Functional Analysis 1

Home , C space, Operator topologies

MATHEMATISCHES INSTITUT PROF. DR. PETER MÜLLER

Summer Term 2013

Lecture Course

Functional Analysis

Typesetting by Kilian Lieret and Marcel Schaub

If you ﬁnd mistakes, I would appreciate getting a short mail from you to

marcel.schaub [at] campus.lmu.de.

Thanks! Version of April 10, 2014 Contents

1 Topological and metric spaces5 1.1 Topological spaces: basics...... 5 1.2 Limits and continuity...... 7 1.3 Metric spaces...... 8 1.4 Example: sequence spaces `p ...... 14 1.5 Compactness...... 16 1.6 Example: spaces of continuous functions...... 19 1.7 Baire’s Theorem...... 23

2 Banach and Hilbert spaces 26 2.1 Vector spaces...... 26 2.2 Banach spaces...... 27 2.3 Linear operators...... 30 2.4 Linear functionals and dual space...... 35 2.5 Hilbert spaces...... 37

3 Measures, integration and Lp-spaces 47 3.1 Measures...... 47 3.2 Integration...... 49 3.3 Lp-spaces...... 53 3.4 Decomposition of Measures...... 66

4 The cornerstones of functional analysis 69 4.1 Hahn-Banach theorem...... 69 4.2 Three consequences of Baire’s theorem...... 72 4.3 (Bi)-Dual spaces and weak topologies...... 76

5 Bounded operators 85 5.1 Topologies on the space of bounded linear operators...... 85 5.2 Adjoint operators...... 88 5.3 The spectrum...... 91 5.4 Compact operators...... 95 5.5 Fredholm alternative for compact operators...... 98

3 Introductory remarks

Functional analysis is...

a child of linear algebra and analysis • a theory of inﬁnite-dimensional vector spaces • Functional analysis has lots of applications

partial diﬀerential equations (PDE’s) • approximation theory • numerical maths • probability theory • quantum mechanics (functional analysis is its language!) •

4 1 Topological and metric spaces

1.1 Topological spaces: basics 1.1 Deﬁnition. Let X be a set. (X) is a topology iﬀ T ⊆ P (1) ∅,X . ∈ T (2) is closed under arbitrary unions (i.e. if I is an arbitrary index set and for every T α I let a set A be given. Then ∈ α ∈ T [ A α ∈ T α I ∈ holds.).

(3) is closed under ﬁnite intersections (i.e. if n N and A1,...,An , then T ∈ ∈ T n \ A k ∈ T k=1

holds.).

(X, ) is called topological space (often just X) • T A (X) is open iff A . • ∈ P ∈ T Let , be topologies on X. is finer than iff and coarser than iff • T1 T2 T1 T2 T1 ⊇ T2 T2 . T1 ⊆ T2 1.2 Examples. (a) Indiscrete topology: = ∅,X . T { } (b) Discrete topology: = (X). T P (c) Euclidean (or standard) topology on Rn, n N: A Rn is open iff x A ε > 0 ∈n ⊆ ∀ ∈ ∃ such that Bε(x) A, where Bε(x) := y R : x y < ε is the Euclidean ball of ⊆ { ∈ | − | } radius ε > 0 about x Rn. ∈ Induced topology on subsets

1.3 Definition. Let (X, ) be a topological space, A (X) (not necessarily open!). T ∈ P (A) is the relative topology on A iff TA ⊆ P := B A : C with B = C A . TA { ⊆ ∃ ∈ T ∩ } 1.4 Remark. (a) is topology on A. TA (b) If A/ and B then it may happen that B/ . ∈ T ∈ TA ∈ T Example. Let X = R with standard topology, A = [0, 1]. Then B := [0, 1/2[ A ∈ T but B/ . ∈ T 1.5 Definition. Let X be a topological space, A X, x X. ⊆ ∈ (a) A is closed iff Ac := X A . \ ∈ T

5 (b) U X (not necessarily open) is a neighbourhood of x iff A such that x A and ⊆ ∃ ∈ T ∈ A U. ⊆ (c) X is a Hausdorff space iff for all x, y X, x = y, there exist neighbourhoods U of x ∈ 6 x and Uy of y such that Ux Uy = ∅. ∩ (d) x is a limit point of A (or accumulation point) iff for all neighbourhoods U of x

U A = ∅. ∩ 6 Note: Every point of A is also a limit point according to this deﬁnition.

(e) x is an interior point of A iff there exists a neighbourhood U of x such that U A. ⊆ (f) x is a boundary point of A iff for every neighbourhood U of x: U A = ∅ and ∩ 6 U Ac = ∅. ∩ 6 Boundary of A: ∂A := x X : x boundary point of A . { ∈ } (g) Interior of A: A˚ := A ∂A closure of A: A := A ∂A = x X : \ ∪ { ∈ x limit point of A . } (h) A is dense in X iff X = A.

1.6 Lemma. Let X be a topological space, A X. ⊆ (a) A is open x A : x is an interior point of A. ⇐⇒ ∀ ∈ (b) A is closed A = A. ⇐⇒ (c) A, ∂A are closed.

Proof. See Problem 1.

1.7 Definition. Let (X, ) be a topological space, a family of open sets. T B ⊆ T (a) is a base for iff consists of unions of sets from . B T T B (b) is a subbase for iff finite intersections of sets from form a base. B T B (c) is a neighbourhood base at x iff every N is a neighbourhood of x and for N ⊆ T ∈ N every neighbourhood U of x there exists N with N U. ∈ N ⊆ 1.8 Remark. (a) Let (X). Then there exists a topology on X such that is a S ⊆ P T S subbase for and is the coarsest topology containing . Jargon: is generated T T S T by . S (b) Example: Rn with standard topology. Let x Rn. ∈ (i) B1/k(x): k N is a neighbourhood base at x. { ∈ } n (ii) B1/k(q): k N, q Q is a base for the standard topology (see the proof of { ∈ ∈ } Thm. 1.19 later).

6 1.9 Definition. Let J = ∅ be an arbitrary index set. For every j J let (Xj, j) be a 6 ∈ T topological space. is the product topology on the Cartesian product space T [ X := f : J X with f(j) X j → j ∈ j j J j J ×∈ ∈ iff has the base T Aj : Aj j j J, Aj = Xj for at most finitely many j’s . j J ∈ T ∀ ∈ 6 ×∈ 1.10 Remark. If J is finite, then the condition “A = X for at most finitely many j’s” j 6 j can be dropped.

1.11 Deﬁnition. Let X be a topological space.

(a) X is separable iff A X countable with A = X. ∃ ⊆ (b) X is 1st countable iff every x X has a countable neighbourhood base. ∈ (c) X is 2nd countable iff there exists a countable (sub-)base for the topology. [Note: countable base countable sub base (see Problem T2).] ⇐⇒ 1.12 Theorem. Let X be a topological space. Then X is 2ndcountable = X is 1stcountable and separable. ⇒ Proof. Let be a countable base for the topology. B Let x X and := B : x B . Then is a neighbourhood base and • ∈ Nx { ∈ B ∈ } Nx countable (hence 1st countable).

∅ = B choose xB B and let A := xB : ∅ = B . We claim A is •∀ 6 ∈ B ∈ { 6 ∈ B} countable (trivial) and A = X. For all x X and neighbourhoods U of x there ∈ exists C such that x C U. C is a union from sets in , so there exists ∈ T ∈ ⊆ B B such that x B U. On the other hand x B means x U and x A ∈ B ∈ ⊆ B ∈ B ∈ B ∈ implies A U = ∅. ∩ 6 1.2 Limits and continuity

1.13 Definition. Let X be a topological space and (xn)n N X be a sequence. ∈ ⊆ (xn)n converges to x X iff for every neighbourhood U of x there exists n0 N and for ∈ ∈ all n n0: xn U. ≥ ∈ n Notations: limn xn = x or xn →∞ x. →∞ −−−→ 1.14 Remark. (a) convergence is harder for finer topologies.

(b) X Hausdorff = limits are unique (see Problem T3). ⇒ 1.15 Definition. Let (X, ) and (Y, ) be topological spaces and let f : X Y . TX TY −→ n n (a) f is sequentially continuous iff x →∞ x (in X) implies f(x ) →∞ f(x) (in Y ). n −−−→ n −−−→ (b) f is continuous iff for every A : f 1(A) (f 1(A) is the inverse image). ∈ TY − ∈ TX − (c) f is open iff A : f(A) = y Y : x A : y = f(x) . ∀ ∈ TX { ∈ ∃ ∈ } ∈ TY

7 (d) f is a homeomorphism iﬀ f is bijective, open and continuous (bijection compatible with topological structure).

1.16 Theorem. Let X,Y be topological spaces and f : X Y a map. Then −→ (a) f is continuous = f is sequentically continuous. ⇒ (b) f is sequentially continuous and X is ﬁrst countable = f is continuous. ⇒ n Proof. (a) Let x →∞ x in X. Let V Y be a neighbourhood of f(x). W.l.o.g.1 assume n −−−→ ⊆ V open (see below). Set U := f 1(V ). U is open (because f is continuous) and x U − ∈ = U is a neighbourhood of x and we can apply the deﬁnition of convergence to U ⇒ = n0 N : n n0 : xn U. But that also means n0 N: n n0 : f(xn) V . ⇒ ∃ ∈ ∀ ≥ ∈ ∃ ∈ ∀ ≥ ∈ If V is not open there exists an open subset V V with x V and one can repeat 0 ⊆ ∈ 0 the same argument with V0 instead of V . (b) Proof by contradiction: Suppose f is not continuous, i.e. there exists an open subset V Y s.t. U := f 1(V ) is not open: x U s.t. for all neighbourhoods N of x: ⊆ − ∃ ∈ C N * U = N U = ∅. (1) ⇒ ∩ 6 ˜ Tk Let Nk k N be a countable neighbourhood base at x. Consider Nk := j=1 Nj. Then { } ∈ N˜ N˜ k and (2) k+1 ⊆ k ∀ N˜ is a neighbourhood of x k. (3) k ∀ ˜ ˜ C So Nk k N is a neighbourhood base of x.(1) and (3) imply that k N xk Nk U { } ∈ ∀ ∈ ∃ ∈ ∩ l l k : x N˜ U C = x →∞ x. •∀ ≥ l ∈ k ∩ ⇒ l −−−→ f(x ) f(U)C V C = f(x ) V k. But f(x) V so f(x ) cannot converge • k ∈ ⊆ ⇒ k 6∈ ∀ ∈ k to f(x) .

1.3 Metric spaces

Known notions and notations: Let X = ∅ be a set. 6 Metric d: X X [0, [ with the known properties: • × −→ ∞ – d(x, y) 0 x, y X with d(x, y) = 0 x = y. ≥ ∀ ∈ ⇐⇒ – d(x, y) = d(y, x) x, y X. ∀ ∈ – d(x, y) d(x, z) + d(z, y) x, y, z X. ≤ ∀ ∈ (X, d) is called a metric space.

Y X = d Y Y is the so-called induced metric on Y . • ⊆ ⇒ | × Open metric ball of radius ε > 0 about x X: • ∈ B (x) := y X : d(x, y) < ε . ε { ∈ } Cauchy sequence, completeness2 • 1without loss of generality 2Completeness is not a topological notion! See Problem 4.

8 Figure 1: Triangle inequality

For A X, x X deﬁne • ⊆ ∈

– diam(A) := supa,a0 A d(a, a0) diameter of A. ∈ – dist(x, A) := infa A d(x, a) distance of x to A. ∈ 1.17 Lemma. Let X be a metric space. Then X is a topological space, which is 1st countable and Hausdorﬀ w.r.t.3 the metric topology, i.e. the one given by the base

B1/k(x) k N . { }x∈X ∈ st Proof. B1/k(x) k N is a neighbourhood base at x which is clearly 1 countable. •{ } ∈ X is Hausdorﬀ: Let x, y X be two distinct points. Then ε := d(x, y) > 0. For • ∈ arbitrary points u B (x) and v B (y) the triangle inequation yields (see ∈ ε/2 ∈ ε/2 Figure1): ε = d(x, y) d(x, u) + d(u, v) + d(v, y) < ε/2 + d(u, v) + ε/2 ≤ Thus d(u, v) > 0 and Bε/2(x) Bε/2(y) = ∅ and X is Hausdorﬀ. ∩ 1.18 Corollary. All topological notions are available in a metric space X. In particular for (x ) X, x X,A X: n n ⊆ ∈ ⊆ (a) (xn)n converges to x ε > 0 n0 N: n n0 : xn Bε(x) ⇐⇒ ∀ ∃ ∈ ∀ ≥ ∈ limn d(xn, x) = 0. ⇐⇒ →∞ (b) x A (yn)n A: limn yn = x. ∈ ⇐⇒ ∃ ⊆ →∞ (c) If Y is a topological space and f : X Y , then −→ f is continuous f is sequentially continuous. ⇐⇒ Proof. (a), (b): simple exercises. (c): See Theorem 1.16.

1.19 Theorem. X a metric space. Then X separable implies X 2nd countable.

Proof. There is a countable and dense subset A X (A = X). We claim that B1/n(a) n N ⊆ { }a∈A is a base of the (metric) topology: Let C X open. x C ε(x) > 0 such that∈ ⊆ ∀ ∈ ∃ Bε(x) C. Thus ⊆ [ C = Bε(x)(x). x C ∈ Denseness of A in X implies: x C a(x) A and n(x) N such that x B1/n(x)(a(x)) ∀ ∈ ∃ ∈ ∈ ∈ ⊆ Bε(x)(x) (see Figure2). But then we can also write [ C = B1/n(x)(a(x)). x C ∈ 3with respect to

9 Figure 2: Finding n(x)

NB. This also proves Remark 1.8 (b): B (q) k is a base of the Euclidean topology 1/k Nn { }q ∈Q ∈ in Rn. 1.20 Lemma. Let X be a complete metric space and A X. Then: A closed A complete. ⊆ ⇐⇒

Proof. See Analysis II.

1.21 Example. Consider the space of all continuous functions f : [0, 1] C: −→ C([0, 1]) := f : [0, 1] C: f is continuous { −→ } with two diﬀerent metrics

(a) Supremum metric

d (f, g) := f g , where f := sup f(x) ∞ k − k∞ k k∞ x [0,1] | | ∈ The metric space (C([0, 1]), d ) is complete and separable (see Thms. 1.41 and 1.42 ∞ later).

(b)1 -metric: d (f, g) := R 1 dx f(x) g(x) deﬁnes a metric on C([0, 1]): 1 0 | − | (i) d (f, g) 0 is obvious. Suppose f = g. Then there exists at least one point 1 ≥ 6 x [0, 1] with f(x ) = g(x ) and we ﬁnd ε > 0 so small that f(x ) g(x ) > ε. 0 ∈ 0 6 0 | 0 − 0 | But f, g are continuous, so there exists δ > 0 such that for any x [0, 1] with ∈ x x < δ we have f(x) f(x ) < ε/3 and g(x) g(x ) < ε/3. Combining | − 0| | − 0 | | − 0 | these inequalities we obtain

f(x) g(x) f(x0) g(x0) f(x) f(x0) g(x0) g(x) > ε/3 | − | ≥ | {z− }| −| −{z }| −| {z− }| >ε <ε/3 <ε/3

for all x [0, 1] with x x < δ. Thus ∈ | − 0| Z 1 Z 1

d1(f, g) = dx f(x) g(x) dx 1]x0 δ,x0+δ[(x) f(x) g(x) 0 | − | ≥ 0 − | {z− }| >ε/3 > ε/3 δ > 0. ·

10 Figure 3: fn

(ii) The symmetry d1(f, g) = d1(g, f) is obvious. (iii) Triangle inequality: Z 1 d1(f, h) = dx f(x) h(x) d1(f, g) + d1(g, h). 0 | {z− }| ≤ f(x) g(x) + g(x) h(x) ≤| − | | − |

Thus, d1 is indeed a metric.

(C([0, 1]), d1) is separable because d1(f, g) d (f, g) and (C([0, 1]), d ) is sep- • ≤ ∞ ∞ arable. (C([0, 1]), d ) is not complete: Consider f from Figure3. • 1 n (fn)n is a Cauchy sequence: Let n, m N, m n: ∈ ≥ Z 1 Z 1/2 1 d1(fn, fm) = dx fn(x) fm(x) = dx fn(x) fm(x) . 0 | − | 1/2 1/n | −{z }| ≤ n − 1 ≤ But Z 1 Z 1/2 1 dx fn(x) 1[1/2,1](x) dx fn(x) 1[1/2,1](x) 0 | − | ≤ 1/2 1/n | −{z }| ≤ n − 1 ≤ and 1 (x) / C([0, 1]). Suppose f C([0, 1]) with d (f , f) 0 as n . [1/2,1] ∈ ∃ ∈ 1 n → → ∞ Then Z 1 dx f(x) 1[1/2,1](x) = 0 0 | − | which cannot be true (consider a neighbourhood of x = 1/2) = f does not ⇒ n converge in C([0, 1])!

1.22 Definition. Let (X, d ), (Y, d ) be metric spaces and T : X Y . X Y −→ T is called an isometry iff x, x X : d (x, x ) = d (T (x),T (x )). • ∀ 0 ∈ X 0 Y 0 X,Y are isometric iff there exists a bijective isometry T : X Y . • −→ 1.23 Remark. Let T : X Y be an isometry. Then −→ (a) T is injective and continuous,

(b) X and the range of T ,

ran T := T (X) = y Y : x X with y = T (x) , { ∈ ∃ ∈ } are isometric.

11 1.24 Theorem. Let X be a metric space. Then there exists a complete metric space X˜ and an isometry i : X X˜ such that i(X) is dense in X˜ and X˜ is unique up to isometric spaces. −→ Proof. Consists of 4 steps: (a) Construct X˜. (b) Construct the isometry i and W := i(X) is dense in X˜. (c) X˜ is complete (d) Uniqueness.

(a) Deﬁne an equivalence relation on the set of all Cauchy sequences in X: ∼ (xn)n (yn)n : lim d(xn, yn) = 0 n ∼ ⇐⇒ →∞ ( is clearly reﬂexive, symmetric and transitive). We look at the equivalence classes ∼ we obtain in this way , so consider

X˜ := equivalence classesx ˜ of Cauchy sequences in X . { } We write (x ) x˜ for a representative (x ) of the equivalence classx ˜. Deﬁne a n n ∈ n n metric on X˜:

d˜(˜x, y˜) := lim d(xn, yn)x, ˜ y˜ X,˜ (xn)n x,˜ (yn)n y˜ n →∞ ∈ ∈ ∈ We now have to check that d˜... (1) ...is well deﬁned: (i) Existence of the limit:

d(x , y ) d(x , x ) + d(x , y ) + d(y , y ) n n ≤ n m m m m n d(x , y ) d(x , y ) d(x , x ) + d(y , y ) n n − m m ≤ n m m n And the same inequality holds with m and n exchanged. Thus we have:

d(xn, yn) d(xm, ym) d(xn, xm) + d(ym, yn)( ) | | {z } − | {z } | ≤ ∗ =:αn =:αm

Because of (xn)n and (yn)n being Cauchy sequences we have ε > 0 N N ∀ ∃ ∈ such that n, m N: ∀ ≥ d(xn, xm) < ε/2 and d(yn, ym) < ε/2

So αn αm < ε and (αn)n R is a Cauchy sequence, thus convergent | − | ⊆ (because R is complete). (ii) Independence of representatives: Let (x ) (x ) and (y ) (y ) . We n n ∼ n0 n n n ∼ n0 n have to show lim d(xn, yn) = lim d(x0 , y0 ) n n n n →∞ →∞ Repeat derivation of ( ) with exchanging x x and y y and ∗ m ←→ n0 m ←→ n0 obtain n d(x , y ) d(x0 , y0 ) d(x , x0 ) + d(y , y0 ) →∞ 0 by deﬁnition of | n n − n n | ≤ n n n n −−−→ ∼

12 (2) ...fulﬁlls the axioms of a metric: d˜ 0 • ≥ X

d˜(˜x, y˜) = 0 lim d(xn, yn) = 0 for some representatives n ⇐⇒ →∞ (x ) (y ) ⇐⇒ n n ∼ n n x˜ =y ˜ ⇐⇒ Symmetry and triangle inequality are clear from the properties of d. • (b) Deﬁne X X˜ i : −→ b ˜b 7−→ where ˜b is the unique equivalence class with (b, b, b, . . .) ˜b. Set W := i(X). The map ∈ i is an isometry because

d˜(˜b, a˜) = d˜(i(a), i(b)) = lim d(a, b) = d(a, b). n →∞ It remains to show that W is dense in X˜ with respect to d˜. Letx ˜ X˜ and ε > 0. ∈ Pick any representative (xn)n x˜.(xn)n being a Cauchy sequence implies N N ∈ ∃ ∈ n, m N: d(x , x ) < ε. Let ˜b W deﬁned by (x , x ,...) ˜b ∀ ≥ n m ∈ N N ∈

= d˜(˜b, x˜) = lim d(xN , xn) < ε n ⇒ →∞ So W is dense. (k) ˜ ˜ (k) (c) Let (˜x )k N) be a Cauchy sequence in X; W is dense in X: k N z˜ W such ∈ ∀ ∈ ∃ ∈ that 1 d˜(˜x(k), z˜(k)) < k Let (z , z , z ,...) z˜(k) be the corresponding constant representative for each k. For k k k ∈ every k, l N we get (due to the isometry property of i): ∈

(k) (k) (k) (l) (l) (l) d(zk, zl) = d˜(i(zk), i(zl)) d˜(˜z , x˜ ) +d˜(˜x , x˜ ) + d˜(˜x , z˜ ) |{z} |{z} ≤ | {z } | {z } z˜(k) z˜(l) <1/k <1/l

Thus (z , z , z ,...) X is a Cauchy sequence and hence a representative of some 1 2 3 ⊆ x˜ X˜. We show now, that ∈ lim d˜(˜x(k), x˜) = 0. n →∞ Let ε > 0, then d˜(˜x(k), x˜) d˜(˜x(k), z˜(k)) + d˜(˜z(k), x˜) < ε ≤ | {z } | {z } <1/k =limn→∞ d(zk,zn)

for all k large enough, as (zn)n is Cauchy.

