‖Du‖pLp .

Lemma 10.14. Let Ω ⊂ Rn be open and bounded and let L be as above. Moreover,

95


we assume that L is elliptic with constant µ > 0, i.e.

〈ξ, a(x)ξ〉 ≥ µ|ξ|2

for all x ∈ Ω, ξ ∈ Rn. Then the associated bilinear form B is coercive on W 1,20 (Ω)

withB(u, u) ≥ λ‖u‖2W 1,2,

where λ = µd2+1 and d = diam(Ω).

Proof. It follows from the ellipticity assumption that

B(u, u) =∫〈aDu,Du〉 ≥ µ

∫|Du|2.

By Theorem 9.13 we get

‖u‖2W 1,2 =‖u‖2L2 + ‖Du‖2L2

≤(d2 + 1)‖Du‖2L2

and henceB(u, u) ≥ µ

d2 + 1‖u‖2W 1,2 .

Theorem 10.15. Let Ω ⊂ Rn be open and bounded, let a ∈ L∞(Ω,Mn(R)) beelliptic with constant µ > 0. Then

L : W 1,20 (Ω)⇒ (W 1,2

0 (Ω))′,

(Lv)(u) = B(u, v) =n∑

α,β=1

∫aαβ∂αu∂βv

is invertible and‖L−1‖ ≤ d2 + 1

µ.

Moreover, if a is symmetric on Ω and ϕ ∈ (W 1,20 (Ω))′, then the solution u ∈W 1,2

0 (Ω)of Lu = ϕ is the unique minimum of the quadratic functional

Q(v) = 12B(v, v)− ϕ(v) = 1

2

∫〈aDv,Dv〉 − ϕ(v).

Proof. This follows immediately from Theorem 9.10 and Lemma 10.14.

Special cases:

96


1. The Dirichlet problem with a right hand side in L2.

We let f ∈ L2(Ω) and we want to solve the problem−div(aDv) = f, in Ωv = 0, on ∂Ω

where a ∈ L∞(Ω,Mn(R)) is elliptic with constant µ > 0. This is the so calledclassical formulation. In order to get the weak formulation of this problem,we multiply the partial differential equation by u ∈ W 1,2

0 (Ω) and we formallyintegrate by parts in order to get ∫

Ω〈aDv,Du〉 =∫

Ω fu

v ∈W 1,20 (Ω)

Next we define the map L2(Ω) → (W 1,20 (Ω))′, f 7→ ϕf , ϕf (u) =

∫Ω fu for all

u ∈W 1,20 (Ω). We note that

|ϕf (u)| ≤ ‖f‖L2‖u‖L2 ≤ ‖f‖L2‖u‖W 1,2 , i.e. ‖ϕf‖ ≤ ‖f‖L2 .

In particular we can restate the weak formulation as follows:Lv = ϕf , in (W 1,2

0 (Ω))′

v ∈W 1,20 (Ω)

. (10.1)

The following Lemma is a direct consequence of Theorem 10.15.Lemma 10.16. Let Ω ⊂ Rn be open and bounded and let f ∈ L2(Ω). Thenthe problem (10.1) has a unique solution v ∈W 1,2

0 (Ω) with

‖v‖W 1,2 ≤d2 + 1µ‖f‖L2 .

2. The Dirichlet problem with a right hand side in divergence form.

We let a ∈ L∞(Ω,Mn(R)) be elliptic with constant µ > 0 and we let F ∈L2(Ω,Rn). We want to solve the problem

−div(aDv) = divF, in Ωv = 0, on ∂Ω

,

Here the weak formulation is given by ∫Ω〈aDv,Du〉 = −

∫Ω〈F,Du〉

v ∈W 1,20 (Ω)

.

97


for all u ∈ W 1,20 (Ω). We define the map L2(Ω) → (W 1,2

0 (Ω))′, F → ΨF withΨF (u) = −

∫〈F,Du〉 and we note that

|ΨF (u)| ≤ ‖F‖L2‖Du‖L2 ≤ ‖F‖L2‖u‖W 1,2

and hence‖ΨF ‖ ≤ ‖F‖L2 .

Therefore we can again restate the weak formulation asLv = ΨF , in (W 1,2

0 (Ω))′

v ∈W 1,20 (Ω).

(10.2)

and the following Lemma is again a direct consequence of Theorem 10.15.Lemma 10.17. Let Ω ⊂ Rn be open and bounded and let F ∈ L2(Ω,Rn).Then the problem (10.2) has a unique solution v ∈W 1,2

0 (Ω) with

‖v‖W 1,2 ≤d2 + 1µ‖F‖L2 .

3. The Dirichlet problem with a nonzero boundary condition.

We let a ∈ L∞(Ω,Mn(R)) be elliptic with constant µ > 0 and we let f ∈ L2(Ω),v0 ∈W 1,2(Ω). Now we want to solve the problem

−div(aDv) = f, in Ωv − v0 ∈W 1,2

0 (Ω).

For this we make the ansatz v = v0 + w with w ∈ W 1,20 (Ω) and we note that

w is then a solution of the problem−div(aDw) = f + div(aDv0), in Ωw ∈W 1,2

0 (Ω).

It the follows from Lemma 10.16 and Lemma 10.17 that there exists a uniquesolution w ∈W 1,2

0 (Ω) of this problem which satisfies the estimate

‖w‖W 1,2 ≤d2 + 1µ

(‖f‖L2 + ‖a‖L∞‖Dv0‖L2).

Hence we conclude that v = v0 + w is the unique solution of our originalproblem and it satisfies the estimate

‖v‖W 1,2 ≤ c(d, µ, a)(‖f‖L2 + ‖v0‖W 1,2).

98

11 Compact and Fredholm operators

One of our motivations for studying compact and Fredholm operators comes againfrom the theory of elliptic boundary value problems. Namely, we have seen inTheorem 10.15 that the operator

L0 : W 1,20 (Ω)→W 1,2

0 (Ω)′, (L0v)(u) =∫

Ω〈aDv,Du〉,

where a ∈ L∞(Ω,Mn(R)) is elliptic, is an isomorphism. But the second order ellipticoperator

L = L0 +K : W 1,20 (Ω)→W 1,2

0 (Ω)′,

withKv = −div(bv) + 〈c,Dv〉+ qv,

and b, c : Ω→ Rn, q : Ω→ R or in the weak formulation

(Kv)(u) =∫

(〈bv,Du〉+ 〈c,Dv〉+ quv)

is in general not an isomorphism.

Example: Let Lu = −u′′−u, hence L0(u) = u′′ andKu = −u on Ω = (−π/2, π/2).

1. The function v0(x) = cos(x) is in W 1,20 (Ω).

For every 0 < δ < 12 we let ηδ : Ω→ [0, 1] be a smooth function with sptηδ ⊂

(−π2 + δ, π2 − δ, ηδ(x) = 1 for all x ∈ (−π

2 + 2δ, π2 − 2δ) and ‖Dηδ‖L∞ ≤ Cδ−1.Then ηδv0 ∈ C∞c (Ω) and

‖ηδv0 − v0‖2L2 ≤ 2∫

Ω\(−π2 +2δ,π2−2δ)|v0|2 → 0

as δ → 0. Moreover

‖D(ηδv0 − v0)‖2L2 ≤ 2∫

Ω\(−π2 +2δ,π2−2δ)(|Dv0|2 + |Dηδ|2|v0|2)→ 0

as δ → 0 since |v0(x)| ≤ cδ for all x ∈ Ω\(−π2 + 2δ, π2 − 2δ).

2. L is not injective.

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For this we let v0(x) = cos(x) be as above and we note that for all u ∈ C∞c (Ω)

(Lv0)(u) =[uv′0]π/2−π/2 −∫ π/2

−π/2(v′′0u+ v0u) = 0.

Since C∞c (Ω) is dense in W 1,20 (Ω) this remains to be true for all u ∈ W 1,2

0 (Ω)and therefore Lv0 ≡ 0.

3. L is not surjective.

Let f ∈ W 1,20 (Ω)′ and let v ∈ W 1,2

0 (Ω) be a solution of Lv = f in the weaksense. Then we get∫ π/2

−π/2fv0 = ϕf (v0) = (Lv)(v0) = B(v0, v) = B(v, v0) = (Lv0)(v) = 0

where v0(x) = cos(x) as above. Therefore we must have that f ⊥L2 v0 and allfunctions which are not L2-orthogonal to v0 are not in the image of L.

We want to consider the operator K as a perturbation of the isomorphism L0 andthis is what naturally leads us to the notion of compact and Fredholm operators.

Definition 11.1. Let X,Y be Banach spaces. A map K ∈ L(X,Y ) is called com-pact, if for every bounded sequence xn ⊂ X the image sequence Kxn ⊂ Y hasa converging subsequence. We define

K(X,Y ) := K ∈ L(X,Y ) : K is compact.

Lemma 11.2. Let X and Y be Banach spaces. For K ∈ L(X,Y ) the followingstatements are equivalent

1. K ∈ K(X,Y ).

2. K(M) is relatively compact in Y for all M ⊂ X bounded (indeed M = B1(0)is sufficient).

3. If X is additionally reflexive then xn x implies Kxn → Kx.

Proof. 1. ⇒ 2. We show that K(M) is sequentially compact for allM ⊂ X boundedand by Theorem 3.2 this implies the claim. We let yn ∈ K(M) be a sequence andit follows that there exists a sequence xn ∈ M so that ‖yn − Kxn‖ < 1

n for alln ∈ N. As K is compact we conclude Kxn → y up to a subsequence and thereforey ∈ K(M) and yn → y.

2. ⇒ 1. Let xn ⊂ X be a bounded sequence, i.e. we assume that there exists

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R < ∞ with ‖xn‖ ≤ R for all n ∈ N. Hence Kxn ⊂ K(BR(0)) = RK(B1(0))which is compact by the assumption. Thus Kxn → y up to a subsequence.

1. ⇒ 3. Assume that xn is a sequence in X with xn x. It follows from Theorem8.2 that Kxn Kx in Y and ‖xn‖ ≤ c for some c < ∞. Since K ∈ K(X,Y )we conclude that there exists a subsequence Kxnj → y and it follows again fromTheorem 8.2 that y = Kx. We claim that this already implies that Kxn → Kx.

Indeed, this follows from the following general fact: If xn is a sequence in a metricspace X such that every subsequence has a further subsequence which converges tothe same limit y for all subsequences, then xn → y. Namely, if we assume that xndoes not converge to y, then there exists at least one subsequence xnj which doesnot converge to y and this contradicts the assumption.

3. ⇒ 1. Let xn ⊂ X be bounded. As X is reflexive, it follows from Theorem8.5 that xn x up to a subsequence and by the assumption we conclude Kxn →Kx.

Examples:

1. K ∈ L(X,Y ) is compact if dim Image(K) <∞, since by Theorem 1.2 all normson Image(K) are equivalent and we can then use the Bolzano-Weierstrasstheorem.

2. Let Ω ⊂ Rn be open and bounded and let 0 ≤ β < α ≤ 1 then the inclusionmap C0,α(Ω) ⊂ C0,β(Ω) is compact by Theorem 3.10.

3. id : X → X is compact if and only if dimX <∞ by Theorem 3.4 and Lemma11.2.

Lemma 11.3. Let X and Y be Banach spaces. Then K(X,Y ) is a closed subspaceof L(X,Y ).