13 (d) Suppose, there exists Xˆ with a dense subset V and ˆ X˜ with a dense subset W and bijective isometries j, X j respectively i between X and V , respectively X and V

W . 1 X j i− We show: X˜ and Xˆ are isometric. We know: V := ◦ i j(X) and W := i(X) are isometric with isometry 1 1 W j i . According to the Problem T5, j i can be ˜ ◦ − ◦ − X uniquely extended to an isometry X˜ Xˆ. Thus −→ we have the uniqueness up to isometries. Figure 4: Uniqueness

1.4 Example: sequence spaces `p 1.25 Deﬁnition (`p-spaces). Set

1/p p X p ` := x = (xn)n N : xn C n and x p := xn < ∈ ∈ ∀ k k | | ∞ n N ∈ whenever p [1, [ and ∈ ∞ n o `∞ := x = (xn)n N : xn C n and x := sup xn < ∈ ∈ ∀ k k∞ n N | | ∞ ∈ ( will be a norm for every p [1, ]). k·kp ∈ ∞ 1.26 Lemma. For every p [1, ], d (x, y) := x y , x, y `p deﬁnes a metric on ∈ ∞ p k − kp ∈ `p.

Proof. All properties clear, except for triangle inequality, this follows from Lemma 1.27.

1.27 Lemma. Let p, q [1, ] be conjugated exponents, i.e. 1/p + 1/q = 1 (convention: ∈ ∞ 1/ = 0“). Then ” ∞ (a) Dual pairing and H¨olderinequality: p, q: P is well x ` y ` x, y := n N xnyn deﬁned and ∀ ∈ ∀ ∈ h i ∈ X x, y xnyn x y . h i ≤ | | ≤ k kp k kq n N ∈ (b) Minkowski inequality: Let x, y `p. Then ∈ x + y x + y . k kp ≤ k kp k kp

Proof. From corresponding inequalities on CN and subsequent limit N . For example, → ∞ to prove (a), use N N N X X 1/pX 1/q x y x p y q | n n| ≤ | n| | n| n=1 n=1 n=1 (see e.g. Forster, vol. 1) and perform limit.

1.28 Remark. Minkowski inequality does not remain true for p ]0, 1[. Therefore we ∈ consider `p-spaces only for p 1. ≥ 1.29 Theorem. (a) `p is separable p [1, [. ∀ ∈ ∞

14 (b) `∞ is not separable. Proof. (a) We use the separability of C. For n N let ∈ p Mn := (x1, . . . , xn, 0,...) ` with xj Q + iQ, j = 1, . . . , n . { ∈ ∈ } S p p So Mn is countable and M := n N Mn is also countable. Claim M = ` . Let y ` ∈ ∈ and ε > 0, then there exists N N: ∈ X∞ εp y p < | j| 2 j=N+1

PN p εp and since Q + iQ is dense in C there exists x MN such that xj yj < . ∈ j=1 | − | 2 This implies d (y, x)p = x y p < εp p k − kp (b) See Problem 6.

1.30 Theorem. `p is complete for every p [1, ]. ∈ ∞ (n) p (n) (n) (n) Proof. (a) Case p [1, [. Let (x )n N ` be a Cauchy sequence (x = (x1 , x2 ,...)). ∈ ∞ ∈ ⊂ Let ε > 0. Then there exists N N such that for every n, m N and J1,J2 N: ∈ ≥ ∈ J X2 x(n) x(m) p < εp ( ) | j − j | ∗ j=J1 Step 1: We have to show the existence of a candidate for the limit using the complete- (n) ness of C. Let J1 = J2 = J.( ) implies (xJ )n C is a Cauchy sequence and ∗ ⊂ (n) since C is complete, there is some xJ C such that limn x = xJ for all ∈ →∞ J J N. So our candidate is x := (x1, x2,...). ∈ J2 Step 2: Set J1 = 1 in ( ) and use the Minkowski inequality in C : ∗ J J J X2 1/p X2 1/p X2 1/p x p x x(n) p + x(n) p | j| ≤ | j − j | | j | j=1 j=1 j=1 | {z } x(n) < ≤k kp ∞ (n) ε + x p J2 N. ≤ k k ∀ ∈ (n) p So we obtain x ε + x p < for every n N and therefore x ` . k kp ≤ k k ∞ ∈ ∈ For every n N and m in ( ): ≥ → ∞ ∗ J2 X (n) p p x xj < ε J2 N, | j − | ∀ ∈ j=1 so for sending J in addition, we have 2 → ∞ (n) x x p < ε. |k {z− k} (n) dp(x ,x)

So x(n) x in `p. −→ PJ2 p (b) Case p = : Replace “ j=J ” by “supJ1 j J2 ” and “ ” by “ ”. ∞ 1 ≤ ≤ | · | | · |

15 1.5 Compactness 1.31 Definition. Let X be a topological space and A X. ⊆ S (a) A is compact iff for every open cover A α I Bα with an index set I = ∅ and ⊆ ∈ 6 Bα X open for every α I there exists a finite open subcover, i.e. N N and ⊆ ∈ ∈ α , . . . , α I with A SN B ( Heine-Borel-property“). 1 N ∈ ⊆ n=1 αn ” (b) A is sequentially compact iff every sequence in A has a convergent subsequence with limits in A ( Bolzano-Weierstraß property“). ” (c) A is relatively compact iff A is compact. 1.32 Remark. (a) Def. applies in particular to A = X. In this case we have =“ instead ” of “ in (a). ”⊆ (b) Some books (e.g. Bourbaki) use compactness only for Hausdorff spaces, otherwise the notion is quasi-compact. 1.33 Theorem. Let X be a topological space. (a) If X is 1st countable, then X compact implies X sequentially compact. (b) If X is 2nd countable, then X compact is equivalent to X sequentially compact. 1.34 Theorem (Lindelöf). Let X be a 2nd countable topological space and let X = S be an open cover. Then there exists such that S , α I Aα (αn)n N I X = n N Aαn i.e.∈ a countable subcover. ∈ ⊆ ∈

Proof. Let Bk k N be a countable base of the topology. Set { } ∈ K := k N : α I with Bk Aα . { ∈ ∃ ∈ ⊆ } Since every Aα is a union of Bk’s, we have [ [ Bk = Aα = X (*) k K α I ∈ ∈ So for every k K pick α I such that B A . Then we get ∈ k ∈ k ⊆ αk ( ) [ [ [ ( ) ∗ ∗ X = Bk Aαk Aα = X |{z} ⊆ ⊆ k K Aα k K α I ∈ ⊆ k ∈ ∈ S So k K Aαk = X. ∈ Proof of Theorem 1.33. (a) By contradiction: Assume X is compact but there exists a sequence (xn)n N X without convergent subsequence. ∈ ⊂ Claim: x X there exists a neighbourhood U(x) such that x U(x) for at most ∀ ∈ n ∈ ﬁnitely many n.

Proof of the claim: Consider a countable neighbourhood base Ui i N at x and let Tm { } ∈ Vm := Ui for m N. Suppose the claim was false. Then for every m N there i=1 ∈ ∈ exist in finitely many k N such that xk Vm. This means there exists a subsequence ∈ ∈ (km)m N increasing such that xkm Vm for every m N. So, for every m, m0 N, ⊆ ∈ ∈ ∈ m0 m: xkm0 Vm. This implies limm xkm = x, i.e. convergent subsequence. ≥ ∈ →∞ Now X = S U(y) = Sn U(y ) for some y , . . . , y X (because X is compact). y X i=1 i 1 n ∈ The claim implies∈ that U(y ) contains at most finitely many sequence members x i k ∈ U(y ), so there are at most finitely many k such that x X. i k ∈

16 (b) “= ” follows from (a) and Lemma 1.17. ⇒ We now show “ =” by contradiction: Assume every sequence has a convergent sub- ⇐ sequence, but there exists an open cover without ﬁnite subcover. Since X is 2nd S countable there exists a countable subcover X = j N Cj of this cover by Lindel¨of. ∈ For every n N pick a point ∈ n [ x X C n ∈ \ j j=1

(always possible n N because there does not exist a finite subcover). Let (xn)n X ∀ ∈ k ⊂ be a sequence in X. By hypothesis, it has a convergent subsequence x →∞ x X. nk −−−→ ∈ There exists N N such that x CN , so CN is a neighbourhood of x. xnk being ∈ ∈ convergent means x C for finally all k, but for every k such that n N we nk ∈ N k ≥ have by definition of x that x / C . n nk ∈ N 1.35 Theorem. Let X be a compact topological space and A X. Then ⊆ (a) A closed = A compact ⇒ (b) If X is also Hausdorff, then A compact = A closed. ⇒ S c Proof. (a) Let A α I Uα be an open cover. Since A is closed A is open and ⊆ ∈ [ X = Ac U ∪ α α I ∈ is an open cover. Since X is compact, there exist α , . . . , α I such that 1 n ∈ n [ X = Ac U ∪ αi i=1 and A Sn U is a finite subcover. ⊆ i=1 αi (b) See Problem 8.

1.36 Warning. Bounded and closed do not imply compact in general! Example: `p, p [1, ] and ∈ ∞ B (0) = x `p : x 1 = x `p : x < 1 = B (0) 1 { ∈ k kp ≤ } { ∈ k kp } 1 is bounded and closed but consider e(n) := (..., 0, 1, 0,...) `p (with 1 at the nth position) ∈ n N. It fulﬁlls ∈ ( 1/p (n) (m) (n) (m) 2 p < dp(e , e ) = e e p = ∞ n, m N, n = m, k − k 1 p = ∀ ∈ 6 ∞ so there exists no convergent subsequence and B1(0) is not sequentially compact. Hence, by Theorem 1.33 (a), B1(0) is not compact. 1.37 Theorem. Let X be a metric space. Then

(a) X is compact ks +3 X is sequentially compact

(c)

X is 2ndcountable ks +3 X is separable. (b)

17 Proof. (a)( ) This is Theorem 1.33 (a). ⇒ ( ) Using the implication (c) and the equivalence (b), which will be proven below: ⇐ If X is sequentially compact it is also 2ndcountable and we can apply Theo- rem 1.33 (b).

(b) For any topological space X 2nd countability implies separability (Theorem 1.12). If X is a separable metric space, then it is also 2nd countable (Theorem 1.19).

(c) We show that sequential compactness implies separability by constructing a countable (n) set M with M = X. Fix n N and use the following algorithm to deﬁne points x : ∈ k Choose an arbitrary x(n) X • 1 ∈ (n) (n) (n) (n) If R := X B1/n(x ) = ∅, pick x R , otherwise stop. • 1 \ 1 6 2 ∈ 1 Suppose x(n), . . . , x(n) are chosen. If • 1 k k (n) [ (n) R := X B1/n(x ) = ∅, k \ 1 6 j=1

pick x(n) R(n), otherwise stop. k+1 ∈ k Claim: This algorithm stops after finitely many steps. (n) (n) True, because for k = l we have d(xk , xl ) > 1/n. Thus, if the algorithm did 6 (n) not stop after finitely many steps we would have an infinite sequence (x )l N X l ∈ ⊂ without a convergent subsequence which is a contradiction to X being sequentially compact.

The claim implies the existence of Kn N such that ∈ Kn [ (n) X = B1/n xj . j=1

(n) S Set Mn := xj : j = 1,...,Kn and M := k N Mn. M is countable and M = X. { } ∈ 1.38 Theorem (Tychonoﬀ). Let J = ∅ be an index set and Xα a compact topological 6 space for all α J. Then ∈ n [ o X = f : J X such that f(α) X α −→ α ∈ α α J α J ×∈ ∈ is compact (in the product topology).

Proof. See any textbook on topology, e.g. Kelley, v. Querenburg.

1.39 Deﬁnition. Let X,Y be topological spaces. Deﬁne

C(X,Y ) := f : X Y such that f is continuous { −→ } in particular for Y = C set C(X) := C(X, C). 1.40 Theorem. Let X,Y be topological spaces, X compact, f C(X,Y ). Then ∈ (a) f(X) is compact.

18 (b) If in addition X,Y are Hausdorﬀ and f a bijection, then f is a homeomorphism.

ε > 0 δ δ : x X : f(B (x)) B (f(x)) ∀ ∃ ≡ ε ∀ ∈ δ ⊆ ε

(Note that Bδ(x) and Bε(f(x)) are balls corresponding to potentially diﬀerent metrics. However we will not use diﬀerent notations for them as long as it is clear from the context to which metric they belong).

(d) If even Y = R then f takes on its maximum and minimum. Proof. (b), (c), (d): See Problems 9, 10 and T7 . S (a) Let f(X) α J Fα be openly covered. Then ⊆ ∈ 1 [ [ 1 X f − F = f − (F ). ⊆ α α α J α J ∈ ∈ Because f is continuous f 1(F ) is open α J and we have an open cover of X. − α ∀ ∈ But X is compact which by deﬁnitions means that we ﬁnd N N and α1, . . . , αn such ∈ that X SN f 1(F ) and, thus, f(X) SN F . ⊆ n=1 − αn ⊆ n=1 αn 1.6 Example: spaces of continuous functions General assumptions in this subsection:

X is a compact Hausdorﬀ space, • K means R or C, • C(X, K) is equipped with the uniform (supremum) metric • d (f, g) := sup f(x) g(x) = f g . ∞ x X | − | k − k∞ ∈ (well deﬁned by Theorem 4.40 (a)).

1.41 Theorem. C(X, K) is complete.

Proof. Follows from completeness of the bounded continuous functions Cb(X, K), see Problem 5, and C(X, K) = Cb(X, K), which follows from compactness of X and The- orem 4.40.

1.42 Theorem. X metrisable C(X, K) separable. ⇐⇒ Proof. For ” =” see e.g. Bourbaki, Elements of Mathematics, General Topology, Part 2, ⇐ Sect. X.3.3. Thm. 1.1. Here, we only show ”= ”: ⇒ For m, n N deﬁne ∈ Gmn := f C(X, K): f(B1/m(x)) B1/n(f(x)) x X ∈ ⊆ ∀ ∈

Compactness of X implies that,

19 ... f C(X, K) is automatically uniformly continuous. Fix n and consider Theorem • ∈ 4.40(c) with ε := 1/n. f is uniformly continuous, i.e. there exists δ > 0 such that

x X : f(B (x)) B (f(x)). ∀ ∈ δ ⊆ 1/n

Then we can ﬁnd m N such that 1/m δ and get that f Gmn. This shows ∈ ≤ ∈ [ C(X, K) = Gmn n N. (1) ∀ ∈ m N ∈

. . . for any m N we can ﬁnd p(m) N and a1, . . . , ap(m) X such that X can be • ∈ ∈ ∈ written a union of open balls of radius 1/m:

p(m) [ X = B1/m(aj). (2) j=1

Now, K is separable, i.e. there exists a countable set κν K: ν N that is dense in K. { ∈ ∈ } For fixed m N and ϕ: 1, . . . , p(m) N (or equivalently ϕ Np(m)) define ∈ { } −→ ∈ (ϕ) G := f Gmn : f(ak) κϕ(k) < 1/n k = 1, . . . , p(m) . mn { ∈ | − | ∀ } (ϕ) We only want to consider those ϕ with Gmn = ∅, so we set 6 p(m) (ϕ) Φmn := ϕ N : G = ∅ . { ∈ mn 6 } (ϕ) Φ is not empty (see below). For ϕ Φ pick some g Gmn. Now define mn ∈ mn ϕ ∈ [ L := g : ϕ Φ and L := L mn { ϕ ∈ mn} mn m,n N ∈ p(m) Lmn is countable because Φmn N . Thus, L is countable as well. We want to show ⊆ that L is dense in C(X, K). Pick an arbitrary f C(X, K). Because of (1) we have: ∈ (ϕf ) n N m N such that f Gmn. There also exists ϕf Φmn such that f Gmn : We ∀ ∈ ∃ ∈ ∈ ∈ ∈ only have to choose ϕf (k) such that f(ak) κϕ (k) < 1/n. But κν : ν N is dense in | − f | { ∈ } K, so this is obviously possible. For each x X choose k 1, . . . , p(m) such that x B (a ) (this is possible ∈ x ∈ { } ∈ 1/m kx because of (2)). Now we consider g (x) L L. Using that f and g both lie in ϕf ∈ mn ⊆ ϕf (ϕf ) Gmn we can estimate the upper bound of the distance f(x) g : | − ϕf (x)|

f(x) gϕf (x) f(x) f(akx ) + f(akx ) κϕ (kx) + | − | ≤ | − | | − f | | {z } | {z } <1/n by definition of Gmn ϕf <1/n by definition of Gmn 4 + κϕ (kx) gϕf (akx ) + gϕf (akx ) gϕf (x) < | f − | | − | n | {z } | {z } ϕf <1/n by definition of Gmn <1/n by definition of Gmn

Thus, the distance vanishes for n and L is countable and dense in C(X, K), i.e. → ∞ C(X, K) is separable. An alternative way to separability:

20 1.43 Definition. (a)A K-vectorspace A is a K-algebra iff there exists a multiplication A A A which fulfils the distributive laws × −→ (a + b)c = ac + bc a, b, c A, ∀ ∈ c(a + b) = ca + cb a, b, c A, ∀ ∈ λ(ac) = (λa)c = a(λc) a, c A, λ K. ∀ ∈ ∈ (b) A subspace B A is a subalgebra iff B is closed under multiplication. ⊆ (c) A subset B C(X, K) separates points in X iff for all distinct points x, y X there ⊆ ∈ exists f B such that f(x) = f(y). ∈ 6 1.44 Theorem (Stone-Weierstraß). Let B C(X, K) be a subalgebra with the following properties: ⊆

1 B (where 1: X K, x 1 x X)4, • ∈ −→ 7−→ ∀ ∈ B is closed with respect to d , • ∞ B separates points. • If K = C, assume further that B is closed under complex conjugation. • Then B = C(X, K). Proof. e.g. Reed/Simon, vol. 1, Appendix to Sect. IV.3

1.45 Corollary. Let K Rd compact, d N. Then the set of all polynomials is dense ⊆ ∈ (with respect to d ) in C(K, K). In particular, C(K, K) is separable. ∞

Proof. Let B0 be the K-Algebra generated by the monomials K K −→ n x = (x1, . . . , xd) x n N0, α 1, . . . , d 7−→ α ∈ ∈ { }

Set B := B0. Then Stone-Weierstraß gives us B = C(K, K). Now we show separability: Q Q If K = R, let B0 be the Q-Algebra generated by the monomials, if K = C, let B0 be the (Q + iQ)-algebra. Then BQ is countable • 0 Q Since K is compact, B = B0 = C(K, K). • 0 1.46 Definition. Let X be a metric space (not necessarily compact). Let be a family F of continuous functions f : X K. −→ is equicontinuous iff for every ε > 0 and x X there exists a δ > 0 such that for •F ∈ every f : ∈ F f(B (x)) B (f(x)) δ ⊆ ε is uniformly equicontinuous iff for every ε > 0 there exists δ > 0 such that for •F every x X and f : ∈ ∈ F f(B (x)) B (f(x)) δ ⊆ ε 4This property is called “unital”

21 1.47 Remark. (a) If X is even compact, then (by Problem T7): equicontinuous ⇐⇒ uniformly equicontinuous. (b) Examples

X = [0, 1]: x cos(x/n): n N is uniformly continuous • { 7−→ ∈ } X = [0, 1]: x x1/n : n N is not equicontinuous. • { 7−→ ∈ } X = ]0, [: x arctan(nx): n N is equicontinuous, (individually) • ∞ { 7−→ ∈ } uniformly continuous, but not uniformly equicontinuous.

1.48 Theorem. Let X be a metric space (not necessarily compact) and (fn)n N ∈ ⊆ C(X, K) an equicontinuous sequence. Suppose there exists a dense subset D X such ⊆ that for every x D the limit limn fn(x) exists. Then limn fn(x) =: f(x) exists ∈ →∞ →∞ x X and the limit function f C(X, K). ∀ ∈ ∈ Proof. Firstly, we show convergence everywhere. Let x X be arbitrary but ﬁxed and let ∈ ε > 0. By assumption there exists δ > 0 such that for every n N ∈ f (B (x)) B (f (x)). (1) n δ ⊆ ε n Since D = X, there is some y D B (x). At y, the sequence (f (y)) converges and is ∈ ∩ δ n n therefore a Cauchy sequence, so there exists N N such that n, m N ∈ ∀ ≥ f (y) f (y) < ε. (2) | n − m | So we have

fn(x) fm(x) fn(x) fn(y) + fn(y) fm(y) + fm(y) fm(x) | − | ≤| {z− }| | −{z }| | {z− }| <ε by (1) <ε by (2) <ε by (1) < 3ε

So limn fn(x) =: f(x) exists for every x X. →∞ ∈ Secondly, we show the continuity of f. (1) is equivalent to

δ > 0 n N x0 X with d(x, x0) < δ = fn(x) fn(x0) < ε, ∃ ∀ ∈ ∀ ∈ ⇒ | − | so for n there exists δ > 0, such that for every x X with d(x, x ) < δ we have → ∞ 0 ∈ 0 f(x) f(x ) < ε. | − 0 | 1.49 Lemma. In addition to the assumptions in Theorem 1.48 suppose that X is compact. Then we have even limn d (fn, f) = 0. →∞ ∞ Proof. We need to show uniform convergence. Let ε > 0. Since X is compact, there exists δ > 0 such that for every x X and n N: ∈ ∈ f (B (x)) B (f(x)) (1) n δ ⊆ ε (see Remark 1.47) and there is l N and a1, . . . , al X: ∈ ∈ l [ X = Bδ(aj). (2) j=1 We have

fn(x) f(x) fn(x) fn(ajx ) + fn(ajx ) f(ajx ) + f(ajx ) f(x) | − | ≤| −{z }| | {z− }| | {z− }| =:T1 =:T2 T3 where a is the centre of some Ball B such that x B (a ). jx δ ∈ δ jx

22 T1 < ε for every n N by (1) (and the definition of ajx ) • ∈ T < ε for all n sufficiently large (there are only finitely many j’s and pointwise • 2 convergence)

T < ε by taking n in (1) (compare the end of the proof of Thm. 1.48). • 3 → ∞ So ﬁnally we have

ε > 0 N N n N x X : fn(x) f(x) < 3ε. ∀ ∃ ∈ ∀ ≥ ∀ ∈ | − |

1.50 Theorem (Arzel`a-Ascoli). Let X be a compact metric space and (fn)n N C(X, K) an equicontinuous and pointwise bounded sequence, i.e. ∈ ⊂

sup fn(x) < x X. n N | | ∞ ∀ ∈ ∈

Then there exists a uniformly convergent subsequence (fnj )j N. Equivalently, every equicon- ∈ tinuous and pointwise bounded subset C(X, K) is relatively compact. F ⊆ Proof. Since C(X, K) is a metric space the equivalence of the 2 statements follows from Theorem 1.37. We prove the “sequence version”: Since X is compact, due to Theorem 1.37, X is also separable, so there exists a dense subset al X : l N X. Pointwise { ∈ ∈ } ⊆ boundedness gives sup fn(al) < l N, n N | | ∞ ∀ ∈ ∈ (l) so by the Bolzano-Weierstraß theorem for every l there exists a subsequence (nj )j N N ∈ ⊆ such that lim f (l) (al) j nj →∞ (l+1) (l) (j) exists. W.l.o.g. assume (nj )j (nj )j. Deﬁne nj := nj (diagonal sequence trick), (l) ⊆ then (nj)j l (nj )j and limj fnj (al) exists for every l N. Since the set of the al’s ≥ ⊆ →∞ ∈ is dense, Lemma 1.49 gives the claim.

1.51 Remark. Both assumptions, compactness and equicontinuity are essential for The- orem 1.50 to turn uniform boundedness into convergence of a subsequence.

1.7 Baire’s Theorem . . . the mother of 3 (out of 4) fundamental theorems of functional analysis.

1.52 Remark. Let X be a metric space and A ,A X be open and dense. Then A A 1 2 ⊆ 1 ∩ 2 is also open and dense.

Completeness gives more:

1.53 Theorem (Baire). Let X be a complete metric space and n N let An X be open and dense. Then T is dense in . ∀ ∈ ⊆ n N An X ∈ 1.54 Remark. T (a) Openness of n N An is false in general. ∈ (b) Completeness is essential for the theorem: Consider Q with the metric inherited from R. Let qn Q : n N be an enumeration of Q. Deﬁne An := Q qn . This is { ∈ ∈ }T \{ } open and dense in Q but n N An = ∅. ∈

23 Figure 5: Constructing the sequence (xn).

Proof. T Deﬁne D := n N An. Let x0 X be arbitrary and ﬁx and ε > 0. To prove the ∈ ∈ denseness of D in X we have to show that D Bε(x0) = ∅. We do this by constructing a ∩ 6 sequence (x ) that converges against x D B (x ) (see Figure5). n ∈ ∩ ε 0 A dense implies that A B (x ) is non-empty and we can pick x A B (x ) and 1 1 ∩ ε 0 1 ∈ 1 ∩ ε 0 ε > 0 with ε ]0, ε/2[ such that 1 1 ∈ B (x ) A B (x ). ε1 1 ⊆ 1 ∩ ε 0 Proceeding analogously with A and B (x ) we pick x A B (x ) and ε ]0, ε /2[ 2 ε1 1 2 ∈ 2 ∩ ε1 1 2 ∈ 1 such that B (x ) A B (x ) A A B (x ). ε2 2 ⊆ 2 ∩ ε1 1 ⊆ 1 ∩ 2 ∩ ε 0 We proceed in the same fashion and get two sequences:

n (a)( εn)n with 0 < εn < 2− ε n N, ∀ ∈ (b)( x ) X with B (x ) A B (x ) A A B (x ) n n ⊂ εn+1 n+1 ⊆ n+1 ∩ εn n ⊆ 1 ∩ · · · ∩ n ∩ ε 0 In particular we have:

N N n N : xn BεN (xN ). ( ) ∀ ∈ ∀ ≥ ∈ ∗

That is, (xn)n is Cauchy and since X is complete, (xn) converges to some x X. But T ∈ because of ( ) we have x BεN (xN ) N N and (b) yields x n N Bεn+1 (xn+1) ∗ ∈ ∀ ∈ ∈ ∈ ⊆ D Bε(x0). Thus, D Bε(x0) = ∅. ∩ ∩ 6 1.55 Definition. Let X be a topological space and A X. ⊆ 1. A is a Gδ(-set) iff A is a countable intersection of open sets. 2. A is nowhere dense iff A has no interior points.

3. A is meagre (or of 1st category) iﬀ A is a countable union of nowhere dense sets.

4. A is non-meagre (or of 2nd category) iﬀ A is not meagre.

24 1.56 Example. S Q is meager in R (just consider the countable union Q = q Q q ). ∈ { } 1.57 Lemma. Let X be a topological space and A X. Then ⊆ (a) A is nowhere dense (A)c dense, ⇐⇒ (b) A is meagre and B A = B is meagre, ⊆ ⇒ (c) meagre T meagre. An X n N = n N An ⊆ ∀ ∈ ⇒ ∈ Proof. (b) and (c) are clear by the very deﬁnitions. Statement (a) follows from the equivalence B has no interior points Bc dense. ⇐⇒ To show ”= ”, consider x X. If x B then x BC . If x B it is by hypothesis no ⇒ ∈ 6∈ ∈ ∈ interior point of B and thus must be a limit point of BC .

We now show 3 equivalent reformulations of Baire’s theorem:

1.58 Lemma. Let X be a topological space. Then the statements (i)–(iv) are equivalent:

(i) open and dense T dense. An X n N = n N An ⊆ ∀ ∈ ⇒ ∈ (ii) is a dense T is a dense . An X Gδ n N = n N An Gδ ⊆ ∀ ∈ ⇒ ∈ (iii) A X,A = ∅ and A is open = A non-meagre. ⊆ 6 ⇒ (iv) A X meagre = AC dense. ⊆ ⇒ 1.59 Corollary. Let X be a complete metric space. Then (i)–(iv) in Lemma 1.58 hold. In particular if X = ∅ then X is non-meagre. 6 Proof of Lemma 1.58. (i) (ii) by deﬁnition of G . ⇔ δ S (i) (iii) Let ∅ = A X be open. Suppose A is meagre, that is A = n N An with ⇔ 6 ⊆ ∈ C An nowhere dense n N. According to 1.57 (a), this is equivalent to (An) dense ∀ ∈ T C and open n N and because of implication (i), the intersection (An) =: B ∀ ∈ n N is dense as well. But then BC = S A A has no interior points.∈ Thus A has n N n ⊇ no interior points either and this is clearly∈ a contradiction to A being open.

(iii) (iv) Problem 13. ⇒ (iv) (i) Problem 13. ⇒

25 2 Banach and Hilbert spaces

2.1 Vector spaces

General assumption: X = 0 is a K-vector space, where K = R or C. 6 { } 2.1 Definition. Let ∅ = M X. 6 ⊆ M is linearly independent iff all non-empty finite subsets F M are linearly inde- • ⊆ pendent, i.e. the following implication holds: X α f = 0 = α = 0 f F. f ⇒ f ∀ ∈ f F |{z} ∈ K ∈ M is linearly dependent iff M is not linearly independent. • B X is a Hamel basis (or algebraic basis) iff • ⊆ – B is linearly independent – every x X can be represented as a finite linear combination from elements in ∈ B (B ”spans” X).

X has ﬁnite dimension if there exists a Hamel basis B with B < . dim X := B • | | ∞ | | is called the dimension of X.

X has infinite dimension iff X does not have finite dimension. • 2.2 Remark. The dimension is well defined: B is the same for every Hamel basis in a | | given space.

2.3 Example. Consider

cc := x = (xj)j N : xj C j N and xj = 0 for only ﬁnitely many j’s . { ∈ ∈ ∀ ∈ 6 }

The index c stands for “compact support”. Let em := (..., 0, 1, 0,... ) with a 1 at the n-th position. Claim: B := en : n N is a Hamel basis for cc. { ∈ } Remark. Even though `1 is separable there exists no countable hamel basis for `1.

2.4 Theorem. Every vector space X = 0 has a Hamel basis. 6 { } Proof. Uses Zorn’s lemma. See later.

2.5 Corollary. X has infinite dimension iff for every n N there exists Mn X such ∈ ⊆ that M = n and M is linearly independent. | n| n Proof. Existence of a Hamel basis B with B = . | | ∞ p d 2.6 Example. Infinite dimensional vector spaces: cc, ` , C(X) (where ∅ = X R 6 ⊆ open).

26 2.2 Banach spaces 2.7 Deﬁnition. Let X be a vector space. A mapping X [0, [, x x is a norm −→ ∞ 7−→ k k iﬀ

(1) x > 0 0 = x X, k k ∀ 6 ∈ (2) λx = λ x λ K x X, k k | |k k ∀ ∈ ∀ ∈ (3) x + y x + y x, y X k k ≤ k k k k ∀ ∈ (X, ) is called a normed space. If only (2) and (3) hold, is called a seminorm. k · k k · k 2.8 Remark. Let X be a normed space. Then d(x, y) := x y is a metric on X. Thus k − k all topological notions and results from the theory of metric spaces are available.

2.9 Example. `p is a normed space p [1, ]. • ∀ ∈ ∞ C(X, K) (where X a compact Hausdorﬀ space) is a normed space with • f := f := sup f(x) k k k k∞ x X | | ∈

Rn, Cn are normed spaces for n N (Euclidian norm or p-norm). • ∈ 2.10 Lemma. Let X be a normed space. Then

the mapping X x x is continuous. • 3 7−→ k k Addition and multiplication are continuous as well:

k k k Let x →∞ x and y →∞ y. Then x + y →∞ x + y. • k −−−→ k −−−→ k k −−−→ k k k Let αk →∞ α in K and xk →∞ x. Then αkxk →∞ αx. • −−−→ −−−→ −−−→ k x Proof. Let (x ) X with x →∞ x. This is equivalent to x x →∞ 0 • k k ⊆ k −−−→ k k − k −−−→ (Corollary 1.18). To show that is sequentially continuous, we have to prove that k k · k xk x →∞ 0. This follows from the following inequalities: k k − k k −−−→ x x x x x + x = x x + x x k kk − k − kk ≤ k − k kk k k ≤ k kk k − kk = lim sup xk x lim inf xk ⇒ k k k ≤k k ≤ k k k →∞ →∞ The following sets are a base of the metric topology of X:

x + B1/k(0) : x X, k N = B1/k(x) = y X : x y 1/k , { ∈ ∈ } { } { ∈ k − k ≤ } Here we used the following notation for sets A, B: A + B = a + b : a A, b B and { ∈ ∈ } a + B := a + B. Warning: In general not every metric comes from a norm. { } Addition is continuous: • k xk + yk x y xk x + yk y →∞ 0. k − − k ≤| k {z− k} |k {z− k} −−−→ k→∞ k→∞ 0 0 −−−→ −−−→

27 Multiplication is continuous: • α x αx = α x αx + α x α x = k k k − k k k k − k − k k = α (x x) + x(α α) α (x x) + (α α)x k k k − k − k ≤ k k k − k k k − k ≤ k αk xk x + αk α x →∞ 0. ≤|{z} | | |k {z− k} | {z− }| k k −−−→ bounded in k k→∞ k→∞ 0 0 −−−→ −−−→ 2.11 Deﬁnition. A complete normed space is called a Banach space.

2.12 Examples. All spaces in Example 2.9 are Banach spaces. • Consider C([0, 1]) with L1 norm: f := R 1 f(x) dx is not a Banach space. This • k k1 0 | | was already discussed in Example 1.21 (b).

2.13 Theorem. Every normed space X can be completed, so that X is isometric to a dense linear subspace W of the Banach space Xˆ, which is unique up to isometric isomorphisms.

Proof. Analogous to the proof of Theorem 1.24. Note that the isometry is even a linear bijection (thus a isomorphism) in this case (see Section 2.3).

2.14 Definition. Let X be a normed space. A set en X : n N is a (Schauder) basis { ∈N ∈ } in X iff for all x X there exists a sequence (xn)n K such that ∈ ⊆ N X lim x xnen = 0. N − →∞ n=1 P Notation: x = n N xnen ”infinite linear combinations”, ”convergent series”. ∈

2.15 Example. Let p [1, [. Then (en)n N with en := (0,..., 1, 0,... ) (where the 1 is ∈ ∞ ∈ p p at the nth position) is a (Schauder) basis of ` : For x = (xn)n N ` we have ∈ ∈ N p X X∞ p N x xnen = xn →∞ 0. − | | −−−−→ n=1 p n=N+1

Note that this construction fails for `∞! 2.16 Lemma. Let X be a normed space. Then X has a (Schauder) basis = X is separable. ⇒ Proof. Let K0 := Q for K = R, respectively K0 := Q + iQ for K = C. Deﬁne

  N X  AN := xnen : xn K0 . ∈ n=1  S Then the union A := N N AN is dense in X and countable. ∈ 2.17 Remark. The implication ” =” in Lemma 2.16 does not hold (Enﬂo, 1973). ⇐ 2.18 Lemma. Let X be a Banach space, A X a (linear) subspace. Then ⊆ A is closed A is complete. ⇐⇒

28 Proof. See Analysis II, like proof of Lemma 1.20. 2.19 Theorem. Let X be a normed space and F X a finite-dimensional subspace. ⊆ Then F is complete and closed. Proof. Choose a basis e , . . . , e in F .(F, ) is a normed space and isometric to { 1 n} k · k (Kn, ) with ||| · ||| n X n α := αjej α = (α1, . . . , αn) K . ||| ||| ∀ ∈ j=1 Kn is complete with respect to the Euclidean norm. But all norms in finite-dimensional spaces are equivalent5. Thus (Kn, ) is complete and because of the isometry, (F, ) ||| · ||| k · k is complete as well. F is closed by Lemma 2.18 with X = A = F . As a preparation for Theorem 2.21 we prove the following lemma: 2.20 Lemma (Riesz, 1918). Let X be a normed space and U ( X a closed subspace. Then for all λ ]0, 1[ there exists x X U such that ∈ λ ∈ \ x = 1 and x u λ u U. k λk k λ − k ≥ ∀ ∈ Proof. Since U is closed, we have d := dist(x, U) = infu U d(x, u) > 0 for all x X U ∈ ∈ \ (see Problem 9(c)!). Since λ < 1 there exists u U such that λ ∈ d 1 λ d x u , hence γ := . ≤ k − λk ≤ λ x u ≥ d k − λk Define x := γ(x u ) X U. The first property x = γ x u = 1 holds by λ − λ ∈ \ k λk | | · k − λk definition of γ and x u = γ(x u ) u = γx (u + γu ) = k λ − k k − λ − k k − λ k u = γ x u + γd λ u U. | | · − λ γ ≥ ≥ ∀ ∈ | {z } U ∈ Warning 1.36 illustrates the more general 2.21 Theorem. Let X be a normed space. Then B (0) = x X : x 1 compact dim X < 1 { ∈ k k ≤ } ⇐⇒ ∞ Proof. ( ) Let dim X < . Proof of Theorem 2.19: X is isometric to (Kn, ). The ⇐ ∞ ||| · ||| statement then follows by Heine-Borel and the equivalence of norms. ( ) Suppose that dim X = . We show that this implies that B (0) is not sequentially ⇒ ∞ 1 compact by constructing a sequence (xn)n in B1(0) without a convergent subsequence: Pick an arbitrary x X with x = 1. Let U := span x be the subspace • 1 ∈ k 1k 1 { 1} spanned by x1. U1 is closed in X. Riesz Lemma, applied with λ = 1/2 shows the existence of x X U such • 2 ∈ \ 1 that x = 1 and x x 1/2. Let U := span x , x . k 2k k 2 − 1k ≥ 2 { 1 2} If we continue this procedure we get a sequence (xn)n N B1(0) that satisfies • ∈ ⊆ x x 1/2 for all n = m. k n − mk ≥ 6 (xn)n clearly has no convergent subsequence. Thus B1(0) is not compact.

5 n This is a theorem from linear algebra: For norms k·k and |||·||| on K we can ﬁnd constants c, c˜ ∈]0, ∞[ such that c · k · k ≤ ||| · ||| ≤ c˜ · k · k.

29 2.3 Linear operators 2.22 Definition. Let X,Y be vector spaces (over the same field), X X a subspace 0 ⊆ and T : X Y . 0 −→ T is linear (a linear operator) iff T (αx+βy) = αT (x)+βT (y) α, β K, x, y X0. • ∀ ∈ ∈ Notation: T x := T (x). T is a (vector space) homomorphism.

dom(T ) := X is the domain of T . • 0 ran T := T (X ) is the range of T . • 0 ker(T ) := x X : T x = 0 is the kernel of T . • { ∈ 0 } 2.23 Examples. (a) The identity operator 1 := 1 : X X, x x is obviously a X −→ 7−→ linear operator.

(b) Consider X = Y = C([0, 1]).

Then the diﬀerentiation operator T : X = C1([0, 1]) C([0, 1]) with f f , • 0 −→ 7−→ 0 so (T f)(x) = f 0(x) and this is a linear operator. x (T f)(x) := R f(t)dt has dom T = X and is called the anti-derivative. • 0 (T f)(x) := xf(x) is the multiplication by x. • 2.24 Lemma. Let T be a linear operator. Then

(a) ran(T ) and ker(T ) are vector subspaces

(b) dim ran(T ) dim dom T ≤ 1 1 (c) ker(T ) = 0 there exists an inverse T − of T such that T − : ran(T ) { }1 ⇐⇒ 1 −→ dom(T ) and T − T = dom(T ). Proof. Copy from linear algebra.

2.25 Remark. If T 1 exists, it is linear. • − Even if we have T : X X with ker(T ) = 0 (then T 1 exists) this does not • −→ { } − imply TT 1 = 1, but only TT 1 = 1 . In other words, ker(T ) = 0 does not − − ran(T ) { } imply T is bijective (only injective).