Proof. Let T ∈ K(X,Y ) and let xn ⊂ X be bounded, i.e. ‖xn‖ ≤ c for all n ∈ Nand some c <∞. There exist maps Kj ∈ K(X,Y ) so that ‖T −Kj‖ < 1/j and bythe diagonal sequence argument we get that Kjxnn∈N converges for all j ∈ N upto a subsequence. Then we estimate

‖Txn − Txm‖ ≤‖Txn −Kjxn‖+ ‖Kjxn −Kjxm‖+ ‖Kjxm − Txm‖≤‖T −Kj‖(‖xn‖+ ‖xm‖) + ‖Kjxn −Kjxm‖

≤2cj

+ ‖Kjxn −Kjxm‖ < ε,

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if j,m, n are sufficiently large. Therefore the sequence Txn is a Cauchy sequenceand hence it converges Txn → y which shows that T ∈ K(X,Y ).

Theorem 11.4 (Schauder). Let X,Y, Z be Banach spaces and let S ∈ L(Y,Z),T ∈ K(X,Y ). Then ST ∈ K(X,Z). The same is true if S ∈ K(Y, Z) and T ∈L(X,Y ). Moreover, if K ∈ K(X,Y ), then the Banach adjoint K ′ is also compact,i.e. K ′ ∈ K(Y ′, X ′).

Proof. In order to show the first statement we let xn ⊂ X be bounded. ThenTxn → y ∈ Y up to a subsequence and hence S(Txn) → Sy up to a subsequence.The other case is treated similarly.

For the proof of the second statement we use Lemma 11.2 and we letM := K(B1(0)) ⊂Y be compact. Let ψn ∈ Y ′ be a sequence with ‖ψn‖ ≤ c for all n ∈ N. We have toshow that K ′ψn = ψn K ∈ X ′ has a converging subsequence.

Since M is compact we have that M ⊂ BR(0) for some R <∞ and hence

supy∈M|ψn(y)| ≤ ‖ψn‖R ≤ cR

which implies that ψn|M is uniformly bounded. Moreover, for all y1, y2 ∈ M wehave

|ψn(y1)− ψn(y2) ≤‖ψn‖‖y1 − y2‖ ≤ C‖y1 − y2‖

and thus ψn|M is uniformly Lipschitz continuous and therefore also equicontinu-ous. Theorem 3.8 (Arzela-Ascoli) then implies that ψn|M is a Cauchy sequence inC0(M), at least up to the choice of a subsequence. From this we conclude that forall x ∈ B1(0) and all n,m large enough |ψn(Kx)− ψm(Kx)| < ε and thus

‖ψn K − ψm K‖ ≤ ε

which shows that K ′ψn is a Cauchy sequence in X ′.

Definition 11.5. Let X and Y be Banach spaces. A map T ∈ K(X,Y ) is calleda Fredholm operator, and we use the notation T ∈ Fred(X,Y ), if and only ifImage(T ) ⊂ Y is closed, dim(kerT ) <∞ and dim coker(T ) <∞, where coker(T ) :=Y/ Image(T ).

Moreover, we define the index of T by ind(T ) := dim(kerT )− dim(cokerT ) ∈ Z.

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Remark: If X and Y are finite dimensional and T ∈ L(X,Y ) then

ind(T ) = dim(kerT )− dim(Y ) + dim(ImageT ) = dimX − dimY

is independent of T .

Examples:

1. If T is an isomorphism then ind(T ) = 0.

2. • The map lp → lp, (x1, x2, ...) 7→ (0, x1, x2, ...) has ind = −1.

• The map lp → lp, (x1, x2, ...) 7→ (x2, x3, ...) has ind = +1.

Theorem 11.6. Let X and Y be Banach spaces and assume that the map T ∈L(X,Y ) has a closed image. Then Image(T ′) is closed as well and we have

1. ker(T ′) ∼= (cokerT )′.

2. coker(T ′) ∼= (kerT )′.

Proof. Since the image of T is closed we can use Theorem 2.8 in order to concludethat cokerT = Y/ ImageT is a Banach space with norm

‖[y]‖ = infx∈X‖y + Tx‖

and we define the projection map

π : Y → Y/ ImageT, π(y) = [y].

In order to show claim 1 we prove that the adjoint map

π′ : (cokerT )′ → ker(T ′) ⊂ Y ′

is an isomorphism. We note that T ′ π′ = (π T )′ = 0 and hence the image of π′ isindeed contained in ker(T ′). Now we define the map

ρ : ker(T ′)→ (cokerT )′, (ρψ)[y] := ψ(y)

for all ψ ∈ ker(T ′) and all y ∈ Y . Note that the map ρψ is well-defined sinceψ(Tx) = T ′(ψx) = 0 as ψ ∈ ker(T ′). Moreover, we have for all x ∈ X that

|ρψ[y]| = |ψ(y + Tx)| ≤ ‖ψ‖‖y + Tx‖

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and by taking the infimum over all x ∈ X we get

|ρψ[y]| ≤ ‖ψ‖‖[y]‖

and thus ρψ ∈ (cokerT )′.

We calculate

(π′(ρψ))(y) =(ρψ)(πy) = ρψ([y]) = ψ(y)

and thusπ′ ρ = idker(T ′) .

On the other hand, for g ∈ (cokerT )′, we have

(ρπ′g)([y]) = (π′g)(y) = g(π(y)) = g([y]),

which shows thatρ π′ = id(cokerT )′

and π′ is an isomorphism.

In order to show claim 2. we consider the map

σ : X ′ → (kerT )′, σϕ = ϕ|kerT ,

for all ϕ ∈ X ′ and we note that σ is continuous with ‖σ‖ ≤ 1.

We claim thatImage(T ′) = ker(σ)

and we note that this implies that Image(T ′) is closed.

(i) First we observe that

σ(T ′ψ) = σ(ψ T ) = (ψ T )|ker(T ) = 0

for all ψ ∈ Y ′ and therefore Image(T ′) ⊂ ker(σ).

(ii) For the other inclusion we define the map T : X/ ker(T ) ∼→ Image(T ), T ([x]) =Tx and we note that T is well-defined, bijective and continuous. By Theorem5.10 there exists S ∈ L(Image(T ), X/ ker(T )) with S T = idX/ ker(T ). Let nowϕ ∈ ker(σ) ⊂ X ′ (⇔ ϕ|ker(T ) = 0) and define

ϕ ∈ (X/ kerT )′, ϕ([x]) = ϕ(x).

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Moreover, we let ψ = ϕ S ∈ (Image(T ))′ and we extend ψ to ψ ∈ Y ′ with thehelp of Theorem 4.2. It follows that

(T ′ψ)(x) = ψ(Tx) = ψ(Tx) =(ϕ S T )([x])=ϕ([x])=ϕ(x).

Hence T ′ψ = ϕ, i.e. ker(σ) ⊂ Image(T ′), and the claim is proved.

Finally we show that the map

σ : X ′/ Image(T ′) = coker(T ′)→ (kerT )′, σ([ϕ]) := ϕ|kerT = σ(ϕ)

is an isomorphism.

First we note that it follows from the claim that σ is well-defined, continuous andinjective. Moreover, σ is also surjective since every λ ∈ (kerT )′ can be extended toΛ ∈ X ′ with the help of Theorem 4.2 and then we calculate

σ([Λ]) = Λ|kerT = λ.

Hence it follows from Theorem 5.10 that σ is an isomorphism.

Lemma 11.7. Let X and Y be Banach spaces and let T ∈ Fred(X,Y ). ThenT ′ ∈ Fred(Y ′, X ′) and ind(T ′) = − ind(T ).

Proof. This result follows directly from Theorem 11.6.

Lemma 11.8. Let X and Y be Banach spaces and let T ∈ L(X,Y ) with Image(T )closed. Then the equation Tx = y is solvable if and only if ψ(y) = 0 for all ψ ∈ker(T ′).

Proof. We note that y ∈ ImageT is equivalent to π(y) = [y] = 0, where π : Y →Y/ ImageT = cokerT is the projection defined in the proof of Theorem 11.6. Itfollows from Lemma 4.4. that this is then equivalent to the fact that g([y]) = 0 forall g ∈ (cokerT )′ and this in turn is equivalent to the fact that (ρψ)([y]) = 0 forall ψ ∈ ker(T ′), where ρ : ker(T ′)

∼=→ (cokerT )′, (ρψ)[y] = ψ(y) is the isomorphismconstructed in the proof of Theorem 11.6.

Lemma 11.9. Let V be a subspace of the Banach space X. If either

1. dimV <∞, or

2. V is closed and dim(X/V ) ≤ ∞,

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then V has a closed complement, i.e. there exists W ⊂ X closed with X = V ⊕W .

Proof. We first assume that dimV < ∞. Then the space (V, ‖ · ‖X) is complete(see Theorem 1.2) and thus V is closed (see exercise sheet 1). Next we choose abasis v1, ..., vn of V and a dual basis ϕ1, ..., ϕn of V ′, i.e. ϕi(vj) = δij for all1 ≤ i, j ≤ n. We extend ϕi to ϕi ∈ X ′ with the help of Theorem 4.2 and we definethe map P : X → V by

Px :=n∑i=1

ϕi(x)vi.

It follows that‖Px‖ ≤ n( sup

i=1,...,n‖ϕi‖‖vi‖)‖x‖

and hence P ∈ L(X,V ). Moreover, we have

Pvj =n∑i=1

ϕi(vj)vi = vj

and thereforeP |V = idV and P 2 = P.

Since for every x ∈ X we can write

x = Px+ (x− Px) ∈ V ⊕ kerP

we have that kerP is a closed complement of V.

In the case dim(X/V ) < ∞ we choose a basis [x1], ..., [xn] of X/V and we notethat span x1, ..., xn is a closed complement of V .

Lemma 11.10. Let V be a closed subspace of the Banach space X and let W ⊂ Xa finite-dimensional subspace. Then V +W is a closed subspace of X.

Proof. Without loss of generality we assume that V ∩W = 0. Let xk = vk +wk ∈V +W and assume that xk → x ∈ X. Then we claim that ‖wk‖ ≤ c for all k ∈ N andsome c <∞. Namely, if ‖wk‖ =: Rk →∞, then xk/Rk → 0, wk/Rk → w ∈W with‖w‖ = 1 and therefore V 3 v ← vk/Rk = xk/Rk −wk/Rk → −w ∈W contradictingthe fact that V ∩W = 0. Since wk is bounded and xk converges, we concludethat vk is bounded and we can extract subsequences so that wk → w ∈ W andvk → v ∈ V , which implies x = v + w ∈ V +W .

Theorem 11.11. Let X and Y be Banach spaces. Then Fred(X,Y ) ⊂ L(X,Y ) isopen and the index is locally constant on Fred(X,Y ).

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Proof. Let T ∈ Fred(X,Y ). Then it follows from Lemma 11.9 that there exist closedsubspaces V ⊂ X and W ⊂ Y with dimW <∞ and

X =V ⊕ kerT andY =W ⊕ Image(T ).

Since V andW are closed they are also complete and hence the space (V×W, ‖(v, w)‖ :=‖v‖X + ‖w‖Y ) is a Banach space. Next we define the map

L(X,Y )→ L(V ×W,Y ), S 7→ S with S(v, w) = Sv + w

and we note that this map is continuous. We split the rest of the proof into foursteps.

• Step 1: T is invertible.

For this we note that T |V : V → ImageT is bijective and continuous and itfollows from Theorem 5.10 that T |V is invertible. We note that T can beexpressed as

T =(T |V 0

0 idW

)

where the first row corresponds to elements in V and the second row to ele-ments in W . Similarly the first line represents elements in Image(T ) and thesecond line in W . From this form it is easy to see that the inverse can berepresented as follows

T−1 =(

(T |V )−1 00 idW

).

• Step 2: If ‖S − T‖ < ‖T−1‖−1 then the map S is invertible.

We note that

(S − T )(v, w) = (Sv + w)− (Tv + w) = (S − T )v

and therefore‖S − T‖ ≤ ‖S − T‖ < ‖T−1‖−1.

The claim now follows from Theorem 5.2.

• Step 3: If S is invertible then S ∈ Fred(X,Y ).