2.26 Example (illustrating Remark 2.25). Let X = ` and x = (x , x ,...) X. ∞ 1 2 ∈ Consider the shift operator deﬁned by T x := (0, x1, x2,...), i.e. ( 0 n = 1, (T x)n := xn 1 n 2. − ≥

Then we have dom T = `∞ and ran T = y = (y1, y2,...) `∞ : y1 = 0 . Deﬁne 1 1 { 1 ∈ 1 } (T − x)n := (x2, x3,...). Then dom T − = `∞ and T − T = 1. But TT − x = x holds only for x ran(T ). Even though dom(T ) = X and ker(T ) = 0 , we have ran(T ) X. This ∈ { } is not possible if Y = X and dim X < . ∞

30 2.27 Definition. Let X,Y be normed spaces, T : X dom(T ) Y . T is bounded iff ⊇ −→ its operator norm is finite, i.e.

T x T dom(T ) Y T := sup k k = sup T x < . k k −→ ≡ k k x dom(T ) x x dom(T ) k k ∞ ∈ x=0 k k ∈ x =1 6 k k 2.28 Examples. (cf. Example 2.23)

(a) 1:(X, ) (X, ) is bounded with 1 = 1. k · k −→ k · k k k (b) Let X = C([0, 1]) with . k·k∞ d T = dx is unbounded. For n 2 take fn : x sin(nx). Then fn = 1 and • ≥ 7−→ k k∞

T fn = sup n cos(nx) = n k k x [0,1] | | ∈ so that fn0 T k k∞ = n. k k ≥ fn k k∞ The anti-derivative fulﬁlls • Zx

T f = sup f(t)dt 1 f k k∞ x [0,1] ≤ k k∞ ∈ 0

for every f C([0, 1]) and f = 0. Because of ∈ 6 T f k k 1 f ≤ k k we obtain T 1. Moreover, k k ≤ Zx T 1 = sup dt = sup x = 1, k · k∞ x [0,1] x [0,1] ∈ 0 ∈

and since 1 = 1 we have T = 1. k k∞ k k The multiplication operator by x has T = 1. • k k 2.29 Theorem. Let X,Y be normed spaces, T : X dom(T ) Y a linear operator. Then the following statements are equivalent. ⊇ −→

(a) T is continuous.

(b) T is continuous in some x dom(T ) 0 ∈ (c) There exists c ]0, [ such that T x c x for every x dom(T ). ∈ ∞ k k ≤ k k ∈ (d) T is bounded.

Proof. (a) (b) obvious. ⇒

31 (b) (c) Suppose T is continuous at x . Claim: It follows continuity at 0. This is true ⇒ 0 because let (x ) dom(T ) be a sequence which converges to 0. Then n n ⊂ n (x + x ) →∞ x , n 0 n −−−→ 0

and by linearity and continuity in x0 we obtain

n T x + T x = T (x + x ) →∞ T x . n 0 n 0 −−−→ 0 n So T x →∞ 0. From this we know that for ε = 1 there is a δ > 0 such that n −−−→ whenever x˜ δ forx ˜ dom(T ) we have T x˜ 1. For general 0 = x dom(T ), xk k ≤ ∈ k k ≤ 6 ∈ letx ˜ := x δ, so that x˜ δ. Then we have k k k k ≤

δ 1 T x = T x˜ 1 = T x δ− x , x k k k k ≤ ⇒ k k ≤ k k k k i.e. c = 1/δ.

n (d) (a) Let (x ) dom(T ) with x →∞ x then ⇒ n n ⊆ n −−−→ n T x T x = T (x x) T x x →∞ 0 k n − k k n − k ≤ k k · k n − k −−−→ 2.30 Lemma. Let T be linear and dim dom(T ) < . Then T is bounded. ∞ Proof. See linear algebra.

2.31 Deﬁnition. Let X,Y be vector spaces and T : X dom(T ) Y . Let U dom(T ) ⊇ −→ ⊆ and W dom(T ). ⊇ Restriction of T to U: T : U Y, x T x := T x |U −→ 7−→ |U Extension of T to W : T˜ : W Y with T˜ x = T x x dom(T ), i.e. T˜ = T . −→ ∀ ∈ |dom(T ) 2.32 Theorem (Bounded linear extension). Let X be a normed space and Y a Banach space. Let T : X dom(T ) Y be a bounded linear operator. Let Z be the completion of ⊇ −→ dom(T ). Then there exists a bounded linear extension T˜ : Z Y of T with T˜ = T . −→ k k k k If X is identiﬁed with a subspace of its completion X˜, i.e. X = W in Theorem 2.13, then T˜ is unique.

NB. (a) If X is a Banach space, then the uniqueness holds anyway.

(b) If dom(T ) X, this gives extension to closure as a special case. ⊆ n Proof. Let x Z. Then there exists a sequence (x ) dom(T ) with x →∞ x in X˜. ∈ n n ⊆ n −−−→ Since (xn)n is a Cauchy sequence, (T xn)n is a Cauchy sequence in Y because

T x T x = T (x x ) T x x . k n − mk k n − m k ≤ k k · k n − mk n Using that Y is a Banach space, there exists y Y such that T x →∞ y in Y . Deﬁne ∈ n −−−→ T˜ x := y (then T˜ = T ). We have to check several things: |dom(T )

32 m Well-deﬁnedness, i.e. independence of approximating sequence: Letx ˜ →∞ x in • m −−−−→ X˜. Then we have T x T x˜ T x x˜ . k n − mk ≤ k k · k n − mk In the limit n, m we get y limm T x˜m = 0. → ∞ k − →∞ k n n Linearity: Let x →∞ x, x →∞ x . Then • n −−−→ n0 −−−→ 0

T˜(αx + α0x) = lim T (αxn + α0x0 ) = lim αT xn + lim α0T x0 = αT˜ x + αT˜ x0. n n n n n →∞ →∞ →∞ Norm: as T˜ is an extension, we have T˜ T , because • k k ≥ k k T˜ x T˜ x T˜ = sup k k sup k k = T . k k x dom(T˜) x ≥ x dom(T ) x k k ∈ ∈ x=0 k k x=0 k k 6 6 On the other hand, since is continuous k·k T˜ x = lim T˜ x k k n k nk xn dom(→∞ T ) ∈ = lim T xn lim T xn n n →∞ k k ≤ →∞ k k · k k = T lim xn n k k →∞ k k = T x . k k k k T˜ x ˜ ˜ So k x k T for every x dom(T ) and therefore T T . So T = T . k k ≤ k k ∈ k k ≤ k k k k k k Uniqueness: The fact that we defined T˜ x := y is necessary to ensure the continuity • in x. 2.33 Definition. Let X,Y be normed spaces. Define

BL(X,Y ) := T : X Y such that T is linear and bounded { −→ } and set BL(X) := BL(X,X). 2.34 Theorem. BL(X,Y ), X Y is a normed space. If Y is complete, so is BL(X,Y ). k·k → Proof. First of all, BL(X,Y ) is a K-vector space with zero element • 0: X Y −→ x 0. 7−→ For T1,T2 BL(X,Y ) and α K deﬁne T1 + T2 and αT1 by ∈ ∈

(T1 + T2)x := T1x + T2x

(αT1)x := αT1x

for every x X. ∈

X Y is a norm on BL(X,Y ). • k·k → – T 0 and if T = 0 then T x = 0 for every x X, so T x = 0 k kX Y ≥ X k kX Y k k ∈ for every→ x X and so T =→ 0. ∈

33 – We have (αT )x αT x T x αT X Y = sup k k = sup k k = sup α k k = α T . k k → 0=x X x 0=x X x 0=x X | | · x | | · k k 6 ∈ k k 6 ∈ k k 6 ∈ k k – We have (T + T )x = T x + T x T x + T x . Then k 1 2 k k 1 2 k ≤ k 1 k k 2 k T1x T2x T sup k k + k k T1 + T2 . k k ≤ 0=x X x x ≤ k k k k 6 ∈ k k k k

Let Y be complete. We show completeness of BL(X,Y ). Let (Tk)k N BL(X,Y ) • ∈ ⊆ be a Cauchy sequence. For every x X and k, l N: ∈ ∈ T x T x T T x . ( ) k k − l k ≤ k k − lk · k k ∗

This implies that (Tkx)k N Y is a Cauchy sequence for every x X and since Y ∈ ⊆ ∈ is complete, there is some limit

lim Tkx =: T x Y. k ∈ →∞ This deﬁnes a map

T : X Y −→ x T x := lim Tkx. 7−→ k →∞

T is linear, because let α, α0 K and x, x0 X. Then • ∈ ∈

T (αx + α0x0) = lim Tk(αx + α0x0) k →∞ = lim (αTkx + α0T x0) k →∞ = αT x + α0T x0.

We show T is bounded and that it is the norm limit of the T ’s. According to ( ) • k ∗ we have that for every ε > 0 there exists an N N such that k, l N and x X: ∈ ∀ ≥ ∈ T x T x ε x . k k − l k ≤ k k So applying the limit l gives T x T x ε x and so T T ε. This → ∞ k k − k ≤ k k k k − k ≤ gives two things:

1. T is bounded because T BL(X,Y ) and (T T ) BL(X,Y ) and since k ∈ k − ∈ BL(X,Y ) is a vector space:

T = T (T T ) BL(X,Y ). k − k − ∈ k 2. Tk →∞ T in X Y . −−−→ k·k →

34 2.4 Linear functionals and dual space 2.35 Deﬁnition. Let X be a normed space. A linear functional (on X) is a linear operator l : dom(l) K. dual space of 6 −→ The X is X∗ := BL(X, K) . Notations for the norm on X∗: X K =: k·k → X∗ =: =: . k·k k·k∗ k·k

2.36 Corollary (from Theorem 2.34). (X , ∗ ) is a Banach space (no matter whether ∗ k·kX X is complete).

2.37 Examples. Let X = C([a, b]) (where a < b R) equipped with . ∈ k·k∞ (a) For f X let ∈ Zb (f) := dx f(x) I a This is clearly a linear functional : X K with I −→ Zb (f) dx f(x) f (b a) |I | ≤ | {z }| ≤ k k∞ − a f ≤k k∞

so ∗ (b a) and the function f 1 yields equalities above and so ∗ = b a. kIkX ≤ − ≡ kIkX − (b) For f X and t [a, b] let δt(f) := f(t). This gives a linear functional δt : X K ∈ ∈ −→ which is called the Dirac-δ functional with δt(f) = f(t) f and equality again | | | | ≤ k k∞ by considering f 1. So δ ∗ = 1. ≡ k tkX NB. Pay attention that the boundedness of δt is heavily dependent on the norm on X. For example it’s not bounded, if X is equipped with the 1-norm = R 1 dx . k · k1 0 | · |

Notation. Let X,Y be metric spaces. We write X ∼= Y iﬀ X is isometrically isomorphic to Y .

1 1 p q 2.38 Theorem. Let p [1, [ and + = 1. Then (` )∗ = ` . ∈ ∞ p q ∼ p p Proof. Case 1 < p < : Let x = (xn)n N ` , f (` )∗. We work with the • ∞ p ∈ ∈ ∈ canonical (Schauder) basis en n N for ` (introduced in Example 2.15). Therefore { } ∈ P x can be written as a -convergent series x = xnen with xn K. Since f is k·kp n N ∈ continuous and linear: ∈ X X f(x) = f(x e ) = x f(e ). ( ) n n n n ∗ n N n N ∈ ∈ p For N N (fixed), definex ˜ := (˜xn)n N ` by ∈ ∈ ∈  q f(en)  | | if n N f(e ) = 0, f(en) n xñ := ≤ ∧ 6 0 otherwise.

6Note: These are only bounded (or equivalently continuous) linear functionals, hence the name topological dual – not to be confused with the algebraic dual {f : X −→ K : f linear} that is common in linear algebra!

35 Apply ( ) tox ˜: ∗ N X f(˜x) = f(e ) q ( ) | n | ∗∗ n=1

so that 0 f(˜x) f p ∗ x˜ where ≤ ≤ k k(` ) k kp N 1/p X (q 1)p x˜ = f(e ) − . k kp | n | n=1

1 1 q Since p = 1 q p = q 1 , we obtain p(q 1) = q and − ⇐⇒ − − N N 1/p ( ) X q X q =∗∗ f(e ) f p ∗ f(e ) ⇒ | n | ≤ k k(` ) | n | n=1 n=1 N 1/q X q = f(e ) f p ∗ . ( ) ⇒ | n | ≤ k k(` ) ∗∗∗ n=1 Thus, the map

p q (` )∗ ` : −→ f f(en) I n N 7−→ ∈ is well deﬁned and also

– is linear, I – f = (f(en))n f p ∗ by ( ), kI kq q ≤ k k(` ) ∗∗∗ – is onto7, because if y = (y ) `q, define I n n ∈ X fy(x) := xnyn n N ∈ p for x = (xn)n N ` . This is well defined because Hölder’sinequality yields ∈ ∈ f (x) x y so that | y | ≤ k kp k kq

fy p ∗ y (` ) ≤ k kq p and fy (` )∗. But fy(en) = yn for all n N. ∈ ∈ – f p ∗ (f(en))n , due to ( ) and H¨older as in the previous argument. k k(` ) ≤ q ∗ So is an isometric isomorphism. I The case p = 1 is analogous, but instead of deﬁningx ˜ , use • n f(e ) f ∗ e . n (`1) n 1 | | ≤ k k |k {zk} =1

This implies (f(en))n f 1 ∗ , which replaces ( ), and the properties of ≤ k k(` ) ∗∗∗ I follow as above. ∞

1 2.39 Remarks. (a)( c0)∗ ∼= ` (see problem sheet). 7surjective

36 1 (b) The map ` (` )∗, x f , deﬁned by −→ ∞ 7−→ x X 1 f (y) = x y , y `∞, x ` , x n n ∈ ∈ n N ∈

is well-deﬁned (H¨older!),linear, isometric but not onto! In other words, (`∞)∗ is strictly “bigger” than `1. (proof later)

2.5 Hilbert spaces The main new feature is the “geometry” from the scalar product.

2.40 Deﬁnition. Let X be a (K-)vector space. A mapping , : X X X is a scalar h· ·i × −→ product (or inner product) iﬀ

(i) x, x 0 for every x X and if x, x = 0 then x = 0 (non-degenerate). h i ≥ ∈ h i (ii) x, αy + βz = α x, y + β x, z for every α, β K and x, y, z X. h i h i h i ∈ ∈ (iii) x, y = y, x for every x, y X. h i h i ∈ (X, , ) is an inner product space (or pre-Hilbert space). h· ·i 2.41 Lemma (Cauchy-Schwarz (-Bunjakowski) inequality). Let X be an inner product space, let x, y X. Then ∈ x, y x, x 1/2 y, y 1/2 | h i | ≤ h i h i with equality if and only if x, y are linear dependent.

Proof. See linear algebra!

2.42 Lemma. Let X be an inner product space. Then X is a normed space with norm x := x, x 1/2. All notions from topological, metric and normed spaces are available. In k k h i particular, the scalar product , : X X K is continuous. h· ·i × −→ Proof. fulﬁlls all axioms of a norm, see linear algebra. • k·k n n Continuity of , : Let x →∞ x, y →∞ y, then • h· ·i n −−−→ n −−−→ x , y x, y x , y x , y + x , y x, y = | h n mi − h i | ≤ | h n mi − h n i | | h n i − h i | 2.41 = x , y y + x x, y | h n m − i | | h n − i | ≤ n,m xn ym y + y xn x →∞ 0 ≤ k k · k| {z− k} |{z}k k ·| k {z− k} −−−−−→ m→∞ < n→∞ 0 ∞ 0 −−−−→ −−−→

because supn N xn < (as (xn)n converges!). ∈ k k ∞ 2.43 Deﬁnition. A complete inner product space is called Hilbert space.

2.44 Theorem. Let X be an inner product space. Then there exists a Hilbert space H, a dense subset W H and a unitary map U : X W (i.e. U is an isomorphism with ⊆ −→ x, y = Ux, Uy for all x, y X). h iX h iH ∈

37 Idea of the proof. See the proofs of Theorem 2.13 and 1.24. Additional aspect here: Deﬁne a scalar product in H := equivalence classesx ˜ of Cauchy sequences in X : { }

x,˜ y˜ := lim xn, yn H n X h i →∞ h i with (x ) , (y ) X being representatives of the equivalence classesx, ˜ y˜, and n n n n ∈ X W U : −→ x x˜ 7−→ wherex ˜ is the equivalence class of the constant representative (x, x, x, . . . ). 2.45 Examples. 2 P (a) ` is a Hilbert space with the scalar product x, y := n N xnyn, h i ∈ where x = (xn)n, y = (yn)n. (b) C([0, 1]) is an inner product space with the scalar product

Z 1 f, g := dx f(x)g(x) h i 0 but not a Hilbert space (proof analogous to Example 1.21 (b)).

(c) `p for p = 2 is not an inner product space, because of the following theorem: 6 2.46 Theorem (Jordan-von Neumann). Let (X, ) be a normed space. Then X is an k · k inner product space with = , 1/2 if and only if satisfies the parallelogram identity k·k h· ·i k·k x + y 2 + x y 2 = 2( x 2 + y 2) k k k − k k k k k for all x, y X. ∈ Proof. Idea: ”= ” elementary computation. ⇒ ” =” Define inner product by polarisation ⇐  1 2 2  4 x + y x y K = R, x, y := k k − k − k 1 i h i  x + y 2 x y 2 + x + iy 2 x iy 2 K = C. 4 k k − k − k 4 k k − k − k Verify that this definition satisfies axioms of an inner product: 1. Symmetry and definiteness is obvious

2. bi-/sesquilinearity follows from parallelogram identity: (here only K = R) 1 x, y + x, z = x + y 2 x y 2 + x + z 2 x z 2 h i h i 4 k k − k − k k k − k − k 2 2! 1 y + z y + z = x + x 2 2 − − 2 y + z = 2 x, (1) 2 z = 0 in (1) x, y = 2 x, y/2 (2) ⇒ h i (1) and (2) x, y + x, z = x, y + z (3) ⇒ h i h i h i (2) with 2y instead of y : x, 2y = 2 x, y (4) h i h i by induction from (3) and (4): x, ny = n x, y n N0 (5) h i h i ∀ ∈

38 z = y in (3) x, y = x, y (5) holds for all n Z. − ⇒ h i − h − iy ⇒ ∈ Let m Z 0 and use (5) with instead of y (writing λ := n ): ∈ \{ } m m y y x, λy = n x, = λm x, = λ x, y . h i m m h i

This holds for all λ Q and thus for all λ R by continuity. ∈ ∈ 2.47 Definition. Let X be an inner product space and A X. ⊆ (a) x, y X are orthogonal iff x, y = 0. In symbols: x y. ∈ h i ⊥ (b) The Orthogonal complement of A is A := x X : x, a = 0 a A . ⊥ { ∈ h i ∀ ∈ } (c) Let J = ∅ be an index set and for all α J let eα X. eα α J is an orthonormal 6 ∈ ∈ { } ∈ set iff

e = 1 α J •k αk ∀ ∈ e , e = 0 α, β J, α = β. • α β ∀ ∈ 6 (d) eα α J is an orthonormal basis (ONB) or complete orthonormal set iﬀ { } ∈ eα α J is an orthonormal set •{ } ∈ if e , x = 0 for some x X and all α J, then x = 0 (i.e. the zero vector is • h α i ∈ ∈ the only vector orthogonal to all eα’s!). 2.48 Lemma. Let X be an inner product space, A X a subspace. ⊆ (a) A⊥ is a closed subspace in X.