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If S is invertible then we know that kerS ∩V = 0. Thus the projection mapπ : X → X/V maps kerS ⊂ X injectively onto kerS ⊂ X/V and we concludethat

dim kerS ≤ dimX/V = dim kerT <∞.

By Lemma 11.10 we get that V ⊕ kerS ⊂ X is a closed subspace with finitecodimension and Lemma 11.9 then implies that that there exists a closedsubset U ⊂ X with X = V ⊕ kerS ⊕ U and dimU <∞. Next, we know thatS(V ) = S(V × 0) is closed, since S−1(S(V × 0)) = V × 0 and S−1 iscontinuous by Theorem 5.10 and V × 0 is closed. Therefore

ImageS = S(X) = S(V )⊕ S(U)

is closed by Lemma 11.10. It remains to show that dim(cokerS) <∞.

Since S is invertible we know that Y = S(V )⊕W (note that S(V )∩W = 0since otherwise S would not be injective). We consider the projection map

p : Y/S(V )→ Y/S(X), p(y + S(V )) = y + S(X)

and we note that p is well-defined since S(V ) ⊂ S(X). Since p is surjective itfollows that

dim(coker(S)) = dim(Y/S(X)) ≤ dim(Y/S(V )) = dimW <∞.

Note that so far we shown that Fred(X,Y ) is open in L(X,Y ).

• Step 4: If S is invertible then indS = indT .

We recall thatX = V ⊕ kerT = V ⊕ kerS ⊕ U

and thereforedim(kerS) = dim(kerT )− dimU.

Additionally, we have that

dim(Y/S(V )) = dimW = dim(Y/T (X)) = dim(cokerT ).

Consider again the projection p : Y/S(V ) → Y/S(X) defined in step 3. Itfollows that ker(p) = S(X)/S(V ) ⊂ Y/S(V ) and (note that all vector spaces

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considered here are finite-dimensional)

dim(cokerS) = dim(Image(p)) = dim(Y/S(X))= dim(Y/S(V ))− dim(ker(p))= dim(cokerT )− dim(S(U))= dim(cokerT )− dimU

as ker(S) ∩ U = 0. Hence we conclude

indS = indT.

The following important result of Riesz combines the theory of compact and Fred-holm operators.

Theorem 11.12 (on compact operators by Riesz). Let X be a Banach space andlet K ∈ K(X,X). Then T := idX −K ∈ Fred(X,X) with ind(T ) = 0.

Proof. We show this result in five steps.

1. We have dim(ker(T )) <∞.

Let xj ∈ ker(T ) with ‖xj‖ ≤ 1. Since K ∈ K(X,X) it follows that

xj = Kxj → y ∈ X

up to a subsequence. As ker(T ) is closed w get y ∈ kerT with ‖y‖ ≤ 1 andhence the set x ∈ ker(T ) : ‖x‖ ≤ 1 is sequentially compact. By Theorem 3.2and Theorem 3.4 it follows that dim(ker(T )) <∞.

2. By Lemma 11.9 there exists a closed set V ⊂ X so that X = V ⊕ kerT . Weclaim that there exists a constant m > 0 so that ‖Tv‖ ≥ m‖v‖ for all v ∈ V .

We show this estimate by contradiction. Thus we assume that there exists asequence vj ∈ V with ‖vj‖ = 1 and so that

‖Tvj‖ ≤1j‖vj‖

for all j ∈ N. As the map K is compact it follows that Kvj → y ∈ X up tothe choice of a subsequence. Hence, as V is closed, vj = Tvj +Kvj → y ∈ V

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and ‖y‖ = 1. Moreover, we have that

Ty = limj→∞

Tvj = 0

and hence y ∈ kerT . But this contradicts the fact that V ∩ ker(T ) = 0.

3. We claim that Image(T ) is complete and therefore closed.

We let (yj) ∈ ImageT be a Cauchy sequence with yj = Txj . Without loss ofgenerality we assume that xj ∈ V . It follows from Step 2. that

‖xj − xk‖ ≤1m‖Txj − Txk‖ = 1

m‖yj − yk‖ < ε

for j, k large enough. Hence (xj) is a Cauchy sequence in X and xj → x ∈ Xwhich gives yj = Txj → Tx =: y.

4. We have dim cokerT <∞.

Consider the adjoint map T ′ = (idX −K)′ = idX′ −K ′. From Theorem 11.4we know that K ′ ∈ K(X ′, X ′) and thus we can apply Step 1. and Theorem11.6 in order to conclude that

dim(coker(T )) = dim(coker(T ))′ = dim(ker(T ′)) <∞.

5. We have indT = 0.

Let T (t) = idX −tK with 0 ≤ t ≤ 1 be a continuous curve in L(X,X). Bythe first four steps we get that T (t) ∈ Fred(X,X) for all t ∈ [0, 1] and henceind(T (t)) ∈ Z is independent of t by Theorem 11.11. Therefore we get

indT = ind(T (1)) = ind(T (0)) = ind(idX) = 0.

Lemma 11.13. Let X and Y be Banach spaces. Then T ∈ Fred(X,Y ) if and onlyif there exist S1, S2 ∈ L(Y,X) with

idX −S1T ∈ K(X,X), and idY −TS2 ∈ K(Y, Y ).

Additionally, one can choose S = S1 = S2 and such a map S ∈ L(Y,X) is called aparametrix for T .

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Proof. "⇒":

Let T ∈ Fred(X,Y ). It follows from Lemma 10.9 that there exist closed subsetsV ⊂ X and W ⊂ Y so that

X = V ⊕ ker(T ) and Y = Image(T )⊕W.

As in the proof of Theorem 11.11 we consider the map

T : (V ×W, ‖(v, w)‖ = ‖v‖X + ‖w‖Y )→ Y, T (v, w) = Tv + w.

The argument from step 1 in proof of Theorem 11.11 implies that T is invertibleand we let S = T−1 : Y → V × W be the inverse map and let PV : V × W ,PV (v, w) = v be the projection onto the first factor. We define S ∈ L(Y,X) byS = PV S : Y → V ⊂ X and we note that for v ∈ V and x ∈ kerT we have

(idX −ST )(v + x) = v + x− PV ST (v, 0) = v + x− v = x.

Thus the map idX −ST is a projection onto the finite-dimensional space kerT andhence idX −ST ∈ K(X,X). Next, we let v ∈ V , w ∈W and we calculate

(idY −TS)(Tv + w) =Tv + w − TPV ST (v, w) = Tv + w − Tv = w

and therefore idY −TS is a projection onto the finite-dimensional space W and weconclude that idY −TS ∈ K(Y, Y ).

"⇐":

We assume that S1T = idX −K1, where K1 ∈ K(X,X). It follows from Theorem11.12 that S1T ∈ Fred(X,X) and hence ker(T ) ⊂ ker(S1T ) is finite-dimensional.Additionally, we let TS2 = id−K2 with K2 ∈ K(Y, Y ) and Theorem 11.12 againimplies that TS2 ∈ Fred(Y, Y ) which yields that Image(TS2) is closed with finitecodimension. By Lemma 11.9 there exists a finite-dimensional closed subspace Z ⊂Y with

Image(T ) = Image(TS2)⊕ Z

and thus it follows from Lemma 11.10 that Image(T ) is closed. Finally, we note that

p : Y/ Image(TS2)→ Y/ Image(T )

is a projection and hence dim(coker(T )) ≤ dim(coker(TS2)) <∞.

In the next Lemma we strengthen the Theorem of Riesz for compact operators.

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Lemma 11.14. Let X and Y be Banach spaces and let T0 ∈ Fred(X,Y ), K ∈K(X,Y ). Then T := T0 +K ∈ Fred(X,Y ) with indT = indT0.

Proof. Let S0 ∈ L(Y,X) be a parametrix for T0, i.e. we have that there existKX ∈ K(X,X) and KY ∈ K(Y, Y ) with

idX −S0T0 = KX and idY −T0S0 = KY ∈ K(Y, Y )

We conclude that

idX −S0T = idX −S0T0 − S0K ∈ K(X,X) andidY −TS0 = idY −T0S0 −KS0 ∈ K(Y, Y ).

where we used Theorem 11.4. Hence S0 is also a parametrix for T and it followsfrom Lemma 10.13 that T ∈ Fred(X,Y ). The claim about the index follows as inthe proof of step 5 in Theorem 11.12.

Theorem 11.15. Let X,Y, Z be Banach spaces and let T1 ∈ Fred(X,Y ), T2 ∈Fred(Y,Z). Then T2T1 ∈ Fred(X,Z) and ind(T2T1) = ind(T1) + ind(T2).

Proof. We first show that T := T2T1 ∈ Fred(X,Z). By Lemma 11.13 there existS1 ∈ L(Y,X) and S2 ∈ L(Z, Y ) so that

idX −S1T1 = K1 ∈ K(X,X), idY −T1S1 = K1 ∈ K(Y, Y )idY −S2T2 = K2 ∈ K(Y, Y ), idZ −T2S2 = K2 ∈ K(Z,Z).

For S := S1S2 we calculate

idX −ST = idX −S1S2T2T1

= idX −S1(idY −K2)T1

= idX −S1T1 + S1K2T1

=K1 + S1K2T1 ∈ K(X,X),

where we used Theorem 11.4, and

idZ −TS = idZ −T2T1S1S2

= idZ −T2(idY −K1)S2

= idZ −T2S2 + T2K1S2

=K2 + T2K1S2 ∈ K(Z,Z)

again by Theorem 11.4. Therefore S is a parametrix for T and hence Lemma 11.13implies that T ∈ Fred(X,Z).

112


In order to show the claim about the index of T , we look at the following sequencesbetween finite-dimensional vector spaces

0 // ker(T1) ⊂ ker(T2T1) T1 //// ker(T2) ∩ Image(T1) // 0

0 // Image(T2)Image(T2T1) ⊂

ZImage(T2T1)

p // ZImage(T2)

// 0

0 // Image(T1)+ker(T2)Image(T1) ⊂ Y

Image(T1)T2 // // Image(T2)

Image(T2T1)// 0

0 // ker(T2) ∩ Image(T1) ⊂ ker(T2) π // Image(T1)+ker(T2)Image(T1)

// 0

.

These four sequences are exact since at each point we have ker = Image. Recall thata sequence

0 // 0 f0 // V1f2 // V2 // . . .

fm// Vm+1 = 0 // 0

is called exact, if Image fj = ker fj+1 for all j = 1, ...,m. It then follows that

m∑j=1

(−1)i dimVi = 0

since dimVj = dim(ker(fj)) + dim(Image(fj)) = dim(ker(fj)) + dim(ker(fj+1)).

In our situation we have that

0 = dim(ker(T1))− dim(ker(T2T1)) + dim(ker(T2) ∩ Image(T1))

0 =− dim( Image(T2)Image(T2T1)) + dim( Z

Image(T2T1))− dim( Z

Image(T2))

0 = dim(Image(T1) + ker(T2)Image(T1) )− dim( Y

Image(T1)) + dim( Image(T2)Image(T2T1))

0 =− dim(ker(T2) ∩ Image(T1)) + dim(ker(T2))− dim(Image(T1) + ker(T2)Image(T1) ),

where we note that we are always free to add the trivial vector space to each sequenceand hence we can decide with which sign we start the dimension formula for the exactsequence. Adding these for equations yields that

dim(ker(T1))−dim(coker(T1)) + dim(ker(T2))− dim(coker(T2))= dim(ker(T2T1))− dim(coker(T2T1))

113

and hence

indT1 + indT2 = indT2T1

as claimed.

Next we are coming back to elliptic boundary value problems and we want to usethe abstract results we just obtained in order to study the operator L = L0 + K,which we introduced at the beginning of this chapter. Recall that for Ω ⊂ Rn openand bounded the operator L0 : W 1,2

0 (Ω)→W 1,20 (Ω)′ is defined by

L0v = −div(aDv),

where a ∈ L∞(Ω,Mn(R)) is elliptic with constant µ > 0. By Theorem 10.15 L0 isan isomorphism. Moreover, the operator K : W 1,2

0 (Ω)→W 1,20 (Ω)′ is defined by

Kv = −div(bv) + 〈c,Dv〉+ qv,

q ∈ L∞(Ω) and b, c ∈ L∞(Ω,Rn). Our goal is to show that K is compact sinceLemma 11.14 then implies that L ∈ Fred(W 1,2

0 (Ω),W 1,20 (Ω)′) with indL = 0.

We start with a Lemma.

Lemma 11.16. Let 1 ≤ p <∞ and let u ∈W 1,p(Rn). Then we have that

1. ‖u τh − u‖Lp ≤ |h|‖Du‖Lp for all h ∈ Rn and where τh(x) = x + h for allx, h ∈ Rn.

2. ‖u− uρ‖Lp ≤ ρ‖Du‖Lp, where uρ is defined as in Theorem 10.4.

Proof. It follows from Theorem 10.9 that we can assume without loss of generalitythat u ∈ C∞c (Rn).

1. We estimate for all h ∈ Rn∫Rn|u(x+ h)− u(x)|p dx =

∫Rn

∣∣∣∣∫ 1

0

d

dsu(x+ sh) ds

∣∣∣∣p dx=∫Rn

∣∣∣∣∫ 1

0〈Du(x+ sh), h〉

∣∣∣∣p dx≤|h|p

∫Rn

∫ 1

0|Du(x+ sh)|p ds dx

=|h|p‖Du‖pLp ,

which shows statement 1.

114

2. For every ρ > 0 we get with the help of the transformation formula, Hölder’sinequality and the Theorem of Fubini that∫

Rn|u(x)− uρ(x)|p dx =

∫Rn

∣∣∣∣∫Rnηρ(x− y)(u(x)− u(y)) dy

∣∣∣∣p dx=∫Rn

∣∣∣∣∫Rnη(z)(u(x)− u(x− ρz)) dz

∣∣∣∣p dx≤∫Rn

(∫η(z)|u(x)− u(x− ρz)|p dz

)dx

=∫B1(0)

η(z)(∫

Rn|u(x)− u(x− ρz)|p dx

)dz

≤ρp‖Du‖pLp

by statement 1.

Theorem 11.17 (Rellich embedding theorem). Let Ω ⊂ Rn be open and boundedand let 1 ≤ p <∞. Then the inclusion W 1,p

0 (Ω) ⊂ Lp(Ω) is compact.

Proof. We have to show that every sequence uk ∈ W 1,p0 (Ω) with ‖uk‖W 1,p ≤ C has

a subsequence such that uk → u in Lp(Ω). Without loss of generality we assumeagain that uk ∈ C∞c (Ω). Let η be as in Theorem 10.4 and extend uk by 0 to all ofRn.

For j ≥ 0 we have

Dj(ηρ ∗ uk)(x) =∫Djxηrho(x− y)uk(y) dy

=ρ−(j+n)∫

(Djxη)

(x− yρ

)uk(y) dy.

Therefore

‖Dj(ηρ ∗ uk)‖L∞ ≤ c(j)ρ−j−nρn−np ‖uk‖Lp ≤ c(j, n)ρ−(j+n/p)

andspt(ηρ ∗ uk) ⊂ x : dist(Ω, x) ≤ ρ.

For ρ > 0 we conclude, using Theorem 3.8, that there exists a subsequence such that(etaρ ∗ uk) → uρ in C1(Rn). Next we choose a sequence ρi → 0 and successivelysubsequences k1

j ⊃ k2j ⊃ . . . for ρ1, ρ2,... such that the C1-convergence remains

true.

The diagonal sequence satisfies (ηρ ∗ uk) → uρ in C1(Rn) for all ρ ∈ ρ1, ρ2, ....

115


Moreover, we have that spt(uρ) ⊂ x : dist(x,Ω) ≤ ρ and

‖uρ‖Lp = limk→∞

‖ηρ ∗ uk‖Lp ≤ C and

‖Duρ‖Lp = limk→∞

‖ηρ ∗Duk‖Lp ≤ C.

For ρ, ρ′ ∈ ρ1, ρ2, ... we get

‖uρ − uρ′‖Lp = limk→∞

‖ηρ ∗ uk − ηρ′ ∗ uk‖Lp

≤ lim supk→∞

(‖ηρ ∗ uk − uk‖Lp + ‖ηρ′ ∗ uk − uk‖Lp)

≤ lim supk→∞

(ρ‖Duk‖Lp + ρ′‖Duk‖Lp) ≤ C(ρ+ ρ′),

where we used Lemma 11.16. Hence uρ is a Cauchy sequence in Lp(Rn) anduρ → u ∈ Lp(Rn). We have that spt(u) ⊂ Ω and ‖u − uρi‖Lp ≤ Cρi. Therefore weget

‖u− uk‖Lp ≤‖u− uρi‖Lp + ‖uρi − ηρi ∗ uk‖Lp + ‖ηρi ∗ uk − uk‖Lp≤Cρi + ‖uρi − ηρi ∗ uk‖Lp → 0

as i, k →∞.

Remark: If Ω ⊂ Rn is open and bounded with C1-boundary, then the inclusionW 1,p(Ω) ⊂ Lp(Ω) is also compact.

Lemma 11.18. Let Ω ⊂ Rn be open and bounded. Then the operator K : W 1,20 (Ω)→

(W 1,20 (Ω))′,

(Kv)(u) =∫

Ω(〈Du, bv〉+ u(〈c,Dv〉+ qv)),

with b, c ∈ L∞(Ω,Rn) and q ∈ L∞(Ω) is compact.

Proof. We denote by I : W 1,20 (Ω)→ L2(Ω) the inclusion map and we note that I is

compact by Theorem 11.17. Moreover, we define the map

J : L2(Ω)→ (L2(Ω))′, (Jv)(u) =∫uv

and we recall that J is continuous with ‖J‖ = 1. It follows from Theorem 11.4 that

I ′ J : L2(Ω)→ (W 1,20 (Ω))′

116


is compact. Next we define the operators

L2(Ω)→W 1,20 (Ω)′, (Bv)(u) =

∫〈b,Du〉v

W 1,20 (Ω)→L2(Ω), Cv = 〈c,Dv〉L2(Ω)→L2(Ω), Qv = qv

and we observe that all of them are continuous. Moreover, we have that

K = B I + I ′ J C + I ′ J Q I

and it follows again from Theorem 11.4 that K is compact.

Our main result for elliptic boundary value problems is the following

Theorem 11.19. Let L = L0 +K : W 1,20 (Ω)→ (W 1,2

0 (Ω))′ be as above,. Then L ∈Fred(W 1,2

0 (Ω),W 1,20 (Ω)′) and indL = 0. Thus the so called Fredholm alternative

holds: L is injective if and only if L is surjective.

Proof. We know from Theorem 10.15 that L0 and from Lemma 11.18 that K iscompact. Hence the result is a direct consequence of Lemma 10.14.

Next we define the so called formal adjoint operator of L. For this we consider thediagram

W 1,20 (Ω)

J

L∗:=L′J

&&W 1,2

0 (Ω)′′L′//W 1,2

0 (Ω)′

,

where J : W 1,20 (Ω)→W 1,2

0 (Ω)′′ is the canonical embedding. We have that

(L∗v)(u) = (L′Jv)(u) = (Jv)(Lu) = (Lu)(v)

for all u, v ∈W 1,20 (Ω). Moreover,

(Lu)(v) =∫

Ω(〈Dv, aDu〉+ 〈Dv, bu〉+ v〈c,Du〉+ qvu)

and thus(L∗v)(u) =

∫Ω

(〈Du, a∗Dv〉+ 〈Du, cv〉+ u〈b,Dv〉+ quv),

orL∗v = −div(a∗Dv)− div(cv) + 〈b,Dv〉+ qv.

117


The following theorem is a variant of Lemma 11.8 for the special case of operatorsconsidered here.

Theorem 11.20. Let Ω ⊂ Rn be open and bounded, let L, L∗ be as above and letϕ ∈W 1,2

0 (Ω)′. Then the following two statements are equivalent.

1. Lv = ϕ has a solution v ∈W 1,20 (Ω).

2. ϕ(u) = 0 for all u ∈ ker(L∗).

Proof. See the exercises.

118

12 Spectral theory for compact operators

In this chapter we let X be a Banach space over K and we mostly consider the caseK = C.

Definition 12.1. Let A ∈ L(X,X).

1. The resolvent set is given by

%(A) = λ ∈ K : λ id−A is invertible

λ ∈ %(A) is called regular with respect to A.

2. The set K \%(A) =: σ(A) is called the spectrum of A and λ ∈ σ(A) is calledspectral value. σ(A) is decomposed into three parts:

a) The point spectrum:

σp(A) = λ ∈ σ(A) : λ id−A is not injective.

b) The continuous spectrum:

σc(A) = λ ∈ σ(A) : λ id−A is injective but not surjective, Image(λ id−A)is dense in X.

c) The residual spectrum:

σr(A) = λ ∈ σ(A) : λ id−A is injective but not surjective, Image(λ id−A)is not dense in X.

It follows from Theorem 5.10 that if λ ∈ K \σ(A), then λ id−A is bijectiveand hence λ ∈ %(A).

3. The resolvent function RA : %(A)→ L(X,X) is defined by

RA(λ) = (λ id−A)−1.

In the next theorem we derive some general facts about the spectrum and the resol-

119

vent set.

Theorem 12.2. Let X 6= 0 be a Banach space over C and let A ∈ L(X,X). Then

1. dist(λ, σ(A)) ≥ ‖RA(λ)‖−1 for all λ ∈ %(A) and thus %(A) is open. Moreover,for all Λ ∈ L(X,X)′ the map λ ∈ %(A) ⊂ C 7→ ΛRA(λ) ∈ C is holomorphic.

2. sup|λ| : λ ∈ σ(A) = limn→∞

‖An‖1/n ≤ ‖A‖ < ∞ which implies that σ(A) 6= ∅is compact. The number sup|λ| : λ ∈ σ(A) is called the spectral radius ofA.

Proof. 1. Let λ0 ∈ %(A), i.e. λ0 id−A is invertible. Then it follows from Theorem5.2 that for λ ∈ C the map

λ id−A = λ0 id−A− (λ0 − λ) id = (λ0 id−A)(id−(λ0 − λ)RA(λ0))

is invertible if |λ − λ0| < ‖RA(λ0)‖−1, and therefore dist(λ0, σ(A)) ≥ ‖RA(λ0)‖−1.Additionally, for |λ− λ0| < ‖RA(λ0)‖−1, we get

RA(λ) =∞∑n=0

(−1)n(λ− λ0)nRA(λ0)n+1.

Hence we get for all Λ ∈ L(X,X)′ that

Λ(RA(λ)) =∞∑n=0

(−1)nΛ(RA(λ0)n+1)(λ− λ0)n

and hence this function is holomorphic for all Λ ∈ L(X,X)′ and all λ ∈ %(A).

2. We define R := sup|λ| : λ ∈ σ(A) and we note that for |λ| > ‖A‖ we have thatthe map

λ id−A = λ

(id−A

λ

)is invertible by Theorem 5.2 and thus

RA(λ) =( 1λ

) ∞∑n=0

λ−nAn. (12.1)

This shows that σ(A) ⊂ λ : |λ ≤ ‖A‖.

Next we claim that λ ∈ σ(A) implies λn ∈ σ(An). This follows from the fact that

λn id−An =n∑j=1

(λn−j+1Aj−1 − λn−jAj) =

(λ id−A)∑n

j=1 λn−jAj−1

(∑nj=1 λ

n−jAj−1)

(λ id−A)

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Now, if λ id−A is not surjective we use the first inequality, and if λ id−A is notinjective we use the second equality in order to get that λn ∈ σ(An).