(b) Every orthonormal set eα α J is linearly independent. { } ∈ Proof. (a) Tutorial sheet. Pn (b) Suppose λjeαj = 0 for some λ1, . . . , λn K, some α1, . . . , αn J with αj = j=1 ∈ ∈ 6 αk j = k, and n N. Then we have ∀ 6 ∈ n n D X E X 0 = e , 0 = e , e λ = λ e , e = λ k = 1, . . . , n. αk αk αj j j h αk αj i k ∀ j=1 j=1 | {z } δjk

2.49 Lemma. Let X be an inner product space, x X and eα α J an orthonormal set. Then: ∈ { } ∈

2 X 2 X 2 x sup eα, x =: eα, x . (Bessel’s inequality) k k ≥ J0 J | h i | | h i | ⊆ α J0 α J J0 < ∈ ∈ | | ∞ P If X is even a Hilbert space and if eα α J is even an ONB, then x = α J eα, x eα, { } ∈ h i where at most countable many terms are = 0, and ∈ 6 X x 2 = e , x 2. (Parseval’s equality) k k | h α i | α J ∈

39 Proof. Let J J be a ﬁnite subset. Then we can write x as follows: 0 ⊆ X X x = e , x e + x e , x e . h α i α − h α i α α J0 α J0 |∈ {z } | ∈ {z } =:u =:v

Using that eα α J0 is an orthonormal set, we get: { } ∈ * + X X X u, u = e , x e , e , x e = e , x e , x e , e h i h α i α β β h α i β α β α J0 β J0 α,β J0 X∈ ∈ X ∈ = e , x e , x = e , x 2, h α i h α i | h α i | α J0 α J0 ∈ ∈ X u, v = u, x u = u, x u, u = e , x e , x u, u = 0. h i h − i h i − h i h α i h α i − h i α J0 ∈ Thus we can estimate x 2: k k 2 X 2 x = u + v, u + v = u, u + v, v eα, x J0 J, J0 < , k k h i h i |h {z }i ≥ | h i | ∀ ⊆ | | ∞ 0 α J0 ≥ ∈ and taking the supremum over all finite subsets J J, we get Bessel’s inequality. 0 ⊆ Digression on uncountable sums: Suppose X X cα := sup cα < where 0 cα < . J0 J ∞ ≤ ∞ α J ⊆ α J0 ∈ J0 < ∈ | | ∞ Then cα = 0 for all but countable many α. To see that, we define the following sets: S := α J : c 1 0 { ∈ α ≥ } Sn := α J : 1/n > cα 1/(n + 1) , n N. { ∈ ≥ } ∈ Claim: Sn < n N0. | | ∞ ∀ ∈ True, because if this was violated for some k , then P c P c = . N α J α α Sk α n o ∈ ∈ ≥ ∈ ∞ But then α J : c > 0 = S S is a countable union of finite sets and thus α n N0 n ∈ ∈ countable. End of digression

This means that there exists a sequence (αn)n N J such that ∈ ⊆ X X e , x 2 = e , x 2 x < , ( ) | h α i | | h αn i | ≤ k k ∞ ∗ α J n N ∈ ∈ where we have used Bessel’s inequality. Digression on series in Banach and Hilbert spaces: P Let Z be a Banach space and (zn)n N Z. Assume that n N zn < . Then • P ∈ ⊆ ∈ k k ∞ n N zn exists in Z. ∈ Pn For, let Sn := zj for n N. Since Z is a Banach space, it suﬃces to show that j=1 ∈ (S ) is Cauchy: Let m n. Then n n ≥ m m X X S S = z z = σ σ k m − nk j ≤ k jk m − n j=n+1 j=n+1 Pn with σn := j=1 zj . But by hypothesis (σn)n N converges in R and, hence, is k k ∈ Cauchy.

40 Let Z be a Hilbert space and let (zn)n N Z be a sequence of pairwise orthogonal • P 2 ∈ ⊆ P vectors. Then: n N zn < n N zn exists in Z. ∈ k k ∞ ⇐⇒ ∈ For, consider again the sequence of partial sums (S ) . Let m n, then n n ≥ m m m X 2 X X S S 2 = z = z , z = z 2 = σ σ k m − nk j h j ki k jk m − n j=n+1 j,k=n+1 j=n+1

Pn 2 with σn := j=1 zj . Thus, (Sn)n is Cauchy if and only if (σn)n N is Cauchy. k k ∈ End of digression

( ) and the above convergence criterion for sequences in Hilbert spaces implies that ∗ X x0 := e , x e h αn i αn n N ∈ exists in X. We need to show x = x0:

We have x x0 eαn n N: − ⊥ ∀ ∈ X eαm , x x0 = eαm , x eαm , x0 = eαm , x eαn , x eαm , eαn = 0 m N. − h i − h i − h i h i ∀ ∈ n N | {z } ∈ δmn

If α αn n N we have: 6∈ { } ∈ X ( ) e , x x0 = e , x e , x e , e = e , x =∗ 0. α − h α i − h αn i h α αn i h α i n N | {z } ∈ =0 | {z } =0

Taken together, we have x x e α J ONB= x = x . − 0 ⊥ α ∀ ∈ ⇒ 0 Moreover,

n 2 2 X x = lim eα , x eα = k k n h j i j →∞ j=1 * n n0 + X X = lim eαj , x eαj , lim eαk , x eαk = n h i n0 h i →∞ j=1 →∞ k=1 * n n0 + X X = lim e , x e , e , x e = n h αj i αj αk αk n0→∞ j=1 k=1 →∞ n n0 X X = lim e , x e , x e , e = n h αj ih αk i h αj αk i n0→∞ j=1 k=1 | {z } →∞ δjk X ( ) X = e , x 2 =∗ e , x 2. |h αj i| |h α i| j N α J ∈ ∈ The above proof showed

2.50 Corollary. Let X be a Hilbert space, eα α J an ONB. { } ∈ P 2 P (a) If (cα)α J K with α J cα < , then α J cαeα is well deﬁned in X. ∈ ⊂ ∈ | | ∞ ∈ 41 (b) If J is countable, then eα α J is a Schauder basis. { } ∈ 2.51 Theorem. Every Hilbert space X = 0 has an ONB. Moreover, 6 { } X separable there exists a countable ONB. ⇐⇒ Proof. ” =” Suppose there exists a countable ONB. By Corollary 2.50 (b), this is a ⇐ Schauder basis. Thus, X is separable by Lemma 2.16.

”= ” Let xn : n N be dense in X. W.l.o.g. assume xn = 0 n N. We construct an ⇒ { ∈ } 6 ∀ ∈ ONB using the the Gram-Schmidt procedure: Deﬁne e := x1 . 1 x1 • k k Throw away all x ’s that are linearly dependent of e . Let n > 1 be the smallest • n 1 1 index of the remaining elements. Set e˜ e˜ := x e , x e and e := 2 . 2 n1 − 1 n1 1 1 e˜ k 2k Throw away all x ’s that are linearly dependent of span(e , e ). Let n > n be the • n 1 2 2 1 smallest index of the remaining elements. Set e˜ e˜ := x e , x e e , x e and e := 3 . 3 n2 − 1 n2 1 − 2 n2 2 3 e˜ k 3k Continue this procedure. • This terminates if and only if dim X < . It is clear, that en n N is an orthonormal set. ∞ { } ∈ span e is dense in X because each x is a ﬁnite linear combination of e ’s. It remains { n}n n n to show that this is a basis: Assume there is y X such that y en n. Denseness ∈ ⊥ ∀ k = (yn)n X such that yk span en : n N for all k N and yk →∞ y in X. ⇒ ∃ ⊆ ∈ { ∈ } ∈ −→ Thus y, y = 0 k and h ki ∀ 2 y = y, y = lim y, yk = 0. k k h i k h| {z }i →∞ =0 The existence of an ONB in the non-separable case will be proven later (when Zorn’s Lemma is available).

2.52 Theorem. Let X be a Hilbert space. If dim X = n, then X is isomorphic to Kn. If dim X = and X is separable, then X is (unitarily equivalent) isomorphic to `2. ∞ nd Proof. We show only the 2 case here: Fix a countable ONB en n N (exists by Theo- { } ∈ rem 2.51 because X is separable) and consider the linear operator

X `2 U : −→ x ( en, x )n N. 7−→ h i ∈ U is well deﬁned by Parseval’s equality. • surjective by Corollary 2.50 (a), • preserves the scalar product, because Parseval implies x 2 = P e , x 2 = • k k n | h n i | Ux 2 x X. This implies x, y = Ux, Uy 2 x, y X by polarisation k k` ∀ ∈ h iX h i` ∀ ∈ (see proof of Thm. 2.46).

42 = U is unitary. ⇒ 2.53 Deﬁnition (Direct sum). Let (X , , ), (X , , ) be inner product spaces. Then 1 h· ·i1 2 h· ·i2  !   x1  X1 X2 := : x1 X1, x2 X2 ⊕  x2 ∈ ∈  is an inner product space with the scalar product * ! !+ x1 y1 , := x1, y1 1 + x2, y2 2 . x2 y2 h i h i ⊕ (Note: If X ,X are Hilbert spaces, so is X X ) 1 2 1 ⊕ 2 2.54 Examples. (a) X = X = C([0, 1]) with f, g = R 1 dx fg. Then 1 2 h i 0  !   f1  X1 X2 = : fj C([0, 1]), j = 1, 2 , ⊕  f2 ∈ 

is the space of the the 2-vector-valued continuous functions with

* ! !+ Z 1 f1 g1 , = dx f 1g1 + f 2g2 . f2 g2 0 ⊕ (b) Orthogonal decomposition. Let X be a Hilbert space and A X a closed subspace. ⊆ Then (A, , ) is a Hilbert space (w.r.t. to inherited scalar product from X) and the h· ·iA same is true for A⊥. Direct sum:  !   a  A A⊥ = : a A, b A⊥ ⊕  b ∈ ∈ 

with the scalar product * ! !+ a1 a2 , = a1, a2 A + b1, b2 A⊥ = b1 b2 h i h i ⊕ = a , a + b , b = a + b , a + b . h 1 2i h 1 2i h 1 1 2 2i The spaces A A and X can be identiﬁed as Hilbert spaces by a x = a + b, ⊕ ⊥ b ←→ if the orthogonal decomposition of x is unique. Indeed, this holds because of

2.55 Theorem (Projection theorem). Let X be a Hilbert space and A X a closed ⊆ subspace. Then for all x X there exists a unique z A and a unique w A such that ∈ ∈ ∈ ⊥ x = z + w.

The proof relies on

2.56 Lemma. For every x X there exists a unique z A such that dist(x, A) = x z . ∈ ∈ k − k z is the ”closest element” to x in A.

43 Proof. Existence: Let d := dist(x, A) = infy A x y . There exists a sequence (yk)k N ∈ k − k ∈ ⊂ A such that d = lim x yk . ( ) k k − k ∗ →∞ | {z } d ≥ Using the parallelogram identity u + v 2 = 2 u 2 + v 2 u v 2, we get: k k k k k k − k − k y y 2 = (y x) + (x y ) 2 = k n − mk k n − − m k 2 2 2 = 2 yn x + ym x 2x ym yn k − k k − k − k| − {z − k} 2 ym yn 4 x − · − 2 | {z } A | {z ∈ } ≥4d2 m,n 2 y x 2 + y x 2 4d2 →∞ 0. ( ) ≤ k n − k k m − k − −−−−−→ ∗∗ n Thus, (y ) is a Cauchy sequence. A is complete: y →∞ z A and d = x z by ( ) n n n −−−→ ∈ k − k ∗ and continuity of the norm. Uniqueness: assume that there exists (y ) A with the same properties, i.e. n0 n ⊆

lim y0 x = d and y0 = lim y0 . n n n n →∞ k − k →∞ Replacing y by y in ( ), yields m m0 ∗∗ 2 2 z y0 = lim lim yn y0 n m m k − k →∞ →∞ k − k 2 2 2 lim lim 2 yn x + y0 x 4d 0, n m m ≤ →∞ →∞ k − k k − k − ≤ thus z = y0.

Proof of Theorem 2.55. Let z A be as in Lemma 2.56 and set w := x z. Let ξ C ∈ − ∈ and y A. Then ∈ x z 2 x (z + ξy) 2 |k −{z k} ≤ k − | {z } k =:d2 A | {z∈ } w ξy − = w 2 + ξy 2 2 Re w, ξy |k {zk} k k − h i =d2 So for y = 0 we have 6 2 Re w, ξy ξ 2 h i 0 | | − y 2 ≥ k k For ξ = t R we obtain therefore • ∈

2 2 Re w, y t h it 0 t R − y 2 ≥ ∀ ∈ k k This implies w, y = 0. h i

44 For ξ = i, t R: • ∈ 2 2 Im x, y t h it 0 t R − y 2 ≥ ∀ ∈ k k So w, y = 0 for every y A. So w A . h i ∈ ∈ ⊥ Uniqueness: Assume there are z, z A and w, w A with z + w = x = z + w . Then 0 ∈ 0 ∈ ⊥ 0 0

z z0 = w0 w | −{z } | {z− } A A⊥ ∈ ∈ So z = z and w = w because A A = 0 . 0 0 ∩ ⊥ { } 2.57 Theorem (Riesz representation). Let X be a Hilbert space and let ` X . Then ∈ ∗ there exists a unique y X such that ` ∈ `(x) = y , x x X ( ) h ` i ∈ ∗ and ` = y` . k k∗ k k Proof. Note ker ` = ` 1( 0 ) is closed by continuity. − { } If ker ` = X then the theorem is true with y = 0. • ` Suppose now ker ` ( X. Then theorem 2.55 tells that (ker `)⊥ ) 0 and this allows • { } to choose 0 = x (ker `) . Deﬁne 6 0 ∈ ⊥ `(x ) y := 0 x X ` x 2 · 0 ∈ k 0k – so ( ) holds for every x ker ` (x ker `). ∗ ∈ 0 ⊥ – Let x = αx0, α K. Then `(x) = α`(x0) and ∈ * + `(x ) y , x = 0 x , αx h ` i x 2 · 0 0 k 0k `(x ) = 0 α x , x x 2 h 0 0i k 0k = α`(x0)

So ` and y , agree on span(ker `, x ). h ` ·i 0 But span(ker `, x ) = X, because for every x X we have 0 ∈ `(x) `(x) x = x x0 + x0 − `(x0) · `(x0) · | {z } | {z } ker ` span(x0) ∈ ∈ because `(x) `(x) ` x x0 = `(x) `(x0) − `(x0) · − `(x0) · = 0

so ` = y , on X. h l ·i

45 Uniqueness of y : Assume there exists y X with ` = y , . Then for every x X: • ` 0 ∈ 0 · ∈ 0 = `(x) `(x) − |{z} |{z}0 y`,x y ,x h i h i = y y0, x ` − Choose x = y y so that 0 = y y 2, so y = y . ` − 0 k ` − 0k ` 0 Norm: We have • `(x) ` = sup | | y` k k∗ 0=x X x ≤ k k 6 ∈ |k{zk} | y`,x | h kxk i

Note y` = 0, because ker ` ( X in the case considered it is allowed to choose x = y` 6 in the sup above: `(x0) y`, y` ` | | = | h i | = y` k k∗ ≥ y y k k k `k k `k So ` = y` . k k∗ k k 2.58 Corollary. Let X be a Hilbert space. Then the map

: X∗ X A −→ ` y 7−→ ` deﬁned as in the theorem 2.57 is an anti-linear, isometric bijection.

Proof. Let α, β K and `1, `2 X∗ with `j = yj, , j = 1, 2. Then ∈ ∈ ·

α`1 + β`2 = α y1, + β y2, h ·i D h ·iE = αy , + βy , h 1 ·i 2 · D E = αy + βy , 1 2 · So (α` +β` ) = α (` )+β (` ). This means is anti-linear. The property of being A 1 2 A 1 A 2 A A isometric was proved in Theorem 2.57 (so is also injective). Futher, is onto, because A A for y X, ` := y, X (use Cauchy-Schwarz for continuity) and (` ) = y. ∈ h ·i ∈ ∗ A y

46 3 Measures, integration and Lp-spaces8

3.1 Measures9 Measures cannot be defined satisfactorily on the power set, thus: 3.1 Definition. Let X be a set and (X). A ⊆ P is a σ-algebra (or σ-field): A ⇐⇒ X • ∈ A A = Ac • ∈ A ⇒ ⊂ A S An , n N = n N An • ∈ A ∀ ∈ ⇒ ∈ ∈ A T measurable thus ∅ , n N An ; A B if A, B . We say A : A ∈ A ∈ ∈ A \ ∈ A ∈ A ⇐⇒ ∈ A 3.2 Lemma. Let (X), then there exists a smallest σ-algebra σ( ): σ( ) G ⊆ P G G ⊆ G ( σ-algebra generated by ) G 3.3 Definition. Let (X, ) be a topological space. Then (X) := σ( ) Borel T B ≡ B T σ-algebra 3.4 Definition. (a) Let be a σ-algebra on X. A mapping µ : [0, ] is a (positive) A A → ∞ measure : ⇐⇒ µ(∅) = 0 • S P An , n N,An Am = ∅ n = m. = µ( An) = µ(An) • ∈ A ∈ ∩ ∀ 6 ⇒ n N n N (σ-additivity) ∈ ∈

(X, , µ) is called a measure space. A (b) If µ(X) = 1 then µ is called a probability measure, a finite measure, if µ(X) < and S∞ a σ-finite measure, if X = An,An , n N and µ(An) < . n ∈ A ∀ ∈ ∞ (c) If X is Hausdorff, = (X) and µ(K) < for all compacts sets K X then µ is a A B ∞ ⊆ Borel measure. 3.5 Definition. (a) (X) is a semi-ring : H ⊆ P ⇐⇒ ∅ • ∈ H A, B = A B • ∈ H ⇒ ∩ ∈ H Sn A, B = C1, ,Cn ,Cj Ck = ∅ j = k : A B = Ck • ∈ H ⇒ ∃ ··· ∈ H ∩ ∀ 6 \ k=1 (b) A mapping µ : [0, ] is a pre-measure : H → ∞ ⇐⇒ µ(∅) = 0 • S An n N,An Am = ∅ n = m and n N An • ∈ H ∀ ∈ ∩ ∀ 6 ∈ ∈ H ! [ X = µ A = µ(A ) ⇒ n n n N n N ∈ ∈ 8Parts of this section were typewritten by Max Klinger, to whom I express my gratitude. 9Literature: H. Bauer, Measure and integration theory; J. Elstrodt, Maß- und Integrationstheorie; P. R. Halmos, Measure theory.

47 3.6 Theorem (Kolmogorov extension of pre-measures). Let µ be a σ-finite pre-measure on a semi-ring , then it has a unique extension to a H measure on σ( ). H 3.7 Example. X = R, := ]a, b]: a, b R, a b is a semi-ring. H ∈ ≤ µ ]a, b] := b a defines a σ-finite pre-measure on • − H σ( ) = (R) • H B = extension λ of µ to (R): λ is called the Lebesgue (-Borel) measure. It is a Borel ⇒ ∃ B measure in the sense of Def. 3.4 (c).

3.8 Deﬁnition. Let (X, ) be a topological Hausdorﬀ space, its Borel σ-algebra and T B µ a Borel measure on X.

(a) µ is inner regular : B µ(B) = sup µ(K) ⇐⇒ ∀ ∈ B K B K compact⊆ (b) µ is outer regular : B µ(B) = inf µ(U) ⇐⇒ ∀ ∈ B U B U open⊇ (c) µ is regular : µ is inner regular and outer regular ⇐⇒ 3.9 Theorem. Let X be a locally compact10, 2nd countable Hausdorﬀ space. Then every Borel measure on X is regular.

Proof. See e.g. H. Bauer, Maß- und Integrationstheorie, Thm. 29.12.

This theorem (or the more much elementary Thm. 11.42 in P. M¨uller, Analysis III ) yield the following special case

3.10 Corollary. The Lebesgue (-Borel) measure on Rd is regular. 3.11 Definition. Let (X, , µ) be a measure space A A X is a µ-null set : A and µ(A) = 0. • ⊆ ⇐⇒ ∈ A a statement P (x) holds for µ-almost all x X : a µ-null set X: • ∈ ⇐⇒ ∃ N ⊂ P (x) holds x X ∀ ∈ \N (X, , µ) is complete : if A is a µ-null set and B A, then so is B. (The point • A ⇐⇒ ⊂ is B !) ∈ A 3.12 Theorem (Completion). Every measure space (X, , µ) has a completion (X, , µ ), A A∗ ∗ that is, C A, B : A C B and µ(A) = µ(B) =: µ (C) ∈ A∗ ⇐⇒ ∃ ∈ A ⊆ ⊆ ∗ 3.13 Definition (Lebesgue measure). The completion of R, (R), λ is R, (R)∗, λ∗ B B where λ∗ is the Lebesgue measure. 3.14 Definition. Let be a σ-algebra in X and f : X X . Then f is ( , )- A0 0 → 0 A A0 measurable : A : f 1(A ) . ⇐⇒ ∀ 0 ∈ A0 − 0 ∈ A 1 3.15 Lemma. Let 0 = σ 0 . Then: G0 σ 0 : f − (G0) = f is - -measurable. A G ∀ ∈ G ∈ A ⇒ A A0 10locally compact :⇐⇒ ∀ x ∈ X ∃ a compact neighbourhood of x

48 3.16 Corollary. Let (X, ) and (X0, 0) be topological spaces and f : X X0 continuous T T → = f is (X), (X ) -measurable. ⇒ B B 0

3.17 Theorem. Let f, g, fn : X K be measurable. Then the following are measurable, → too: Re f, Im f, f + g, f g, sup f , inf f , lim sup f , lim inf f , lim f (if it exists). · n n n n n n 3.18 Remark. The theorem remains true with K = R := R and := (R) = ∪ {±∞} B B σ( + ). B ∪ { ∞} ∪ {−∞} 3.2 Integration 3.19 Deﬁnition. Let (X, , µ) be a measure space. A (a) f : X [0, [ is a simple function (or step function): N N, α1, , αN > → ∞ ⇐⇒ ∃ ∈ ··· 0 and A , ,A such that 1 ··· N ∈ A N ( X 1, x A f = αn1A , where 1A(x) := ∈ N 0, x A n=1 6∈ is the indicator function of A. (Note: 1 measurable A measurable) A ⇐⇒ Without loss of generality one can assume that α , , α and A ,...,A are pair- 1 ··· N 1 N wise disjoint.