Hence it follows from the above estimate that λ ∈ σ(A) implies λn ∈ σ(An) andtherefore |λ|n ≤ ‖An‖ which gives

|λ| ≤ lim infn→∞

‖An‖1/n.

For s > R and an arbitrary Λ ∈ L(X,X)′ we now consider the holomorphic functionΛ(RA(λ)) and we note that the integral

− 12πi

∫∂Bs(0)

λµΛ(RA(λ))dλ.

is independent of s as long as s > R by Cauchy’s integral theorem. Next, we lets > ‖A‖ and we calculate with the help of (12.1)

− 12πi

∫∂Bs(0)

λµΛ(RA(λ))dλ =∞∑n=0

Λ(An) 12πi

∫∂Bs(0)

dλ

λn+1−µ = Λ(Aµ).

On the other hand we have for all s > R∣∣∣∣∣ 12πi

∫∂Bs(0)

λµΛ(RA(λ))dλ∣∣∣∣∣ ≤sµ+1‖Λ‖ sup

|λ|=s|RA(λ)|.

Together these estimates show that for all s > R and all Λ ∈ L(X,X)′ we have

|Λ(Aµ)| ≤ sµ+1‖Λ‖ sup|λ|=s

|RA(λ)|.

By Lemma 4.4 there exists Λ ∈ L(X,X)′ such that Λ(Aµ) = ‖Aµ‖ and ‖Λ‖ = 1.Hence we get for all s > R

‖Aµ‖ ≤ sµ+1 sup|λ|=s

|RA(λ)|

and thereforelim supµ→∞

‖Aµ‖1/µ ≤ R.

In particular we conclude that

R = limn→∞

‖An‖1/n ≤ ‖A‖.

If we now assume that σ(A) = ∅ then ρ(A) = C and thus λ id−A is invertible forall λ ∈ C. Therefore

− 12πi

∫∂Bs(0)

λµΛ(RA(λ))dλ = 0

121


for all s ≥ 0 and by using the above argument with µ = 1 we get A = 0. Butker(A) = 0 as A− 0 id is invertible and this gives a contradiction.

Theorem 12.3 (Spectral theorem for compact operators). Let X be a Banach spaceover C and let K ∈ L(X,X) compact. Then we have that

1. σ(K)\0 = σp(K)\0.

2. σ(K) is a compact, at most countable subset of C. The only possible accumu-lation point is 0.

3. If dimX =∞ then 0 ∈ σ(K).

Proof. 1. Let λ ∈ σ(K)\0. Then it follows from Theorem 11.12 that

λ id−K = λ

(id−K

λ

)∈ Fred(X,X)

with ind(λ id−K) = 0, which shows that λ ∈ σp(K).

2. Assume that µ 6= 0 is an accumulation point of σ(K). Then it follows thatthere exists a sequence λn ∈ σ(K)\0 with λn → µ and we assume without loss ofgenerality that λn 6= λm for all m 6= n. By statement 1. there exists xn ∈ X\0with Kxn = λnxn and we define for all n ∈ N

Xn := span x1, ..., xn.

It follows that Xn is closed and invariant under K, i.e. KXn ⊂ Xn. Moreover,dimXn = n since if we assume that xn = ∑n−1

i=1 αixi, then 0 = Kxn − λnxn =∑n−1i=1 αi(λi − λn) and thus by induction αi(λi − λn) = 0 which implies αi = 0. By

Lemma 3.3 there exists zn ∈ Xn with ‖zn‖ = 1 and dist(zn, Xn−1) ≥ 12 . We have

zn = αnxn + yn for some yn ∈ Xn−1 and some αn ∈ C and we note that(id− 1

λnK

)zn =

(id− 1

λnK

)yn ∈ Xn−1

and therefore (id− 1

λnK

)zn + 1

λnKzi ∈ Xn−1

for all 1 ≤ i ≤ n− 1. Thus∥∥∥∥zn − (id− 1λnK

)zn −

1λnKzi

∥∥∥∥ ≥ 12

122

for all 1 ≤ i ≤ n− 1 and we conclude that

‖Kzn −Kzi‖ ≥|λn|

2 ≥ |µ|4 > 0

for all 1 ≤ i ≤ n − 1 and all n large enough. This shows that Kzn has noconverging subsequence which contradicts the fact that K ∈ K(X,X). Hence thesets λ ∈ σ(K) : |λ| ≥ 1/N are finite for all N ∈ N.

3. Assume that 0 /∈ σ(K) which implies that K is invertible by Theorem 5.10.Choose a sequence xi ⊂ X with ‖xi‖ ≤ 1 for all i ∈ N. Then Kxi → y up to asubsequence and therefore xi → K−1y. This shows that the set x ∈ X : ‖x‖ ≤ 1is sequentially compact and we conclude from Theorem 3.4 that dimX <∞.

Theorem 12.4 (Normal form for compact operators). Let X be a Banach spaceover C and let K ∈ K(X,X). Then for λ ∈ σ(K)\0 there exists a uniquedecomposition X = N(λ)⊕ F (λ) so that

1. N(λ), F (λ) are closed subspaces of X which are invariant under K with dimN(λ) =codimF (λ) <∞.

2. λ id−K : N(λ)→ N(λ) is nilpotent, i.e. (λ id−K)n|N(λ) = 0 for some n ∈ N,and λ id−K : F (λ)→ F (λ) is an isomorphism.

Moreover, we have that

3. ker(K − λ id) ⊂ N(λ) and

4. N(λ) ⊂ F (µ) for λ 6= µ ∈ σ(K)\0.

Proof. Step 1: Construction.

For λ ∈ σ(K)\0 and all j ∈ N we define

Nj(λ) := ker(λ id−K)j and Fj(λ) := Image(λ id−K)j .

These spaces are invariant under K since

K(λ id−K)j = −(λ id−K)j+1 + λ(λ id−K)j

= −(λ id−K)j(λ id−K) + (λ id−K)jλ id= (K − λ id)jK.

123

Moreover, we have by Theorem 11.12 that

(λ id−K)j =λj(id + compact) ∈ Fred(X,X)

and ind(λ id−K)j = 0. Hence, Fj(λ) is closed for all j ∈ N and dimNj(λ) =codimFj(λ). From the definition and Theorem 12.3 it follows that

0 ( N1(λ) ⊂ N2(λ), . . . .

and if we assume that Nj(λ) ( Nj+1(λ) for all j ∈ N it follows from Lemma 3.3 thatthere exists a sequence zj ∈ Nj(λ) with ‖zj‖ = 1 and dist(zj , Nj−1(λ)) ≥ 1/2. Thus

1λ

((λ id−K)zj +Kzi) ∈ Nj−1(λ)

for all i ≤ j − 1 and hence∥∥∥∥ 1λ

((λ id−K)zj +Kzi)− zj∥∥∥∥ ≥ 1

2

for all 1 ≤ i ≤ j − 1. But this implies that ‖Kzj −Kzi‖ ≥ |λ|2 > 0 for all i < j andis therefore a contradiction to K ∈ K(X,X). Thus there exists a smallest n ∈ N sothat Nn(λ) = Nn+1(λ). Now we assume that Nn+j(λ) ( Nn+j+1(λ) for some j ∈ N.Then there exists x ∈ X with

(λ id−K)n+1(λ id−K)jx = 0 and (λ id−K)n(λ id−K)jx 6= 0.

But this means (λ id−K)jx ∈ Nn+1(λ)\Nn(λ) = ∅ and we get another contradiction.Since codim(Fj(λ)) = dimNj(λ) we have

X ) F1(λ) ) F2(λ) ) . . . ) Fn(λ) = Fn+1(λ) = . . .

Therefore the spaces N(λ) := Nn(λ) and F (λ) := Fn(λ) are well-defined. Nowassume that there exists x ∈ N(λ) ∩ F (λ) and x 6= 0. Then there exists y ∈ X

so that x = (K − λ id)ny and 0 = (K − λ id)nx = (K − λ id)2ny. Since N2n(λ) =Nn(λ) = N(λ) it follows that 0 = (K − λ id)ny = x and we get a contradiction.Finally, we note that the projection

N(λ)→ X/F (λ)

is injective and hence surjective because of dimensional reasons. This shows that

X = N(λ)⊕ F (λ)

and we have proved statement 1.

124

By definition we have that

(λ id−K)n|N(λ) = 0.

Moreover, λ id−K : F (λ) → F (λ) is injective since N1(λ) ∩ F (λ) = 0 because ofN1(λ) ⊂ N(λ). Next, we let y ∈ F (λ) = Fn(λ) = Fn+1(λ) and hence there existsx ∈ X so that

y = (λ id−K)n+1x =(λ id−K)(λ id−K)nx ∈ (λ id−K)(F (λ))

and therefore (λ id−K : F (λ)→ F (λ) is also surjective, which shows statement 2.

Step 2: Uniqueness.

Let X = N ⊕ F be another decomposition so that N and F satisfy statements1. and 2. By statement 2. we get that (λ id−K)|N is nilpotent which impliesby the definition of N(λ) that N ⊂ N(λ). Next we let z ∈ F and we note that(λ id−K)nz ∈ F (λ) and therefore it follows from statement 2. that

F = (λ id−K)nF ⊂ F (λ).

Since X = N ⊕ F it follows that N = N(λ) and F = F (λ).

Step 3: Proof of statement 4.

Let λ, µ ∈ σ(K)\0 with λ 6= µ and let x ∈ N(λ) with x = y + z, where y ∈ N(µ)and z ∈ F (µ). Here we assume that N(µ) = Nm(µ) and F (µ) = Fm(µ) for somem ∈ N. Then

(µ id−K)mx = (µ id−K)mz ∈ F (µ)

and therefore(µ id−K)mN(λ) ⊂ F (µ).

Next we claim that (µ id−K)mN(λ) = N(λ) which will finish the proof. Assumethat there exists x ∈ N(λ)\0 with (µ id−K)x = 0. Then

(λ id−K)x =(µ id−K)x+ (λ− µ)x = (λ− µ)x

and hence λ−µ 6= 0 is an eigenvalue of the map (λ id−K)|N(λ). But this contradictsthe fact that (λ id−K)|N(λ) is nilpotent and therefore (µ id−K)|N(λ) is injective. AsdimN(λ) < ∞ it follows that (µ id−K)|N(λ) is also surjective and this shows theclaim and therefore also statement 4.

Example: A compact operator with no non-zero eigenvalues.

125


Let X = C0([0, 1],C) and note that K ∈ L(X,X), (Kf)(x) :=∫ x

0 f(y)dy is compact(see the exercises). Assume that there exists λ 6= 0 and f ∈ X\0 with

λf(x) =∫ x

0f(y)dy.

It follows from the Fundamental Theorem of Calculus that f ∈ C1((0, 1)) andλf ′(x) = f(x). Hence f(x) = c exp

(xλ

)for some c 6= 0 but this contradicts that

fact that f(0) = 0. Therefore we conclude that σ(K) = 0.

Definition 12.5. Let X be a Hilbert space over C. A map A ∈ L(X,X) is called

• normal if [A,A∗] := AA∗ −A∗A = 0.

• hermitian or self-adjoint if A∗ = A.

• unitary if A∗A = AA∗ = id.

Lemma 12.6. Let X be a Hilbert space over K and let A ∈ L(X,X) be normal.Then

1. kerA = kerA∗ and ‖Ax‖ = ‖A∗x‖ for all x ∈ X.

2. If K = C then maxλ∈σ(A) |λ| = ‖A‖.

Proof. 1. For all x ∈ X we have

‖Ax‖2 = 〈x,A∗Ax〉 = 〈A∗Ax, x〉 = 〈AA∗x, x〉 = ‖A∗x‖2

which finishes the proof of statement 1. 2. By Theorem 12.2 we know that

maxλ∈σ(A)

|λ| = limn→∞

‖An‖1/n ≤ ‖A‖.

Now we claim that for a normal operator A we have ‖An‖ = ‖A‖n for all n ∈ N.We show this by induction and we note that the case n = 1 is obvious n = 1. Forn → n+ 1 we have

‖Anx‖2 =〈Anx,Anx〉=〈An−1x,A∗Anx〉≤‖A∗Anx‖‖An−1x‖=‖An+1x‖‖An−1x‖≤‖An+1x‖‖An−1‖‖x‖≤‖An+1x‖‖A‖n−1‖x‖

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where we used statement 1. in the fourth line and the induction hypotheses in thelast line. By taking the supremum over ‖x‖ ≤ 1 we obtain

‖A‖2n = ‖An‖2 ≤ ‖An+1‖‖A‖n−1

and hence‖A‖n+1 ≤ ‖An+1‖.

Theorem 12.7 (Spectral theorem for compact and normal operators). Let X be aHilbert space over C with dimX =∞ and let K ∈ K(X,X) be normal. Then

1. σ(K) is compact and 0 is the only possible accumulation point, 0 ∈ σ(K) andmax|λ|, λ ∈ σ(K) = ‖K‖.

2. For λ ∈ σ(K)\0 the eigenspace Eλ(K) := ker(λ id−K) is non-trivial, finite-dimensional and Eλ(K) ⊥ Eµ(K) for λ 6= µ.

3. X is the Hilbert sum of the Eλ(K) with λ ∈ σ(K) and Kx = ∑λ∈σ(K)\0 λPλ(x)

where Pλ : X → Eλ(K) is the orthogonal projection onto Eλ(K).

Proof. Statement 1. follows directly from Theorem 12.2, Theorem 12.3 and Lemma12.6.The first part of Statement 2. follows from Theorems 12.3 and 12.4. since λ id−Kis a normal operator as (λ id−K)∗ = λ id−K∗. By Lemma 12.6 we have that

Eλ(K) = ker(λ id−K) = ker(λ id−K∗) = Eλ(K∗)

and for x ∈ Eλ(K), y ∈ Eµ(K) we get

(λ− µ)〈x, y〉 =〈λx, y〉 − 〈x, µy〉=〈Kx, y〉 − 〈x,K∗y〉 = 0

and therefore 〈x, y〉 = 0, which shows that Eλ(K) ⊥ Eµ(K).

3. We use Theorem 9.12 and hence we have to show the maximality of the setsEλ(K)λ∈σ(K). We let

V =

⊕λ∈σ(K)

Eλ(K)

⊥

and our goal is to show that V = 0. First we observe that V is invariant under

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K since for v ∈ V and x ∈ Eλ(K), λ ∈ σ(K), we have

〈Kv, x〉 =〈v,K∗x〉 = 〈v, λx〉 = λ〈v, x〉 = 0

and hence Kv ∈ V . Moreover, V is a complex Hilbert space, K|V ∈ L(V, V ) and(K|V )∗ = K∗|V , which shows that K|V is normal. The map K|V is also compactand by the definition of V it follows that K|V has no eigenvalue. Hence we concludefrom statement 2. that σ(K|V ) = 0. Statement 1. then yields that ‖K|V ‖ = 0,i.e. K|V ≡ 0, and thus V ⊂ ker(K) = E0(K). By the definition of V we also havethat V ⊂ E0(K)⊥ and therefore we conclude V = 0. The rest of the proof nowfollows from Theorem 9.12.

Theorem 12.8 (Spectral theorem for compact and selfadjoint operators). Let Xbe a Hilbert space over K = C or R and let K ∈ K(X,X) with K∗ = K. Thenσ(K) ⊂ R and all three statements from Theorem 12.7 remain true.

Proof. Case 1: K = C

Let λ ∈ σ(K)\0. Then we know from Theorem 12.7 that there exists an eigen-vector x 6= 0, so that

λ‖x‖2 =〈λx, x〉 = 〈Kx, x〉=〈x,Kx〉 = 〈x, λx〉 = λ‖x‖2,

which implies that λ = λ and therefore λ ∈ R.

Step 2: K = R

We complexify the real vector space X and we get a complex vector space XC :=X ×X with the complex multiplication i(x, y) = (−y, x) for all x, y ∈ X. Hence weget for all a, b ∈ R and all x, y ∈ X

(a+ ib)(x, y) = (ax− by, ay + bx)

and one concludes that XC is a complex vector space with X ⊂ XC via the mapx 7→ (x, 0). We also have that (x, y) = (x,−y). Next we define

〈(x1, y1), (x2, y2)〉 := 〈x1, x2〉+ 〈y1, y2〉 − i(−〈x1, y2〉 − 〈y1, x2〉)

and we note that〈(x1, y1), (x2, y2)〉 = 〈(x2, y2), (x1, y1)〉

and‖(x, y)‖2 = ‖x‖2 + ‖y‖2

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Moreover, we have

〈i(x1, y1), (x2, y2)〉 =〈(−y1, x1), (x2, y2)〉=− 〈y1, x2〉+ 〈x1, y2〉+ i(〈y1, y2〉+ 〈x1, x2〉)=i〈(x1, y1), (x2, y2)〉

and all of these properties combined show that (XC = X × X, 〈·, ·〉) is a complexHilbert space.

Now we define the map K : XC → XC by K(x, y) = (Kx,Ky) and we note that Kis complex linear since

K(i(x, y)) = K(−y, x) = (−Ky,Kx) = i(Kx,Ky) = iK(x, y).

Moreover, we have that (K)∗ = (K∗) and hence we can apply the results from step1 to the triple (XC, 〈·, ·〉, K). We get that σ(K) ⊂ R, dimEλ(K) <∞ if λ 6= 0 and

XC =⊕

λ∈σ(K)

Eλ(K).

Since K(x, y) = (Kx,Ky) we conclude that σ(K) = σ(K) and

Eλ(K) = Eλ(K)× Eλ(K) ⊂ X ×X = XC.

It remains to show that

X =⊕

λ∈σ(K)Eλ(K).

For this we let x ∈ X with x ⊥ Eλ(K) for all λ ∈ σ(K) and we note that this implies(x, 0) ⊥ Eλ(K) for all λ ∈ σ(K). Since the set of Eλ(K)’s is maximal, we concludethat (x, 0) = 0 ∈ XC and thus x = 0. This shows that the set of Eλ(K), λ ∈ σ(K),are also maximal and this finishes the proof by Theorem 9.12.

As an application of this result we obtain the

Theorem 12.9 (Spectral theorem for the Dirichlet problem). Let Ω ∈ Rn be openand bounded and let L : W 1,2

0 (Ω)→W 1,20 (Ω) be defined by

Lv := −div(aDv) + qv,

where a ∈ L∞(Ω,Mn(R)) is symmetric and elliptic with constant µ > 0 and q ∈L∞(Ω).

Then there exists a sequence λj ⊂ R with λ1 < λ2 < · · · → ∞, so that

129

1. the eigenspace Eλ(L) := v ∈ W 1,20 (Ω): Lv = λv is finite-dimensional and it

is non-trivial if and only if λ ∈ λj.

2. The space L2(Ω) is the Hilbert sum of the sets Eλj (L), j ∈ N.

Proof. Without loss of generality we assume that q ≥ 0 since otherwise look at theoperator L := L+ ‖q‖L∞ id and this does not change the eigenspaces. We considerthe bilinear form B : W 1,2

0 (Ω)×W 1,20 (Ω)→ R associated to L

B(u, v) =∫

Ω〈Du, aDv〉+ quv = (Lv)(u)

and we note that B is symmetric since a is symmetric. Moreover, it follows fromthe ellipticity of a and the Poincare inequality (Theorem 10.13) that

B(v, v) ≥ µ∫|Dv|2 ≥ c‖v‖W 1,2 .

Hence we can apply the Lax-Milgram Theorem (Theorem 9.11) in order to concludethat L is invertible with ‖L−1‖ ≤ c. Now the map

I : W 1,20 (Ω)→ L2(Ω), v 7→ v

is compact by Theorem 11.16 and the adjoint map is given by

I ′ : L2(Ω)→ (W 1,20 (Ω))′, (I ′f)(u) =

∫Ωfu.

We define G : L2(Ω) → L2(Ω) by G := I L−1 I ′ and we note that for f ∈ L2(Ω)the map Gf ∈ W 1,2

0 (Ω) is the unique weak solution v ∈ W 1,20 (Ω) of Lv = f . For

f1, f2 ∈ L2(Ω) we let Gfi = vi, i = 1, 2, i.e. Lvi = fi and we calculate

〈Gf1, f2〉L2 =〈v1, Lv2〉=(Lv2)(v1)=B(v1, v2)=B(v2, v1)=〈f1, Gf2〉

which shows that G is self-adjoint. Moreover, it follows from Theorem 11.4 that G iscompact since it is the composition of a compact map with continuous maps. Hencewe can apply Theorem 12.8 to G and we conclude that σ(G) = µ1, µ2, . . . ⊂ Rwith µj → 0 as j → ∞. Note that we have infinitely many eigenvalues sincedim(Eµj (G)) <∞ but dim(L2(Ω)) =∞. Now for λ, µ ∈ R we have

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1. If there exists v ∈W 1,20 (Ω)\0 with Lv = λv then we have

λ‖v‖2 = (Lv)(v) = B(v, v) ≥ c‖v‖2W 1,2 > 0.

This shows that λ > 0 and hence L(v/λ) = v which is equivalent to Gv = vλ .

2. If Gv = µv for some v ∈ L2(Ω)\0 then µv ∈W 1,20 (Ω) and L(µv) = v. Hence

µ 6= 0, v ∈W 1,20 (Ω) and Lv = v

µ .

Therefore we conclude that σ(G) = µ1, µ2, . . . ⊂ R+ and σ(L) = 1µ1, 1µ2, . . . from

which the claim follows.

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Motivation: For an open subset Ω ⊂ Rn and a given initial data u0 ∈W 2,2∩W 1,20 (Ω)

we want to construct a solution of the heat equation

∂tu =∆u in Ω× (0,∞)u =0 on ∂Ω× (0,∞)u =u0 on Ω× 0,u : Ω× [0,∞)→ R

The idea is to consider this PDE as an ordinary differential equation in a Banachspace

u : [0,∞)→W 2,2 ∩W 1,20 (Ω)

so that ddtu = ∆uu(0) = u0

We have the formal solution u(t) = et∆u0. In the following we consider the timeevolution u(0)→ u(t) as a time dependent family of operators and this is what leadsus to C0-semigroups.

Definition 13.1. Let X be a Banach space. A map T : [0,∞)→ L(X,X) is calledC0-semigroup if and only if

1. T (t1 + t2) = T (t1)T (t2) for all t1, t2 ≥ 0.

2. T (0) = id.

3. the map Tx : [0,∞)→ X, Tx(t) = T (t)x is continuous for all x ∈ X.

Examples:

1. Let A ∈ L(X,X) and consider for all t ≥ 0 the map

T (t) = exp(tA) :=∞∑k=0

(tA)kk! .

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This series converges absolutely in the operator norm and the first two prop-erties of a C0-semigroup are obvious. Moreover, the map T : R→ L(X,X) iscontinuous.

2. Let X = L2(R) and let (T (t)u)(x) = u(x+ t). Then T is a C0-semigroup since‖T (t)u − u‖L2 → 0 as t → 0, a fact which follows from the proof of Theorem10.4. Note that T : R→ L(X,X) is not continuous since ‖T (t1)− T (t2)‖ = 2for t1 6= t2.