(b) (µ-) integral of a simple function:

N Z X µ(dx)f(x) := α µ(A ) (can be + ). N n ∞ X n=1 R R R Other notations: X dµ(x)f(x), X dµf, X f(x) dµ(x). 3.20 Lemma. Let f : X [0, ]. Then → ∞  (fn)n, a sequence of simple functions  ∃ X [0, [ f is ( , )-measurable → ∞ A B ⇐⇒   with fn fn 1 n and f = lim fn = sup fn − n ≤ ∀ →∞ n 3.21 Deﬁnition. Let (X, , µ) be a measure space. A (a) For f : X [0, ] measurable let → ∞ Z Z dµ f := lim dµ fn, n X →∞ X

be the (µ-) integral of f, where (fn)n is any sequence as in Lemma 3.20.

(b) For f : X R let f+ := max f, 0 , f := max 0, f such that f = f+ f → { } − { − } − − (decomposition into negative and positive parts). Then Z f (µ-) integrable : f measurable and dµ f < . ⇐⇒ X ± ∞

In this case Z Z dµ f := dµ f+ dµ f . X − X −

49 (c) For f : X C measurable let → f integrable : Re f, Im f integrable. ⇐⇒ In this case Z Z Z dµ f := dµ Re f + i dµ Im f. X X X

3.22 Remark. (a) Deﬁnition 3.21 (a) does not depend on the choice of (fn)n. (b) For A X measurable ⊆ Z Z dµ f := dµ f1A, A X in particular for f = 1, Z Z dµ = dµ 1A = µ(A). A X

(c) For f, g integrable and α, β C ∈ Z Z Z dµ (αf + βg) = α dµ f + β dµ g (linearity). X X X For f, g integrable with f g ≤ Z Z dµ f dµ g (monotonicity). X ≤ X

3.23 Example. (a) Lebesgue-Borel measure λd on Rd is deﬁned by

! d d Y λd ]a , b ] := (b a ) α α α − α α×=1 α=1 on rectangles and extended to d := (Rd) by Theorem 3.6, compare Example 3.7. If B B f : Rd C is Riemann-integrable = f is λd-integrable and → ⇒ Z Z ddx f(x) = λd(dx) f(x). Rd Rd Justiﬁes notation λd(dx) ddx. ≡ d d (b) Dirac measure on R : for x0 R let ∈ ( 1, x0 A δx (A) := ∈ 0 0, x A 0 6∈ d d A . Hence, every measurable f : R R is δx0 -integrable and ∀ ∈ B → Z

δx0 (dx) f(x) = f(x0). Rd

3.24 Lemma. Let (X, , µ) be a measure space and f : X R measurable. Then the following are equivalent:A →

(a) f integrable

50 (b) f+, f integrable − (c) f integrable | | (d) u, v : X R integrable: f+(x) u(x), f (x) v(x) for µ-almost all x X. ∃ → ≤ − ≤ ∈

(e) g : X R integrable: f(x) g(x) for µ-almost all x X ∃ → ≤ ∈ 3.25 Theorem. Let (X, , µ) be a measure space and f : X R measurable. Then A → (a) if f integrable and R dµ f < then µ x X : f(x) = = 0 ∞ ∈ ±∞ (b) R dµ f = 0 f(x) = 0 for µ-almost all x X | | ⇐⇒ ∈ The beneﬁt of general integration theory is the interchangeability of integration and limits. 3.26 Theorem (Beppo Levi – monotone convergence). Let

1 1 1 (fn)n N (X, µ) := g : X R: g is µ-integrable (X) ∈ ⊂ L { → } ≡ L ≡ L with 0 fn fn+1 n N. Then ≤ ≤ ∀ ∈ Z Z lim dµ fn = dµ lim fn. n n →∞ →∞ 1 3.27 Theorem (Fatou’s Lemma). Let (fn)n with fn 0 n N. Then ⊂ L ≥ ∀ ∈ Z Z dµ lim inf fn lim inf dµ fn. n n →∞ ≤ →∞ 1 3.28 Theorem (Lebesgue – dominated convergence). Let (fn)n , f : X R measur- 1 ⊂ L → able, limn fn(x) = f(x) for µ-almost all x X and g : fn(x) g(x), x X. →∞ ∈ ∃ ∈ L ≤ ∀ ∈ Then f 1 and ∈ L Z Z lim dµ fn = dµ f. n →∞ Useful notions of convergence in

3.29 Deﬁnition. Let f, fn : X R measurable n N. → ∀ ∈ (a) Convergence (µ-) almost everywhere:  , µ( ) = 0, such that a.e. ∃ N ∈ A N fn f (for µ-a.a. x): −−→ ⇐⇒ lim fn(x) = f(x), x X . n →∞ ∀ ∈ \N (b) Stochastic Convergence (convergence in measure):  > 0 A with µ(A) < µ ∀ ∀ ∈ A ∞ fn f : −→ ⇐⇒ lim µ x A : fn(x) f(x) > = 0 n →∞ { ∈ − } (c) Convergence in L1-norm:

1 fn k·k f : lim fn f = 0, n 1 −−→ ⇐⇒ →∞ k − k where Z 1 f 1 := dµ f L -norm. k k X | |

51 3.30 Theorem. (a) convergence almost everywhere = stochastic convergence ⇒ (b) convergence in L1-norm = stochastic convergence ⇒ A partial converse:

3.31 Theorem. Let µ be σ-ﬁnite. Then µ a.e. fn f = subsequences (fn )k of (fn)n a subsequence (fn )l such that fn f −→ ⇒ ∀ k ∃ kl kl −−→ as l → ∞ Next we turn to product spaces. Here only the product of two spaces. The generalisation to ﬁnitely many factors is obvious.

Notation. (X , , µ ), j = 1, 2, are measure spaces in the following. j Aj j 3.32 Deﬁnition. Product σ-algebra on Cartesian product X X 1 × 2 := σ A A : A ,A (X X ) A1 ⊗ A2 { 1 × 2 1 ∈ A1 2 ∈ A2} ⊆ P 1 × 2

3.33 Theorem (Existence and uniqueness of σ-finite product measures). Let µ1, µ2, be σ-finite. Then measure on , the product measure µ µ , with ∃1 A1 ⊗ A2 1 ⊗ 2 (µ µ )(A A ) = µ (A )µ(A ) A , j = 1, 2. 1 ⊗ 2 1 × 2 1 1 2 ∀ j ∈ Aj 3.34 Example. Lebesgue-Borel measure on Rd = R R. We have (Rd) = d = Nd d Nd × · · · × B B and λ = λ (Lebesgue Borel-measure on R). i=1 B i=1 3.35 Theorem (Fubini-Tonelli). Let (X , , µ ), (X , , µ ) be σ-finite measure spaces 1 A1 1 2 A2 2 and let f : X1 X2 R be ( 1 2, )-measurable. Then × → A ⊗ A B Z x1 µ2(dx2) f(x1, x2) is ( 1, )-measurable, 7→ X2 A B Z x2 µ1(dx1) f(x1, x2) is ( 2, )-measurable. 7→ X1 A B If one of the three integrals Z

(µ1 µ2)(dx1dx2) f(x1, x2) X1 X2 ⊗ × Z Z

µ1(dx1) µ2(dx2) f(x1, x2) X1 X2 Z Z

µ2(dx2) µ1(dx1) f(x1, x2) X2 X1 is ﬁnite, then so are the other two and Z Z Z (µ1 µ2)(dx1dx2) f(x1, x2) = µ1(dx1) µ2(dx2) f(x1, x2) X1 X2 ⊗ X1 X2 × Z Z = µ2(dx2) µ1(dx1) f(x1, x2). X2 X1

52 3.3 Lp-spaces

Notation. In this section: (X, , µ) a measure space, K = R or C, f : X K measurable. A → 3.36 Deﬁnition (Lp-norm). (a) For p ]0, [ set ∈ ∞ 1 Z p f := dµ f p (possibly ), k kp | | ∞

f := inf α > 0: µ x X : f(x) > α = 0 = inf sup f(x) k k∞ { ∈ } x X µN( ∈A)=0 ∈ \N A =: ess sup f(x) µ-essential supremum x X ∈ (b) For p ]0, ] introduce vector space of p-integrable functions w.r.t. µ ∈ ∞ p p p n o (µ) (X, µ) := f : X K: f measurable and f < L ≡ L ≡ L → k kp ∞ p = f : X K : f is µ-integrable { −→ | | } (c) Equivalence relation on p ∼ L f g : f = g µ-a.e. ∼ ⇐⇒ Vector space of equivalence classes of p-integrable functions w.r.t. µ: Lp := p/ . L ∼ 3.37 Lemma. Let f, g : X K be measurable. Then → (a) r, p, q [1, ] with 1 + 1 = 1 : fg f g (gen. H¨older’sinequality) ∀ ∈ ∞ p q r k kr ≤ k kp k kq (b) p [1, ]: f + g f + g (Minkowski’s inequality) ∀ ∈ ∞ k kp ≤ k kp k kp Proof. (a) Case q = : follows from fg f g . ∞ | | ≤ | | k k Remains to treat ∞ Case q < , 1 p < : Without loss of generality assume f, g 0 and f , g > ∞ ≤ ∞ ≥ k kp k kq 0. Claim: r r r r x p 1 + (x 1) = x + x 1, p r |{z} ≤ p − p q ∀ ≥ ≥ u(x) | {z } v(x)

r r 1 r true, because u(1) = v(1) = 1 and u (x) = x p − = v (x). 0 p ≤ p 0 Now, write x = α in claim and multiply inequality by β = β ⇒ r r r r α p β q α + β α β > 0 ( ) ≤ p q ∀ ≥ ∗ also true with p q = condition α β can be dropped. • ↔ ⇒ ≥ also true, if α = 0 or β = 0 = (*) holds α, β 0. • ⇒ ∀ ≥ f p gp (fg)r r f p r gq apply ( ) with α := p , β := q = r r p + q f p g q f p g q p f p q g q ∗ r k k k k ⇒ k k k k ≤ k k k k R dµ fg r r == k kr + = 1. ⇒ f r g r ≤ p q k kp k kq 53 (b) Without loss of generality assume

f, g 0 because f + g f + g • ≥ k kp ≤ | | | | p f + g > 0 (otherwise nothing to prove) •k kp Case p = : note that for α, β > 0 and γ = α + β ∞ µ f > α = 0 and µ g > β = 0 = µ f + g > γ = 0. { } { } ⇒ { } Thus n o f + g = inf γ > 0: µ f + g > γ = 0 k k∞ { } n o n o inf α > 0: µ f > α = 0 + inf β > 0: µ g > β = 0 ≤ { } { } = f + g . k k∞ k k∞ Case q p < : ≤ ∞ Z Z p p 1 p 1 f + g p = dµ f(f + g) − + dµ g(f + g) − k k X X H¨older p 1 p 1 f p (f + g) − + g p (f + g) − ≤ k k q k k q

| {z } p−1 q(p−1)! Z | {z } p dµ (f + g) p X | {z } kf+gkp−1 p 1 f + g − f + g . ≤ k kp k kp k kp

p 2 3.38 Lemma. (L , p) is a normed space p [1, ]. Moreover, L is an inner product space with scalar productk·k ∀ ∈ ∞ Z f, g := dµ(x) f(x) g(x) f, g L2. h i X ∀ ∈ Proof. f 0 clear. k kp ≥ Let f = 0 = f(x) = 0 for µ-a.e. x X by Thm. 3.25 (b), i.e. f = 0 in Lp. k kp ⇒ ∈ αf = α f clear. k kp | | k kp Triangle inequality given by Thm. 3.37 (b) – needs p 1! ≥ , is a scalar product with f, f = f 2. h· ·i h i k k2 3.39 Warning. Notation does not distinguish between equivalence classes and representatives!

3.40 Lemma. If µ(X) < , then Lq is dense in Lp 1 p q and ∞ ∀ ≤ ≤ ≤ ∞ k·kp ≤ 1 1 µ(X) p − q . k·kq

Proof. From 3.37 (a) with g = 1, we get f r f p 1 q , and cyclic permutation k k ≤ k k |k{zk} 1 − 1 µ(X) r p (r, p, q) (p, q, r). Thus, it suﬃces to prove L dense in Lp, p 1. Let f Lp. → ∞ ∀ ≥ ∈ Without loss of generality assume f 0 (otherwise decompose Re f, Im f in positive and ≥

54 negative parts). Now fn := min f, n L∞ n N = { } ∈ ∀ ∈ ⇒ Z Z p p p f fn p = dµ f min f, n dµ f k − k X − { } ≤ f n Z Z { ≥ } n = dµ f p dµ f p →∞ 0, X − f

3.41 Theorem (Riesz-Fischer). Let (X, , µ) be a measure space. Then Lp is a Banach A space p [1, ]. In particular, L2 is a Hilbert space. ∀ ∈ ∞ Proof. All that remains to be shown is completeness p [1, ]. ∀ ∈ ∞ Case p = : Let (fn)n N L∞ be Cauchy. Since ∞ ∈ ⊂ c n, m N null set Nn,m , µ(Nn,m) = 0: fn(x) fm(x) fn fm , x Nn,m ∀ ∈ ∃ ∈ A − ≤ k − k∞ ∀ ∈ : S c = N = n,m N Nn,m is a null set and fn(x) n is Cauchy in K, uniformly x N . ( ) ⇒ ∈ ∀ ∈ ∗ ( c K complete limn fn(x), x N = X x f(x) := →∞ ∈ well deﬁned, measurable and ⇒ 3 7→ 0, x N ∈

f fn sup f(x) fn(x) k − k∞ ≤ x N c − ∈ = sup lim fm(x) fn(x) c m x N →∞ − ∈ sup sup fm(x) fn(x) ≤ x N c m n − ∈ ≥ | {z } fm fn ≤k − k∞

( ) =∗ lim f fn = 0. n ⇒ →∞ k − k∞ p Case 1 p < : Let (fn)n L be Cauchy. ≤ ∞ ⊂ k = subsequence (nk)k N: fnk fnk+1 p < 2− k N. ⇒ ∃ ⊂ k − k ∀ ∈ X∞ p X∞ k Now, h := fn1 + fnk+1 fnk L because h p fn1 + 2− < . | | | − | ∈ k k |{z}≤ p ∞ k=1 triangle k=1 | {z } 1 X∞ Thm. 1.25 X∞ In particular, f f Lp ======f (x) f (x) < for µ-a.a. x X | nk+1 − nk | ∈ ⇒ | nk+1 − nk | ∞ ∈ k=1 k=1 = fnk (x) K is Cauchy for µ-a.a. x X, because ⇒ k ⊂ ∈

k2 1 − ∞ X X k1 fn (x) fnk (x) fnj+1 (x) fnj (x) ... →∞ 0 k2 − 1 ≤ − ≤ | | −−−−→ j=k1 j=k1

K complete ======f : X K measurable with f(x) = limk fnk (x) for µ-a.a. x X. →∞ ⇒ ∃ → Pk 1 ∈ Moreover, fnk h k N since fnk = fn1 + − fnj+1 fnj = f h a.e. ≤ ∀ ∈ j=1 − ⇒ | | ≤

55 p p 1 = fnk , f L . ⇒ | | ∈ p k Let gk := fn f = gk →∞ 0 a.e. and k − ⇒ −−−→ p p 1 0 gk fnk + f 2 h L k N ≤ ≤ | | ≤ | | ∈ ∀ ∈ dom. cvg. R = limk dµgk = 0 = subsequence (fnk )k converges to f in p. ⇒ →∞ X ⇒ k·k (f ) Cauchy = (f ) converges to f in . n n ⇒ n n k·kp 3.42 Theorem (Clarkson inequalities). Let p ]1, 2[, let f, g Lp and 1 + 1 = 1. Then: ∈ ∈ p q q q q 1 f + g f g 1 p 1 p − + − f p + g p , (1<) 2 p 2 p ≤ 2 k k 2 k k p p f + g f g 1 p 1 p + − f p + g p . (2<) 2 p 2 p ≥ 2 k k 2 k k For p [2, [, the inequalities are reversed. We refer to them as (1 ) and (2 ). ∈ ∞ > > Proof. The case p = 2 reduces to the parallelogram identity. For the case of general p, see e.g. R. A. Adams, Sobolev spaces, or Hirzebruch/Scharlau.

3.43 Theorem. Let p ]1, [. Then Lp is uniformly convex, i.e. ε > 0 δ > 0 f, g Lp ∈ ∞ ∀ ∃ ∀ ∈ with f = g = 1 k kp k kp

f + g f g p ε = 1 δ. k − k ≥ ⇒ 2 p ≤ − Proof. Case 1 < p 2: From (1 ) we get ≤ < = 1 · p f q f + g q q 1 2− ε . X 2 p ≤ − g Case 2 p < : From (2>) we get ≤ ∞ f + g p f + g p p 2 1 2− ε . X 2 p ≤ − 3.44 Theorem (Riesz representation for (Lp) ). Let (X, , µ) be a measure space. Let ∗ A p [1, [ and 1/p + 1/q = 1. If p = 1, assume in addition that µ is σ-finite. Then the mapping∈ ∞ Lq (Lp) Z J : −→ ∗ where ` : Lp g dµ fg f ` f 3 7→ 7−→ f X q p is an isometric isomorphism, L ∼= (L )∗. q Proof. J is clearly linear because of the linearity of the integral. Now let f L . `f is well- 1 p ∈ defined because fg L by Hölder’sinequality for every g L and `f (g) f q g p. p ∈ ∈ | | ≤ k k k k So `f (L )∗ and ∈ `f p ∗ f . ( ) (L ) ≤ k kq ∗ Thus, J is well defined and bounded.

Claim: `f p ∗ = f (= J is isometric and hence injective). (L ) k kq ⇒ Proof of the claim. W.l.o.g. assume f = 0 (the case f = 0 is clear!). 6

56 Case 1 < p < : Let • ∞ q 2 g := f f − ( ) | | ∗∗ with the convention g(x) := 0 if f(x) := 0. Then

p (q 1)p q(1 1/q)p q 1 g = f − = f − = f L . | | | | | | | | ∈ So g Lp and g = f q/p. Furthermore ∈ k kp k kq R q q `f (g) dµ f f q q(1 1/p) = = k k = f − = f . | | q/p q q g p g p f k k k k k k k k k kq This gives

`f p ∗ f , (L ) ≥ k kq and the claim follows with ( ). ∗ Case p = 1: Since µ is σ-ﬁnite we have • [ X = Xn,Xn Xn+1 n N ⊆ ∀ ∈ n N ∈

with µ(Xn) < for every n N. Let ε > 0 and define ∞ ∈ M := x X : f(x) f ε . { ∈ | | ≥ k k∞ − } By definition of the essential supremum, µ(M) > 0. Then M := M X obeys n ∩ n 0 < µ(M ) < n ∞ for all sufficiently large n. Let

f 1 g := 1 f µ(M ) Mn | | n (again with the convention g(x) := 0 if f(x) := 0). Then g L1 with g = 1 and ∈ k k1 Z 1 Z `f (g) = dµ fg = dµ f f ε. X µ(Mn) Mn | | ≥ k k∞ − Since this is true for every ε > 0, this gives

`f (L1)∗ f , ≥ k k∞ and the claim follows with ( ). ∗ p It remains to prove that J is surjective. Let ξ (L )∗. We have to show that there is q ∈ some f L such that ξ = ` . W.l.o.g. assume ξ p ∗ = 1. ∈ f k k(L ) Case p > 1: Idea: “Determine maximiser of ξ, i.e. find g Lp such that ξ(g) = 1 = ∈ ξ p ∗ and determine f from g as in ( ).” k k(L ) ∗∗ p k By definition of the sup a sequence (gk)k L , gk = 1 k N,with ξ(gk) →∞ 1. ∃ ⊆ k kp ∀ ∈ | | −→ W.l.o.g. assume ξ(gk) > 0 for finally every k (otherwise multiply gk by an appropriate overall phase factor).

Claim:(gk)k is a Cauchy sequence.