3. Let X = L∞(R) and (T (t)u)(x) = u(x + t). Then properties 1 and 2 of thedefinition of a C0-semigroup hold and ‖T (t)‖ = 1 but T is not a C0-semigroup.

Lemma 13.2. Let X be a Banach space and let T be a C0-semigroup on X. Then

1. the limit

limt→∞

( log ‖T (t)‖t

)= ω0 ∈ [−∞,∞).

exists and

2. for all ω > ω0 there exists c = c(ω) <∞ so that for all t ≥ 0

‖T (t)‖ ≤ ceωt.

Proof. For all t > 0 we define

ω(t) := log ‖T (t)‖t

and we let τ > 0. For every t > 0 there exists n ∈ N0 and t0 ∈ [0, τ) so thatt = nτ + t0. It follows from the definition of a C0-semigroup that

ω(t) =ω(nt+ t0) = log(‖T (nτ + t0)‖

t

)= log

(‖T (τ)nT (t0)‖t

)≤n log ‖T (τ)‖+ log ‖T (t0)‖

t

and therefore, by letting t∞, we get

lim supt∞

ω(t) ≤ log ‖T (τ)‖tτ

= ω(τ) <∞.

Next we choose a sequence τk ∞ so that ω(τk)→ lim inft→∞ ω(t) and we concludethat the limit lim

t→∞ω(t) ∈ [−∞,∞) exists.

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13 Semigroups

In order to show statement 2. we note that there exists t1 > 0 so that ω(t) ≤ ω forall t ≥ t1 by statement 1. Hence it follows that

‖T (t)‖ ≤ eωt

for all t ≥ t1. Next, we recall that the map [0, t1]→ X, t 7→ T (t)x is continuous forall x ∈ X and therefore

sup0≤t≤t1

‖T (t)x‖ <∞

for all x ∈ X. By Theorem 5.7 there exists a constant c <∞ so that

sup0≤t≤t1

‖T (t)‖ ≤ c <∞

and thus‖T (t)‖ ≤ cmax1, e−ωt1eωt

for all t ≥ 0.

Examples:

1. Let X = L2(R) and (T (t)u)(x) = u(x+ t). Then we know that ‖T (t)‖ = 1 forall t ≥ 0 and thus ω0 = 0.

2. Let I = (0, 1) and X = L2(I). Then the map

(T (t)u)(x) =u(x+ t), if x+ t < 10, otherwise

is a C0-semigroup with T (t) = 0 for all t ≥ 1 and therefore ω0 = −∞.

3. Let X be a Hilbert space over C and let K ∈ K(X,X) be normal with eigen-values λj , j ∈ N∪0. Then T (t)x := etKx = ∑∞

j=1 etλjPjx (see Theorem

12.7) is a C0-semigroup with

‖T (t)x‖ =

∥∥∥∥∥∥∞∑j=1

etλjPjx

∥∥∥∥∥∥=

∞∑j=1

e2tRe(λj)‖Pjx‖21/2

≤etmaxj(Reλj)‖x‖

and therefore ω0 = maxj(Re(λj)).

Definition 13.3. Let X be a Banach space and let T be a C0-semigroup on X. We

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define

Ax := limh0

T (h)x− xh

if the limit exists. Moreover, we define the domain of A by

D(A) = x ∈ X : Ax exists

and we note that it is a linear space and A : D(A) → X is a linear map. The mapA is called the infinitesimal generator of the C0-semigroup T .

Example: Let X = L2(R) and let (T (t)u)(x) = u(x+ t). Then

(T (h)− id)uh

= ∆hu := u(x+ h)− u(x)h

is the so called difference quotient of u. Let u ∈ D(A), i.e. Au = limh→0

∆hu exists,and let ϕ ∈ C∞c (R). Then we calculate∫

(∆hu)ϕ =−∫u(∆−hϕ)→ −

∫uϕ′

as h 0. Hence we get that ∫(Au)ϕ = −

∫uϕ′

and therefore u ∈W 1,2(R) and Au = ddxu.

Conversely, if we assume that u ∈W 1,2(R), we get∥∥∥∥∆hu−du

dx

∥∥∥∥L2≤ sup

0≤t≤h‖u′(·+ t)− u′(·)‖L2 → 0

as h→ 0 by the proof of Theorem 10.4. Hence u ∈ D(A) and A(u) = ddxu.

We therefore conclude that

A = d

dx: W 1,2(R) = D(A)→ L2(R).

Lemma 13.4. Let X be a Banach space and let T be a C0-semigroup on X. Let Abe an ininitesimal generator of T (t). Then we have for all t > 0

1. T (t)x = limh→0

1h

∫ t+ht T (s)x ds.

2.∫ t

0 T (s)x ds ∈ D(A) and A∫ t

0 T (s)x ds = T (t)x− x for all x ∈ X.

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13 Semigroups

3. T (t)x ∈ D(A) and AT (t)x = T (t)Ax if x ∈ D(A).

4. ddtT (t)x = T (t)Ax for all x ∈ D(A).

Proof. 1. This follows from the Fundamental Theorem of Calculus.

2. We note that for all h > 0 we have

T (h)− idh

∫ t

0T (s)x ds = 1

h

∫ t

0(T (s+ h)− T (s))x ds

= 1h

(∫ t+h

tT (s)x ds−

∫ h

0T (s)x ds

)→ T (t)x− x

as h 0 since the map T (·)x is continuous by assumption.

3. Let x ∈ D(A) and h > 0, then

T (h)− idh

T (t)x =T (t)T (h)− idh

x→ T (t)Ax

as h 0 and thus T (t)x ∈ D(A) and AT (t)x = T (t)Ax.

4. Let h > 0 and x ∈ D(A). Then

T (t+ h)x− T (t)xh

=T (t)T (h)− idh

x→ T (t)Ax

as h 0 and

T (t)x− T (t− h)xh

=T (t− h)T (h)− idh

x

=T (t− h)Ax+ T (t− h)(T (h)− id

h−A

)x

→T (t)Ax

as h 0. Hence the claim follows.

Theorem 13.5. Let X be a Banach space and let T be a C0-semigroup on X withinfinitesimal generator A. Then D(A) is dense in X and A is closed, i.e. the graphof A, G(A) ⊂ X ×X, is closed.

Proof. By Lemma 13.4, statement 1, we get for every x ∈ X

x = T (0)x = limh→0

1h

∫ h

0T (s)x ds

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13 Semigroups

and the right hand side is in D(A) by Lemma 13.4, statement 2. Hence D(A) isdense in X.

Next we let xn be a sequence in D(A) with xn → x and Axn → y as n → ∞. Inorder to show that G(A) is closed we have to verify that x ∈ D(A) and y = Ax. ByLemma 13.4, statement 4, we have for every h > 0

T (h)xn − xn =∫ h

0

d

dtT (t)xndt =

∫ h

0T (t)Axn dt

and by Lemma 13.2 we conclude that∥∥∥∥∥∫ h

0(T (t)Axn − T (t)y) dt

∥∥∥∥∥ ≤ sup0≤t≤h

‖T (t)‖‖Axn − y‖h ≤ Ch‖Axn − y‖ → 0

as n→∞. This implies that T (h)x− x =∫ h

0 T (t)y dt and thus

T (h)− idh

x = 1h

∫ h

0T (t)ydt→ y

as h 0. Therefore we conclude that x ∈ D(A) and Ax = y.

Definition 13.6. Let X be a Banach space and let A : D(A)→ X be linear. Thenwe define the resolvent set of A by

%(A) = λ ∈ K : λ id−A : D(A)→ X is bijective and (λ id−A)−1 is bounded.

Moreover, the resolvent function RA : %(A)→ L(D(A), X) is defined by

RA(λ) = (λ id−A)−1.

Note that here we do not assume that A is bounded.

Lemma 13.7. Let X be a Banach space and let T be a C0-semigroup on X in-finitesimal generator A and let ω0 = lim

t→∞

(log ‖T (t)‖

t

). Then we have for all λ ∈ C

with Re (λ) > ω0 that

1. λ ∈ %(A) and RA(λ)x =∫∞

0 e−λtT (t)x dt for all x ∈ D(A).

2. if x ∈ D(A) then RA(λ)x ∈ D(A) and ARA(λ)x = RA(λ)Ax.

Proof. Let ω0 < ω < Re (λ) and note that by Lemma 13.2 we get the estimate‖T (t)‖ ≤ ceωt for all t > 0. Define

I(λ)x =∫ ∞

0e−λtT (t)x dt

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13 Semigroups

and note that

‖I(λ)‖ ≤ c∫ ∞

0eω−Re (λ)t dt ≤ c

Re(λ)− ω . (13.1)

Now let x ∈ D(A) and calculate with the help of Lemma 13.4

I(λ)Ax =T∫ ∞

0e−λtT (t)Axdt

=∫ ∞

0e−λt

d

dtT (t)x dt

=[e−λtT (t)x]t=∞t=0 + λ

∫ ∞0

e−λtT (t)x dt

=− x+ λI(λ)x

and therefore

I(λ)(λ id−A)x = x ∀x ∈ D(A). (13.2)

This implies that λ id-A is injective and I(λ) is a left inverse of λ id−A. Next, forx ∈ X and h > 0 we calculate

T (h)− idh

I(λ)x = 1h

(∫ ∞0

e−λtT (t+ h)x dt−∫ ∞

0e−λtT (t)x dt

)=eλh

h

∫ ∞h

e−λsT (s)x ds− 1h

∫ ∞0

e−λsT (s)x ds

=eλh − 1h

I(λ)x− eλh

h

∫ h

0e−λsT (s)x ds

→λI(λ)x− x

as h 0 since the function s 7→ e−λsT (s)x is continuous for all x ∈ X. Hence weget that

(λ id−A)I(λ)x = x. (13.3)

By (13.1)-(13.3) it follows that for all λ ∈ C with Re (λ) > ω we have λ ∈ ρ(A) andRA(λ) = I(λ), which proves statement 1. Moreover, we have that RA(λ)x ∈ D(A)for all x ∈ D(A).

If x ∈ D(A) it follows from (13.2) and (13.3) that for all x ∈ D(A)

(λ id−A)RA(λ)x = RA(λ)(λ id−A)x

and thereforeARA(λ)x = RA(λ)Ax

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13 Semigroups

Lemma 13.8. Let X be a Banach space and let A : D(A)→ X be a linear operator.For λ, µ ∈ %(A) we have

1. RA(λ)−RA(µ) = (λ− µ)RA(λ)RA(µ).

2. RA(λ)RA(µ) = RA(µ)RA(λ).

Proof. First we note that statement 2. follows from statement 1. and hence this isthe only thing to prove. For µ, λ ∈ %(A) and y ∈ X there exists x ∈ D(A) with

y = (µ id−A)x = (λ id−A)x+ (µ− λ)x

and hence(µ− λ)RA(λ)RA(µ)y = (µ− λ)RA(λ)x.

Using these facts we get

(RA(λ)−RA(µ))y =x+ (µ− λ)RA(λ)x− x=(µ− λ)RA(λ)RA(λ)RA(µ)y.

Theorem 13.9 (Hille-Yosida). Let X be a Banach space and let A : D(A) → X

be a closed and linear operator so that D(A) is dense in X and let ω ∈ R, M > 0.Then the following two statements are equivalent:

1. A is the infinitesimal generator of a C0-semigroup T on X with ‖T (t)‖ ≤Meωt

for all t > 0.

2. (ω,∞) ⊂ %(A) and ‖RA(λ)n‖ ≤ M(λ−ω)n for all n ∈ N and all λ > ω.

Proof. 1 ⇒ 2:

For n ∈ N and λ ∈ C with Re (λ) > ω define

In(λ)x := 1(n− 1)!

∫ ∞0

tn−1e−λtT (t)x dt.

As in the proof of Lemma 13.7 one shows that the integral exists and we have theestimate

‖In(λ)‖ ≤ M

(n− 1)!