57 Proof by contradiction (needs uniform convexity): assume this is not the case. Then ε > 0 and subsequences (kj)j N and (k0 )j N with ∃ ⊆ j ⊆

0 gkj gk p ε j N. k − j k ≥ ∀ ∈ Uniform convexity (Thm. 3.43) = δ > 0 such that ⇒ ∃ gk + g 0 j kj 1 δ j N. 2 p ≤ − ∀ ∈ On the other hand, we also have ! ξ ∗=1 gk + g 0 k k j kj 1 1 ξ = ξ g + ξ g 0 kj kj ≥ gk + g 0 p gk + g 0 p k j kj k k j kj k 1 ξ(gk + ξ(g 0 ) ≥ 2(1 δ) j kj − | {z } j→∞ 2 −−−→ 1 = 1 1 δ , proving the claim. ⇒ ≥ − p p So (gk)k is a Cauchy sequence and by completeness of L , there exists g L , such that k ∈ g →∞ g in Lp. Also g = 1 and k −→ k kp

ξ(g) = lim ξ(gk) = 1. k →∞ Deﬁne f := g g p 2 (with the convention f(x) := 0, if g(x) = 0). Note f q = g (p 1)q = | | − | | | | − g p L1 = f Lq with f = g p/q = 1. Thus, | | ∈ ⇒ ∈ k kq k kp Z p p `f : L g˜ dµ fg˜ (L )∗ and 3 7→ X ∈ Z

`f (g) = dµ fg = 1 = f q = `f (Lp)∗ X |{z} k k = g p | |

(see the claim after ( ) for the last equality). Now, ξ = `f follows from ∗ p p Claim: Let ξ , ξ (L ) with ξ p ∗ = 1 = ξ p ∗ . Suppose g L , g = 1, 1 2 ∈ ∗ k 1k(L ) k 2k(L ) ∃ ∈ k kp such that ξ1(g) = ξ2(g) = 1. Then ξ1 = ξ2. Proof by contradiction (needs Clarkson inequalities). Suppose the claim is wrong = ⇒ h˜ Lp such that ξ (h˜) = ξ (h˜). Then the vectors ∃ ∈ 1 6 2 ! ! ! 1 ξ (g) ξ (h˜) = 1 =: u and 1 =: v 1 ξ2(g) ξ2(h˜) are linearly independent in K2. So they span K2. Thus α, β K such that ∃ ∈ ! ! 1 ξ (h) = αu + βv = 1 where h := αg + βh˜ Lp. 1 ξ2(h) ∈ −

= ξ1(g + th) = 1 + t = ξ2(g th) t 0. Since ξj p ∗ = 1, j = 1, 2, we must have ⇒ − ∀ ≥ (L ) 1 + t 1. g th ≤ k ± kp

58 Case p ]1, 2]: Clarkson inequality = • ∈ ⇒ p p p p (g + th) + (g th) (g + th) (g th) α(t) := 1 + t h p = − + − − k k 2 p 2 p (2<) 1 1 g + th p + g th p ≥ 2 k kp 2 k − kp (1 + t)p =: β(t). ( ) ≥ ∗∗∗ We know

– α(0) = 1 = β(0), – α (t) = ptp 1 h p, 0 − k kp p 1 – β0(t) = p(1 + t) − . Since p > 1 we have α (t) < β (t) t 0 suﬃciently small. So α(t) < β(t) t > 0 0 0 ∀ ≥ ∀ suﬃciently small, which contradicts ( ). ∗∗∗ Case p [2, [: Clarkson inequality = • ∈ ∞ ⇒ q q q q (g + th) + (g th) (g + th) (g th) 1 + t h p = − + − − k k 2 p 2 p q 1 (1>) 1 1 − g + th p + g th p (1 + t)q ≥ 2 k kp 2 k − kp ≥

Contradiction as in the case p ]1, 2] above. = Claim is true = J surjective for ∈ ⇒ ⇒ p > 1.

Case p = 1: Idea: “Step 1. reduce to the above case p > 1, if µ is ﬁnite. Step 2. generalise to σ-ﬁnite µ.”

1. Assume ﬁrst µ(X) < . Letp ˜ > 1 and 1/p˜ + 1/q˜ = 1. Recall from Lemma 3.40 p˜ 1 ∞ that L∞ L L (the inclusions being dense) ⊆ ⊆ 1 1/p˜ p˜ p˜ = ξ(˜g) ξ 1 ∗ g˜ µ(X) − g˜ g˜ L = ξ (L )∗. ⇒ ≤ k k(L ) k k1 ≤ k kp˜ ∀ ∈ ⇒ ∈ | {z } | {z } 1 =:Mp˜ q˜ From the already proved Riesz representation forp ˜ > 1 = 1fp˜ L such that ⇒ ∃ ∈ ξ p˜ = `f and fp˜ Mp˜. Thus p1, p2 > 1, g˜ L∞ |L p˜ q˜ ≤ ∀ ∀ ∈ Z ` (˜g) = ` (˜g) i.e. dµ (f f )˜g = 0. fp1 fp2 p1 p2 X −

Chooseg ˜ := (f f )/ f f = f = f =: f is independent of the index. p1 − p2 | p1 − p2 | ⇒ p1 p2 \ q˜ 1/q˜ = f L and f q˜ Mp˜ = [µ(X)] q˜ ]1, [ ⇒ ∈ k k ≤ | {z } ∀ ∈ ∞ q>˜ 1 bounded inq ˜ Problem 30 1/q˜ = f L∞ and f lim [µ(X)] = 1 ⇒ ∈ k k ≤ q˜ ∞ →∞ = ` (L1) . Since Lp˜ dense in L1 = ξ and ` agree on a dense subspace ⇒ f ∈ ∗ ⇒ f Thm.= 2.32 ξ = ` on L1. ⇒ f

59 S˙ 2. Relax finiteness assumption: assume µ is σ-finite. = X = n NXn with µ(Xn) < 1 ⇒ ∈ n N. Let ξ L (X)∗ = ∞ ∀ ∈ ∈ ⇒ 1 1. 1 ξ L (Xn)∗ n N = 1fn L∞(Xn): ξ = `fn on L (Xn) ∈ ∀ ∈ ⇒ ∃ ∈ and ξ 1 ∗ = fn ∞ n N. k kL (Xn) k kL (Xn) ∀ ∈ Define ( X fn(x) x Xn f := fñ, fñ(x) := ∈ 0 x Xn n N ∈ 6∈ = f L (X) because f ∞ ξ 1 ∗ . ⇒ ∈ ∞ k kL (X) ≤ k kL (X) For g L1(X) let g := PN g 1 = ∈ N n=1 Xn ⇒ Z X∞ N →∞ g gN 1 = dµ g 1Xn 0 k − k | | −−−−−−→dom. cvg. X n=N+1 Z Z X X X ˜ = ξ(g) = lim ξ(gN ) = ξ(g 1Xn ) = dµ fng = dµ fng. ⇒ N X X →∞ n N | {z } n N n n N ∈ `fn (g) ∈ ∈

PN ˜ 1 Since n=1 fng fg L N N (H¨older!),dominated convergence gives ≤ | | ∈ ∀ Z∈ ξ(g) = dµ fg = `f (g). X 3.45 Deﬁnition. Let X be a topological space. Then (a) X is locally compact : For all x X there exists a compact neighbourhood ⇐⇒ ∈ -compact S (b) X is σ : There exist Xn X compact n N such that X = n N Xn ⇐⇒ ⊆ ∀ ∈ ∈ (c) space of continuous functions with compact support supp(f) := x X : f(x) = 0 { ∈ 6 } n o C (X) := f C(X) : supp(f) compact c ∈

(d) space of continuous functions ”converging to 0 outside compacts” n o C (X) := f C(X): ε > 0 K X compact with f(x) ε x X K 0 ∈ ∀ ∃ ε ⊆ | | ≤ ∀ ∈ \ ε

3.46 Example. X compact = X locally compact and σ-compact. • ⇒ Rd is locally compact and σ-compact • d C0(R ) is the set of all continuous functions vanishing at . • ∞ 3.47 Theorem. Let X be a locally compact and σ-compact Hausdorﬀ space. Let µ be a regular Borel measure on X and p [1, [. Then Lp(X, µ) is separable. ∈ ∞ p Idea of the proof: 1. Show Cc(X) dense in L (see e.g. Rudin, Real and complex analysis, Thm. 3.14). Needs regularity of Borel measures on X!

2. Show that Cc(X) is separable by reducing it to separability of C(Xn) known!

3.48 Corollary. Let p [1, [. Then Lp(Rd) Lp(Rd, λd) is separable. Here, λd is the ∈ ∞ ≡ Lebesgue-Borel measure on Rd.

60 From now on we only consider X = Rd and µ = λd. 3.49 Theorem (Young’s inequality). Let p, q, r [1, ] such that ∈ ∞ 1 1 1 + + = 2. ( ) p q r ∗ Let f Lp(Rd), g Lq(Rd), h Lr(Rd). Then ∈ ∈ ∈

Z Z

I := dx dy f(x)g(x y)h(y) dx dy f(x) g(x y) h(y) Rd Rd − ≤ Rd Rd | || − || | × × f g h . ≤ k kpk kqk kr Proof. A smart application of H¨older. W.l.o.g. assume f 0, g 0, h 0. Introduce ≥ ≥ ≥ conjugate exponents 1 1 1 1 1 1 + = 1, + = 1, + = 1 p p0 q q0 r r0 Then 1 1 1 1 1 1 ( ) + + = 3 =∗ 1. ( ) p0 q0 r0 − p − q − r ∗∗ (a) Case p, q, r ]1, [: Let ∈ ∞ 0 0 α(x, y) := f(x)p/r g(x y)q/r , 0 − 0 β(x, y) := g(x y)q/p h(y)r/p , − 0 0 γ(x, y) := f(x)p/q h(y)r/q . We have p p 1 1 1 + = p + = p 1 = 1, r0 q0 r0 q0 − p0 q q + = = 1, r0 p0 ··· r r + = = 1. p0 q0 ··· Thus, α(x, y)β(x, y)γ(x, y) = f(x)g(x y)h(y). − Apply the (generalised) H¨olderinequality twice for functions on Rd Rd, using that × r and r are conjugate, respectively ( ), 0 ∗∗ I = αβγ α 0 βγ α 0 β 0 γ 0 , ( ) k k1 ≤ k kr k kr ≤ k kr k kp k kq ∗∗∗ where 0 Z !1/r p q α r0 = dx dy f(x) g(x y) = k k Rd Rd − × 0 Z Z !1/r p q p/r0 q/p0 = dx f(x) dy g(y) = f p g q , Rd Rd k k k k q/p0 r/p0 β 0 = = g h , k kp ··· k kq k kr p/q0 r/q0 γ 0 = = f h k kq ··· k kp k kr so that α 0 β 0 γ 0 = f g h , and the claim follows with ( ). k kr k kp k kq k kpk kqk kr ∗∗∗

61 (b) Case p = (analogous to q = , r = ): = q = r = 1 by ( ). ∞ ∞ ∞ ⇒ ∗ Take out f from I by f and separate the remaining integral. k k∞ 1 1 (c) Case p = 1 (analogous to q = 1 or r = 1): = q + r = 1 by ( ). R ⇒ ∗ Apply H¨olderto obtain d dy g(x y)h(y) g h . R − ≤ k kqk kr

3.50 Theorem. Let 1 1 + 1 2 and f Lp(Rd), g Lq(Rd). Then the convolution ≤ p q ≤ ∈ ∈ of f and g, Z (f g)(x) := dy f(x y)g(y), ∗ Rd − exists for Lebesgue almost every x Rd, is commutative f g = g f and (the equivalence class) obeys ∈ ∗ ∗ r d f g L (R ) with f g r f p g q, ∗ ∈ k ∗ k ≤ k k k k where 1 := 1 + 1 + 1 . r − p q 1 1 r d Proof. Rename r r0 and let r := 1 r0 . Let h L (R ). Because of Young’s inequality 1 1 → 1 1 − ∈ and 1 = 0 = 1 + + , we have − r r − p q Z dx dy f(y) g(x y) h(x) f p g q h r < , Rd Rd | || − || | ≤ k k k k k k ∞ × so Fubini gives Z x h(x) dy f(y) g(x y) L1. 7−→ | | Rd | || − | ∈ (a) Case r = : choose h = 1 (allowed!) = f g L1 and f g f g . ∞ ⇒ ∗ ∈ k ∗ k1 ≤ k kpk kq R r d (b) Case r < : deﬁne `(h) := d dx (f g)(x) h(x) for h L (R ). ∞ R ∗ ∈ ` is linear in h • ` is bounded, because `(h) f g h . • | | ≤ k kpk kqk kr r d = ` L (R )∗ with ` (Lr)∗ f p g q. By Theorem 3.44 (Riesz representation) ⇒ ∈ r0 d k k ≤R k k k k 1 ξ L (R ) such that `(h) = d dx ξ(x)h(x) = ∃ ∈ R ⇒ Z r d dx [(f g)(x) ξ(x)] h(x) = 0 h L (R ). ( ) Rd ∗ − ∀ ∈ ∗ Now, choose h = 1 (f g ξ)/ f g ξ , where B d is of ﬁnite Lebesgue B ∗ − | ∗ − | ∈ B measure, λd(B) < , but otherwise arbitrary. (Here, we use the convention that ∞ h(x) = 0, whenever f g(x) ξ(x) = 0.) Thus, h Lr(Rd) and ( ) gives ∗ − ∈ ∗ Z dx (f g)(x) ξ(x) = 0 B d with λd(B) < B | ∗ − | ∀ ∈ B ∞

Riesz = ξ = f g and f g 0 = ξ 0 = ` r ∗ f g . ⇒ ∗ k ∗ kr k kr k k(L ) ≤ k kpk kq Commutativity:

Z y=:x z Z (f g)(x) = dy f(x y)g(y) =− dz f(z)g(x z) = (g f)(x). ∗ Rd − Rd − ∗

62 3.51 Definition. Let ∅ = Ω Rd be open. 6 ⊆ For k N: space of k-times continuously differentiable functions • ∈ ( ) α1 αd k f C(Ω) such that ∂x1 ... ∂xd f C(Ω), C (Ω) := ∈ · P· d ∈ where α1, . . . , αd N0 with αd = k ∈ j=1 space of arbitrarily often differentiable functions

\ k C∞(Ω) := C (Ω) n N ∈

For k N : space of k-times continuously diﬀerentiable functions with • ∈ ∪ {∞} compact support Ck(Ω) := Ck(Ω) C (Ω) c ∩ c and space of k-times continuously diﬀerentiable functions vanishing at ∞ Ck(Ω) := Ck(Ω) C (Ω). 0 ∩ 0 The main result:

d 3.52 Theorem. Let ∅ = Ω R be open and let p [1, [. Then C∞(Ω) is dense in 6 ⊆ ∈ ∞ c Lp(Ω).

3.53 Remark. Thus

p k·k p (a) Cc∞(Ω) = L (Ω). (b) Lp(Ω) is separable, since C (Ω) C (Ω) is separable. c∞ ⊂ c ∞ k·k (c) Cc∞(Ω) = C0∞(Ω) (exercise). An important technical result: Lp functions are made arbitrarily often diﬀerentiable by a convolution with a molliﬁer.

p d 3.54 Lemma. Let p [1, ], f L (Ω) with compact support in Ω. Let j C∞(R ), R ∈ ∞ ∈ ∈ c j 0, d dx j(x) = 1 and for ε > 0 define the mollifier ≥ R d R R 0 jε : −→ ≥ . x ε dj(x/ε) 7−→ − Then for all sufficiently small ε > 0: f := f j C (Ω) and f f . (Here, f ε ∗ ε ∈ c∞ k εkp ≤ k kp may be trivially extended to Rd). Proof. Norm: use Theorem 3.50 with r = p. Then q = 1, because 1 = 1 + 1 + 1 there. Thus, r − p q f f j = f . k εkp ≤ k kpk εk1 k kp Support: let δ := dist(supp f, ∂Ω). δ > 0 by Problem 9c, since supp f is compact and ∂Ω is closed. By definition of the mollifier ε > 0 such that for all ε ]0, ε ]: ∃ 0 ∈ 0 supp jε Bδ/2(0). Since ⊂ Z fε(x) = dy jε(x y)f(y), Rd −

63 we infer [ supp f B (x) Ω ε ⊆ δ/2 ⊂ x supp f ∈ and supp f is compact. Derivatives: let e Rd, e = 1 be a unit vector and λ R 0 . Then x Ω ∈ | | ∈ \{ } ∀ ∈ Z 1 1 Fλ(x) := fε(x + λe) fε(x) = dy jε(x y + λe) jε(x y) f(y) λ − λ Rd − − − 1 Z Z 1 d = dy dt jε(x y + tλe) f(y) λ d dt − R 0 | {z } λe ( jε)(x y+tλe) · ∇ − Z Z 1 ! = dy dt e ( jε)(x y + tλe) f(y). Rd 0 · ∇ − | {z } D

1 p Since D jε < and f L (because f L and of compact support), a | | ≤ k|∇ |k∞ ∞ ∈ ∈ double application of dominated convergence shows that Z e ( fε)(x) = lim Fλ(x) = dy e ( jε)(x y) f(y) · ∇ λ 0 d · ∇ − → R exists x Ω. The existence of higher derivatives follows by induction in the same ∀ ∈ manner.

3.55 Lemma. Let p [1, [, let A (Rd) be bounded. Then ∈ ∞ ∈ B

lim 1A jε 1A p = 0. ε 0 k − ∗ k & Proof of Theorem 3.52. Step 1: let ε > 0, f Lp(Ω). For n N deﬁne ∈ ∈ 1 K := x Ω : dist(x, ∂Ω) and x n . n ∈ ≥ n | | ≤ S Ω is open, thus Ω = n N Kn. Also Kn Kn+1 for all n N. By dominated convergence ∈ ⊆ ∈

lim f f 1Kn = lim f(1 1Kn ) = 0. n p n p −→∞ − −→∞ −

This implies that there exists n N such that f f 1Kn p < ε/3. ∈ k − k Step 2: we know from Problem T14 that step functions are dense in Lp: there exists d L N, c1, . . . , cL K and A1,...,AL (R ) with Al Kn l = 1,...,L such that for ∈ PL ∈ ∈ B ⊂ ∀ g := l=1 cl 1Al we have

supp g supp(f1 ) and g f 1 < ε/3. ⊆ Kn k − Kn kp Step 3: mollify step function: δ > 0 such that ∃ j g C (Ω) by Lemma 3.54. • δ ∗ ∈ c∞ j g g < ε/3 by Lemma 3.55, because g has ﬁnitely many steps. •k δ ∗ − kp

64 In summary we have: f jδ g p < ε. k − | {z∗ } k C∞(Ω) ∈ c Proof of Lemma 3.55. Step 1: we claim it suﬃces to prove the lemma for p = 2. d p d For, consider g L∞(R ) with bounded support (= g L (R ) p [1, [). ∈ ⇒ ∈ ∀ ∈ ∞ p p 2 2 p 2: g p g − g 2. • ≥ k k ≤| k k{z∞ } k k < ∞ p [1, 2]: by applying H¨olderwith p = 2/p 1 and q conjugated to p • ∈ 0 ≥ 0 0 1 1 Z Z p0 1 p p 2 d − p0 g p = g 1supp g g λ (supp g) . k k | | ≤ | | | {z } | {z } < 2/p0 ∞ g = g p k k2 k k2

Step 2: Z 1A(x) (jε 1A)(x) = dy jε(y) 1A(x) 1A(y x) − ∗ Rd | {z } − − √jε(y√jε(y) ! 1 1 Z 2 2 2 h i jε 1 dy jε(y) 1A(x) 1A(x y) . ≤| k {zk } Rd | −{z − } =1 =:G(x,y)

Thus,

2 Z Z 2 1A jε 1A dx dy jε(y) G(x, y) − ∗ 2 ≤ Rd Rd Z Z h i = dx dy 1A(x) + 1A(x y) 2 1A(x)1A(x y) Rd Rd − − · − Z Z = 2 dy jε(y) dx G(x, y) Rd A z:=y/ε Z Z = 2 dz j(z) dx G(x, εz) . Rd A | {z } Iε(z)

Since I (z) 2λd(A) and j L1, dominated convergence gives | ε | ≤ ∈ Z Z 2 h i lim 1A jε 1A 2 2 dz j(z) lim dx 1A(x) 1A(x εz) . ε 0 k − ∗ k ≤ d ε 0 − − & R & A Step 3: We need to prove Z Z z 0 J(z) := dx 1A(x z) → dx. A − −−−→ A Since J(z) R dx z it is enough to show ≤ A ∀ Z lim inf J(z) dx. ( ) z 0 ≥ ∗ → A

65 Lebesgue measure is outer regular (Def. 3.8 (b) and Cor. 3.10): Z δ > 0 B A, B open and dx < δ. ∀ ∃ ⊇ B A \ Now Z Z J(z) = dx 1B(x z) dx 1B A(x z) A − − A \ − | R {z } d dx 1B\A(x z)<δ ≤ R − Z dx 1B(x z) δ. ≥ A − −

B open Fatou = limz 0 1B(x z) = 1B(x) for all x B = ⇒ → − ∈ ⇒ Z A B Z lim inf J(z) dx lim inf 1B(x z) δ =⊆ dx δ z 0 ≥ z 0 − − − → A → A which holds for all δ > 0. This proves ( ). ∗ 3.4 Decomposition of Measures11 Notation. is a σ-Algebra on a space X; µ, ν are measures on . A A 3.56 Deﬁnition. (a)

µ is absolutely continuous w.r.t.  ν(A) = 0 for A = µ(A) = 0 ν, : ∈ A ⇒ ⇐⇒ (“every ν-null set is a µ-null set”) in symbols, µ ν  (b)  has a density w.r.t.   h : X [0, ] measur- µ h ν   ∃ → ∞ (or Radon-Nikodym deriva-  able and unique up to modiﬁca- : tions on ν-null sets: A tive), ⇐⇒ Z   ∀ ∈ A in symbols, h = dµ/dν   µ(A) = dν h.  A

(c) µ and ν are mutually singular, A : µ(A) = 0 and ν(X A) = 0 : ∃ ∈ A \ in symbols, µ ν ⇐⇒ (“µ is concentrated on X A, ν on A”) ⊥ \ (d) sum of measures [0, ], µ + ν : A → ∞ A µ(A) + ν(A) 7→ is a measure.