∫ ∞0

tn−1e(ω−Re (λ))t dt ≤ M

(Re (λ)− ω)n .

Now we show by induction that In(λ) = RA(λ)n. The case n = 1 can be found in

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Lemma 13.7. Next we let n ≥ 2 and we show that the formula for (n−1) implies thecorresponding one for n. For this we let x ∈ D(A) and using Lemma 13.4, statement4, we calculate

In(λ)Ax = 1(n− 1)!

∫ ∞0

tn−1e−λtT (t)Axdt

=− 1(n− 2)!

∫ ∞0

tn−2e−λtT (t)x dt+ λ

(n− 1)!

∫ ∞0

tn−1e−λtT (t)x dt

=− In−1(λ)x+ λIn(λ)x.

Hence it follows that

In(λ)(λ id−A)x = In−1(λ)x = RA(λ)n−1x = RA(λ)n(λ id−A)x.

By Lemma 13.7 we know that (ω,∞) ⊂ %(A) and hence the operator (λ id−A) :D(A)→ X is surjective and therefore it follows that In(λ) = RA(λ)n.

2 ⇒ 1:

We let λ > ω and start with a formal computation:

RA(λ) = 1λ

(id− 1

λA

)−1≈ 1λ

(id +A

λ+ · · ·

)and hence one expects that

A ≈λ(λRA(λ)− id)

for λ large enough. Therefore we define A(λ) := λ(λRA(λ)− id) ∈ L(X,X) and wenote that for x ∈ D(A) and λ > ω we have

λRA(λ)x− x = RA(λ)(λx− λx+Ax) = RA(λ)Ax

and thus

‖λRA(λ)x− x‖ ≤ ‖RA(λ)‖‖Ax‖ ≤ M

Re (λ)− ω‖Ax‖ → 0λ→∞.

as λ → ∞. For an arbitrary x ∈ X we choose a sequence xn ∈ D(A) with xn → x

and we estimate

‖λRA(λ)x− x‖ ≤‖λRA(λ)(x− xn)‖+ ‖λRA(λ)xn − xn‖+ ‖xn − x‖

≤(Mλ

λ− ω+ 1

)‖xn − x‖+ M

λ− ω‖Axn‖

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13 Semigroups

and therefore

lim supλ→∞

‖λRA(λ)x− x‖ ≤ (M + 1)‖xn − x‖ → 0

as n→∞.

We thus conclude that for x ∈ D(A) we have

A(λ)x = λRA(λ)Ax→ Ax,

as λ→∞. As RA(λ) is a bounded operator, we can define for all t ≥ 0

Tλ(t) = etA(λ) = e−λteλ2RA(λ)t

and we note that Tλ is a C0-semigroup with infinitesimal generator A(λ). We esti-mate for all t ≥ 0

‖Tλ(t)‖ ≤e−λt∞∑n=0

1n! (λ

2t)n‖RA(λ)n‖

≤Me−λt∞∑n=0

1n!

(λ2t

λ− ω

)n=Me−λte

λ2tλ−ω

and thus

‖Tλ(t)‖ ≤Me

(ω

1−ωλ

)t. (13.4)

By Lemma 13.8 we know that RA(λ)RA(µ) = RA(µ)RA(λ) for all λ, µ > ω andtherefore we also get A(λ)A(µ) = A(µ)A(λ) which in turn implies that

A(µ)Tλ(t) = Tλ(t)A(µ).

for all λ, µ > ω. Using this property we calculate with the help of Lemma 13.4

Tλ(t)− Tµ(t) =∫ t

0

d

ds(Tµ(t− s)Tλ(s)) ds

=∫ t

0Tµ(t− s)Tλ(s)(A(λ)−A(µ)) ds

In the following we let λ, µ ≥ 2ω since then ω1−ω

λ, ω

1−ωµ≤ 2ω. Hence we use the

estimate (13.4) in order to conclude for all x ∈ D(A) that

‖Tλ(t)x− Tµ(t)x‖ ≤∫ t

0M2e2ω(t−s)+2ωs‖A(λ)x−A(µ)x‖ ds

≤M2te2ωt‖A(λ)x−A(µ)x‖ → 0

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13 Semigroups

as λ, µ→∞. Note that the convergence is uniform for t ∈ [0, t0] for all t0 <∞. Forx ∈ X arbitrary we choose again a sequence xn ∈ D(A) with xn → x and we getwith the help of (13.4)

‖Tλ(t)x− Tµ(t)x‖ ≤‖Tλ(t)(x− xn)‖+ ‖Tλ(t)xn − Tµ(t)xn‖+ ‖Tµ(t)(x− xn)‖≤2Meωt‖x− xn‖+ ‖Tλ(t)xn − Tµ(t)xn‖ → 0

as n, λ, µ→∞. This convergence is again uniform if t ∈ [0, t0].

Now we define T (t)x := limλ→∞

Tλ(t)x for all x ∈ X and we note that this limit existscontinuous in t. We have that T (0) = id and

T (t1 + t2)x←Tλ(t1 + t2)x = Tλ(t1)Tλ(t2)x=Tλ(t1)T (t2)x+ Tλ(t1) (Tλ(t2)− T (t2))x→0

as λ → ∞ since ‖Tλ(t1)‖ ≤ Me2ωt1 for λ large enough. Moreover, it follows from(13.4) that

‖T (t)‖ ≤Meωt

and thus T is a C0-semigroup on X. Next we show that A is an infinitesimalgenerator of T . It follows from Lemma 13.4 that

Tλ(t)x− x =A(λ)∫ t

0Tλ(s)x ds =

∫ t

0Tλ(s)A(λ)x ds.

For x ∈ D(A), λ large enough and s ∈ [0, t] we estimate

‖Tλ(s)A(λ)x− T (s)Ax‖ ≤‖Tλ(s)(A(λ)x−Ax)‖+ ‖Tλ(s)Ax− T (s)Ax‖≤Me2ωt‖A(λ)x−Ax‖+ ‖(Tλ(s)− T (s))Ax‖→0

as λ→∞ uniformly in s ∈ [0, t]. Hence, for x ∈ D(A), we obtain

T (t)x− x =∫ t

0T (s)Axds

and therefore we get for all x ∈ D(A) that

limt0

T (t)x− xt

= limt0

1t

∫ t

0T (s)Axds = Ax

since the function in the integral is continuous in s. Altogether we have shown thatA is an infinitesimal generator of the C0-semigroup T .

Finally, assume that A : D(A) → X is also an infinitesimal generator of the C0-

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13 Semigroups

semigroup T on X with D(A) ⊂ D(A). Then it follows from Lemma 13.7 and ourassumption that (ω,∞) ⊂ %(A) and (ω,∞) ⊂ %(A). Hence for λ ∈ (ω,∞) we havethat the operator (λ id−A) is surjective and

(λ id−A)D(A) = X

since A = A on D(A). But the map (λ id−A) : D(A) → X is also injective andhence we conclude that D(A) = D(A) and A = A.

Remarks:

1. Under the assumptions of Theorem 13.9 we have that λ ∈ C : Re (λ) > ω ⊂%(A) and ‖RA(λ)n‖ ≤ M

(Re (λ)−ω)n .

2. A C0-semigroup T on X is uniquely determined by its infinitesimal generator.

In order to see this, we assume that T1, T2 are two C0-semigroups with in-finitesimal generator A. It follows from Lemma 13.2 that there exist ω1, ω2 ∈[−∞,∞) so that for ω0 := maxω1, ω2 we have for i = 1, 2, all t ≥ 0 and allω > ω0

‖Ti(t)‖ ≤ ceωt.

Thus, we can apply Lemma 13.7 and we get for all λ > ω0 that

RA(λ)x =∫ ∞

0e−λtTi(t)x dt.

By letting T (t) = T1(t)− T2(t) we conclude that∫ ∞0

e−λtT (t)x dt = 0

for all λ > ω. For φ ∈ X ′ and ω > ω0 we define f(t) = e−ωtφ(T (t)x and wenote that f ∈ L1(0,∞) and ∫ ∞

0e−µtf(t) dt = 0

for all µ ≥ 0. Next we do the substitution t = − log u and we get∫ 1

0uµf(− log u)

udu = 0

for all µ ≥ 0. The function g : (0, 1)→ R, g(u) = f(− log u)u is in L1(0, 1) since∫ 1

0 |g(u)| du =∫∞

0 |f(t)| dt. Moreover, by the Theorem of Weierstrass, we knowthat the polynomials uµ are dense in C0(0, 1) and hence it follows from theFundamental Theorem of the Calculus of Variations that g ≡ 0 which yields

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13 Semigroups

f ≡ 0. By Lemma 4.4 this finally shows that T ≡ 0.

Example:

We look again at the heat equation on a smooth domain Ω ⊂ Rn, i.e. we want tosolve the PDE

∂tu = ∆u, in Ω× (0,∞)u = 0, on ∂Ω× (0,∞)u = u0, on Ω× 0

with u0 ∈W 2,2∩W 1,20 (Ω). We want to use the Theorem of Hille-Yosida and for this

we let X = L2(Ω), A = ∆ and D(A) = W 2,2 ∩W 1,20 (Ω). Since C∞c (Ω) is dense in X

and in D(A) it follows that D(A) is dense in X.

First we claim that (0,∞) ⊂ %(A), which is equivalent to the fact that for allλ ∈ (0,∞) we have that the operator λ id−∆: W 2,2 ∩W 1,2

0 (Ω)→ L(Ω) is bijective.Or in other words, we have to show that for every λ > 0 and every f ∈ L2(Ω) thereexists a unique solution u ∈W 2,2 ∩W 1,2

0 (Ω) of the PDE

−∆u+ λu =f. (13.5)

In order to see this, we note that the associated bilinear form B : W 1,20 (Ω) ×

W 1,20 (Ω)→ R

B(u, v) =∫

(〈Du,Dv〉+ λuv)

is coercive with the estimate

B(u, u) ≥ min(1, λ)‖u‖2W 1,2 .

Hence the existence of a unique solution u ∈ W 1,20 (Ω) of the above PDE follows

from the Lax-Milgram Theorem (Theorem 9.10). The fact that u is additionally inW 2,2(Ω) is shown in standard PDE courses.

Next we have to show that for all λ > 0 we have the estimate ‖RA(λ)n‖ ≤ λ−n. Inorder to do this, we let f ∈ L2(Ω) and we let u ∈ W 2,2 ∩W 1,2

0 (Ω) be the uniquesolution of (13.5) which is equivalent to the fact that u = RA(λ)f . We get∫

|∇u|2 + λu2 = B(u, u) =∫fu ≤ 1

2λ

∫|f |2 + λ

2

∫u2

and thereforeλ

2

∫u2 ≤ 1

2λ

∫f2

which yields‖u‖L2 ≤ λ−1‖f‖L2

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13 Semigroups

or‖RA(λ)‖ ≤ λ−1.

It remains to show that A is closed. For this we let xn ∈ D(A) be a sequencewith xn → x in D(A) and Axn → y in X. Let λ > 0 so that the operator(λ id−A) : D(A) → X is bijective. Hence, for every λ > 0 there exists z ∈ D(A)with (λ id−A)z = λx− y. Using the fact that ‖RA(λ)‖ ≤ λ−1 we estimate

‖xn − z‖ ≤ λ−1‖(λ id−A)(xn − z)‖ ≤ λ−1 (λ‖xn − x‖+ ‖Axn − y)‖)→ 0

as n→∞. Hence we conclude that z = x ∈ D(A) and Ax = y.

Finally, Theorem 13.9 implies thatA generates a C0-semigroup T onX with ‖T (t)‖ ≤1 for all t ≥ 0 and by Lemma 13.4 it follows that u(t) := T (t)u0 is a solution of ourproblem.

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