3.57 Theorem (Radon-Nikodym). Let µ be a ﬁnite measure and let ν be σ-ﬁnite. Then measure µac and measure µs such that ∃1 ∃1

µ = µac + µs, where µac ν and µs ν (hence also µac µs). Moreover, µac has a density w.r.t. ν ⊥ ⊥ which is ν-integrable.

11optional reading

66 In the special case µs = 0, we get 3.58 Corollary. Let µ be a finite measure and let ν be σ-finite. Then µ ν µ has a density w.r.t. ν which is ν-integrable. ⇐⇒ 3.59 Remark. Thm. 3.57 and Cor. 3.58 can be extended to σ-finite measures µ, but then the density will only be ν-integrable over sets of finite µ-measure. Proof of Theorem 3.57 (after von Neumann). First act: assume ν finite. Define ϕ := µ + ν = R dϕ f = R dµ f + R dν f f 0 measurable. ⇒ X X X ∀ ≥ Now, let f L2(X, ϕ) ∈ Z Z Z CSI 1/2 = dµ f dµ f dϕ f 1 f 2;ϕ ϕ(X) < ⇒ X ≤ X | | ≤ X | | · ≤ k k ∞ = f R dµ f L2(ϕ) ∗. ⇒ 7→ X ∈ Z Z Riesz= g L2(ϕ): f L2(ϕ) dµ f = dϕgf. ( ) ⇒ ∃ ∈ ∀ ∈ X X ∗ Choose f = 1 in ( ), where A = µ(A) = R dϕ g, and since 0 µ(A) ϕ(A) A ∗ ∈ A ⇒ A ≤ ≤ 1 Z = 0 dϕ g 1 A with ϕ(A) > 0 ⇒ ≤ ϕ(A) A ≤ ∀ ∈ A = g(x) [0, 1] for ϕ-a.a. x X ⇒ ∈ ∈ (because otherwise choose A := x X : g(x) > 1 or A := x X : g(x) < 0 and get { ∈ } { ∈ } a contradiction). Now, modify g on ϕ-null set such that g(x) [0, 1] x X. Note this ∈ ∀ ∈ does not affect ( ). ∗ ( ) Z Z =∗ dµ f(1 g) = dν fg f L2(ϕ). ( ) ⇒ X − X ∀ ∈ ∗∗ Let n o Aac := x X : g(x) [0, 1[ ∈ ∈ ∈ A n o A := x X : g(x) = 1 = X Aac s ∈ \ ∈ A and define two finite measures [0, [ µκ : A → ∞ for κ = ac, s. B µ(B A ) 7→ ∩ κ Claim: µs ν ⊥ R For, choose f = 1As in ( ) = 0 = A dν g = ν(As) and µs(X As) = 0. ∗∗ ⇒ s |{z} | {z\ } 1 Aac Claim: µac has a density w.r.t. ν (= µac ν) ⇒ For, choose f = (1 + g + g2 + ... + gn)1 L (ϕ) L2(ϕ) in ( ), where B and B ∈ ∞ ⊆ ∗∗ ∈ A n N. ∈ n+1 Z Z X = dµ (1 gn+1) = dµ gj . ⇒ − B B j=1 | {z } =:hn

67 (hn)n is monotone increasing sequence with limit h := limn hn 0 (possibly ). →∞ ≥ ∞ Z Z Z monot. cvg. n+1 = dν h = lim dν hn = lim dµ (1 g ) n n ⇒ B →∞ B →∞ B | −{z } n+1 1A (1 g ) ac − |∈{z[0,1[}

dom. cvg. Z = dµ 1 = µac(B) < B Aac ∞ ∩ 1 B , i.e. µac ν. In particular, for B = X = h L (X, ν). ∀ ∈ A ⇒ ∈ Uniqueness: Assume µ = µac + µs =µ ˜ac +µ ˜s.

= µac µãc = µs µ˜s = µac µãc = 0 = µs µ˜s. ⇒ | {z− } | {z− } ⇒ − − ν ν ⊥ Second act: extend to ν σ-finite. Decompose X = S˙ X , X pairwise disjoint, ν(X ) < , use the first act on each X n N n n n ∞ n and assemble everything.∈

Towards a ﬁner decomposition of measures: 3.60 Deﬁnition. Let (X, , µ) be a measure space. Assume x x X. Let A { } ∈ A ∀ ∈ n o App := x X : µ( x ) > 0 set of atoms of µ. ∈ { } (a) pure point part of µ [0, ] µpp : A → ∞ X B µ B App = µ( x ) 7→ ∩ { } x B App ∈ ∩ (b) continuous part of µ [0, ] µc : A → ∞ B µ B (X App) 7→ ∩ \

(c) µ is only pure point : µ = µpp (i.e. µ = 0) ⇐⇒ c (d) µ is continuous : µ = µ (i.e. µpp = 0) ⇐⇒ c (e) µ is singular continuous w.r.t. another measure ν : µpp = 0 and µ ν ⇐⇒ ⊥ 3.61 Remark. µ is σ-finite = App is at most countably infinite. (See digression on ⇒ uncountable series in the proof of Lemma 2.49) Collecting facts, we get 3.62 Theorem (Lebesgue decomposition). Let µ, ν be σ-finite measures on , x A { } ∈ x X and assume νpp = 0. Then decomposition A ∀ ∈ ∃1 µ = µac + µsc + µpp = µac + µs = µc + µpp, where µac ν, µsc ν and µpp ν. ⊥ ⊥ 3.63 Remark. (a) Theorem 3.62 mostly used with ν = λd (Lebesgue-Borel).

(b) Condition νpp = 0 needed, because otherwise ambiguities in distinguishing µpp and ∃ µsc.

68 4 The cornerstones of functional analysis

4.1 Hahn-Banach theorem 4.1 Definition. Let M be a set and D M M. Let be the associated binary relation: ⊆ × ≺ x y : (x, y) D. ≺ ⇐⇒ ∈ (a) is a partial ordering (or M a partially ordered set) iff for all x, y, z M: ≺ ∈ x x (reflexive) • ≺ if x y and y x, then x = y, (antisymmetry) • ≺ ≺ if x y and y z, then x z (transitive). • ≺ ≺ ≺ (b) x, y M are comparable iff x y or y x. x, y are incomparable iff x, y are not ∈ ≺ ≺ comparable.

(c) is a total ordering iff ≺ is a partial ordering •≺ x, y are comparable for all x, y M. • ∈ (d) Let W M. u M is an upper bound of W iff w u w W . ⊆ ∈ ≺ ∀ ∈ (e) m W is a maximal element of W iff the implication ∈ m w for w M = m = w ≺ ∈ ⇒ holds. (Note: a maximal element need not be an upper bound and vice versa.)

4.2 Example. “ ” is a total ordering on R. • ≤ “ ” is a partial ordering on (X), but not a total ordering. • ⊆ P The following axiom is equivalent to the axiom of choice.

4.3 Axiom (Zorn’s Lemma). Let M = ∅ be a partially ordered set. Suppose every totally 6 ordered subset of M has an upper bound. Then M hat a maximal element.

The next Theorem ﬁnishes the proof of Theorem 2.51.

4.4 Theorem. Every Hilbert space X = 0 has an orthonormal basis. 6 { } Proof. Let M := X orthonormal = {E ⊆ | E } ⇒ M = ∅. • 6 “ is a partial ordering on M. • ”⊆ Let W be a totally ordered subset of M. Then S M (because it is orthonor- • W E ∈ mal due to W being totally ordered) is an upperE∈ bound for W .

=Zorn M has a maximal element . ⇒ M Claim: is an orthonormal basis M Orthonormal is clear since M. • M ∈

69 Suppose: is not complete. Then there is some 0 = x X such that x m for • M 6 ∈ ⊥ every m . Thus ∈ M x 0 := M M x ∪ M ∈ k k (since 0 is orthonormal). So & 0 which contradicts being maximal. M M M M 4.5 Remark. Similar arguments prove Theorem 2.4 (existence of a Hamel basis; see exercise).

4.6 Theorem (Hahn-Banach; real version). Let X be an R-vector space, let p: X R → be convex, i.e. for every x, y X and α [0, 1] ∈ ∈ p(αx + (1 α)y) αp(x) + (1 α)p(y). − ≤ − Let Y X be a subspace and let λ: Y R be linear with λ(x) p(x) for every x Y . ⊆ → ≤ ∈ Then there exists Λ: X R linear such that → Λ = λ • |Y Λ(x) p(x) for every x X. • ≤ ∈ Proof. Step 1: (A preparation for the main step). W.l.o.g. assume Y & X. Then • there is an element z X Y ( z = 0). Let Y˜ := span(Y, z ). For everyy ˜ Y˜ ∈ \ ⇒ 6 { } ∈ there is a unique decompositiony ˜ = y + αz (for y Y and α R). The extension ∈ ∈ λ˜ of λ to Y˜ is uniquely speciﬁed by λ˜(z) because

λ˜(˜y) = λ(y) + αλ˜(z)

must hold because of linearity. To ensure

λ˜(˜y) p(˜y) y˜ Y,˜ ( ) ≤ ∀ ∈ ∗ we consider β , β > 0, y , y Y : 1 2 1 2 ∈

β1λ(y1) + β2λ(y2) = λ(β1y2 + β2y2) β1 β2 = (β1 + β2) λ (y1 β2z) + (y2 + β1z) β1 + β2 − β1 + β2 | {z } β1 β2 p (y1 β2z)+ (y2+β1z) ≤ β1+β2 − β1+β2 p convex β p(y β z) + β p(y + β z) ≤ 1 1 − 2 2 2 1 1 1 = (λ(y1) p(y1 β2z)) (p(y2 + β1z) λ(y2)). ⇒ β2 − − ≤ β1 − Thus, there exists a R: ∈ 1 1 sup λ(y) p(y βz) a inf p(y + βz) λ(y) . ( ) β>0 β − − ≤ ≤ β>0 β − ∗∗ y Y y Y ∈ ∈ Set λ˜(z) := a. Check that this choice ensures ( ). (To do so, use the right inequality ∗ in ( ), ify ˜ = y + αz with α 0 there, otherwise use the left inequality.) ∗∗ ≥

70 Step 2: Use Zorn to construct the extension. Let • n o := linear extensions e of λ with e p on dom(e) . E ≤ = = ∅ because λ . Deﬁne partial ordering on ⇒ E 6 ∈ E ≺ E e e : dom(e ) dom(e ) e = e . 1 ≺ 2 ⇐⇒ 2 ⊇ 1 ∧ 2|dom(e1) 1

Let W be totally ordered. Write W = eα α I . Consider ⊆ E { } ∈ S dom(eα) R e˜: α I −→ ∈ x e (x) 7−→ β where β is any element of I such that x dom(e ) ∈ β – e˜ is well-deﬁned (because W is totally ordered) – e˜ is linear – e˜ is an upper bound for W S – e˜(x) p(x) for every x dom(˜e) = α I dom(eα) ≤ ∈ ∈ =Zorn has a maximal element Λ. Claim: dom(Λ) = X (then the Theorem follows). ⇒ E Suppose not. Then Y := dom(Λ) and there is some 0 = z X Y . Let Y˜ := 6 ∈ \ span(Y, z . Then by step 1 there is some linear extension Λ˜ of Λ to Y˜ which { } ∈ E contradicts Λ being maximal.

4.7 Theorem (Hahn-Banach; complex version). Let X be a C-vector space. Let p: X R with → p(αx + βy) α p(x) + β p(y) ≤ | | | | for every x, y X and α, β C with α + β = 1. Let Y X be a subspace and λ: Y C ∈ ∈ | | | | ⊆ → linear with λ(x) p(x) for every x Y . Then there exists a linear extention Λ: X C | | ≤ ∈ → of λ (i.e. Λ = λ ) with |Y Λ(x) p(x) x X. | | ≤ ∀ ∈ Proof. Define `(x) := Re λ(x) for every x Y . Then `: X R is R-linear and all ∈ → hypotheses of Theorem 4.6 are fulfilled. Then, by Theorem 4.6, there is an R-linear 12 functional L: X R with L Y = ` and L p on X. Note: λ(x) = `(x) i`(ix) for → | ≤ − every x X. Define Λ(x) := L(x) iL(ix), x X. Then ∈ − ∈ Λ is R-linear on X by Theorem 4.6 • Λ = λ • |Y Λ is C-linear, because Λ is R-linear and • Λ(ix) = L(ix) iL( x) = i L(x) iL(ix) = iΛ(x) − − −

Claim: Λ(x) p(x) for every x X. For, ﬁx x X and let θ θ(x) := arg(Λ(x))13. • | | ≤ ∈ ∈ ≡ Then p convex iθ C-linear iθ L=Re Λ iθx iθ Λ(x) = e− Λ(x) = Λ(e− x) = L(e− ) p(e− x) p(x). | | ≤ ≤ 12because `(ix) = Re λ(ix) and λ(ix) = iλ(x), so Re λ(ix) = − Im λ(x) 13This means θ should be the angle of the polar representation of Λ(x) = |Λ(x)| · eiθ

71 4.8 Corollary. Let X be a normed space, Y X be a subspace and f Y . Then there ⊆ ∈ ∗ is F X with F = f and F ∗ = f ∗ . ∈ ∗ |Y k kX k kY Proof. If K = C, apply Theorem 4.7 with p(x) := f ∗ x for x X (fulﬁlls assump- k kY · k k ∈ tions) to λ = f. If K = R, apply Theorem 4.6 with the same p to λ = f. ± = F : X K linear with F (x) f ∗ x = ⇒ ∃ → | | ≤ k kY k k ⇒ F ∗ f ∗ . k kX ≤ k kY But F (x) f(x) F X∗ = sup | | sup | | = f Y ∗ k k 0=x X x ≥ 0=x Y x k k 6 ∈ k k 6 ∈ k k 4.9 Corollary. Let X be a normed space and let 0 = x X. Then there exists some 6 0 ∈ f X with f(x ) = x and f ∗ = 1. ∈ ∗ 0 k 0k k kX Proof. Let Y = span x0 . If y Y then y = αx0 for some unique α K. Deﬁne ϕ on { } ∈ ∈ Y by ϕ(y) := α x . This implies ϕ(x ) = x and ϕ Y with ϕ ∗ = 1 because · k 0k 0 k 0k ∈ ∗ k kY ϕ(y) = y . |Cor. 4.8| k k = f : X K linear with f Y = ϕ (in particular f(x0) = x0 ) and f ∗ = ⇒ ∃ → | k k k kX ϕ ∗ = 1. k kY 4.10 Corollary. Let X be a normed space, Z X be a subspace and x X Z with ⊆ 0 ∈ \ 0 < dist(x ,Z) =: d. Then there exists some f X with f = 0, f(x ) = d and 0 ∈ ∗ |Z 0 f ∗ = 1. k kX Proof. Exercise.

4.2 Three consequences of Baire’s theorem 4.11 Theorem (Banach-Steinhaus, uniform boundedness principle). Let X be a Banach space, Y a normed space and BL(X,Y ). If supT T x < for all x X, then F ⊆ ∈F k k ∞ ∈ supT T < . ∈F k k ∞ Proof. For n N, deﬁne ∈ A := x X : T x n T . n { ∈ k k ≤ ∀ ∈ F} S Then X = n N An by hypothesis. Also An = An for all n N (since and T ∈ ∈ k · k k k continuous). By Cor. 1.59 (a reformulation of Baire’s theorem) there exists n0 N such ∈ that A has non-empty interior, i.e. there exists x A and r > 0 such that B (x ) n0 0 ∈ n0 r 0 ⊆ An0 . Now let z B (0) and T . Using that z +x B (x ) and therefore T (z +x ) n , ∈ r ∈ F 0 ∈ r 0 k 0 k ≤ 0 we can estimate: T z T (z + x ) + T x n + T x . k k ≤ k 0 k k 0k ≤ 0 k 0k Finally, let 0 = x X. Deﬁne z := r x B (0). Plugging this into the last inequality, 6 ∈ 2 x ∈ r we get: k k 2 x 2 x T x = k k T z k k n + T x , k k r · k k ≤ r · 0 k 0k and by taking the supremum over all 0 = x X: 6 ∈ T x 2 T = sup k k n0 + T x0 . k k 0=x X x ≤ r · k k 6 ∈ k k Taking the supremum over all T proves the claim. ∈ F

72 Figure 6: Claim 1

4.12 Theorem (Open mapping theorem). Let X,Y be Banach spaces. Let T BL(X,Y ) ∈ be onto. Then T is open. Proof. We need to show the implication

A X open = T (A) open in Y. ⊆ ⇒ This will follow from 3 subsequent claims. For clarity, we include the ambient space in X the notation of a ball, e.g. Br (x). X Claim 1: It suﬃces to show that there exists r > 0 such that T (Br (0)) has non-empty interior. X Proof: Suppose T (Br (0)) has non-empty interior. Let y = T x be an interior point of T (BX (0)), i.e. there exists r such that BY (y) T (BX (0)) (see Figure 4.2). By continuity r y ry ⊆ r of T , 1 Y X X := T − (B (y)) B (0) B (0) Nx ry ∩ r ⊆ r X is an open neighbourhood of x (the intersection with Br (0) is necessary because T is not necessarily injective). Thus, there exists ρ > 0 such that

BX (x) and T (BX (x)) BY (y). Nx ⊆ ρ ρ ⊇ ry By scaling, translations and linearity, we get r > 0 ∀ 0 X X r0 X r0 X T B 0 (x) = T B 0 (0) + x = T B (0) + x = T B (0) + T x = r r ρ ρ ρ ρ r r = 0 T BX (x) x + T x 0 BY (y) T x + T x = ρ ρ − ⊇ ρ ry − |{z} =y

r0 Y Y Y 0 0 = Bry (0) + T x = B r ry (0) + T x = B r ry (y). ρ ρ ρ

Now, let A X be open and let y0 := T x0 T (A) with x0 A arbitrary. Since A is open, ⊆ X ∈ ∈ there exists r > 0 such that B 0 (x ) A. Using the inclusion above: 0 r 0 ⊆ X X Y Y T (A) T (Br0 (x0)) = T Br0 (x) + x0 x B r0ry (y) + y0 y = B r0ry (y0), ⊇ − ⊇ ρ − ρ

73 i.e. T (A) is open. This proves Claim 1. From now on all balls will be centred about 0 (unless otherwise speciﬁed), and we drop the centre from the notation. Claim 2: There exists ε > 0 such that

BY T (BX ). ( ) ε ⊆ 1 ∗ S X Proof: Since T is onto we can write Y = n N T (Bn ). Y is complete and we can apply ∈ X Baire’ Theorem in the form of Cor. 1.59: There exists n N such that T (Bn ) has non- X ∈ Y X empty interior, i.e. there exists y T (Bn ) and ε0 > 0 such that Bε0 (y) T (Bn ), or Y X ∈ ⊆ equivalently, B 0 T (B ) y. ε ⊆ n − We have

X y = limk T xk with xk Bn k N • →∞ ∈ ∀ ∈ T (BX ) T x = T (BX x ) T (BX ), • n − k n − k ⊆ 2n Y X thus B 0 T (B ) and, by scaling, the claim holds with ε := ε /2n. ε ⊆ 2n 0 Claim 3: T (BX ) T (BX ). 1 ⊆ 2 Proof: Let ε be as in Claim 2. Let y T (BX ). Then there exists x BX such that ∈ 1 1 ∈ 1 ( ) y T x BY ∗ T (BX ). − 1 ∈ ε/2 ⊆ 1/2 Analogously there exists x BX such that 2 ∈ 1/2

( ) y T x T x BY ∗ T (BX ). − 1 − 2 ∈ ε/4 ⊆ 1/4 X Inductively we get: For all n N there exists xn B −(n−1) such that ∈ ∈ 2 n X Y y T x B −n . ( ) − j ∈ ε2 ∗∗ j=1

Since X x < 2 < k nk ∞ n N | {z } ∈ <2−(n−1) P and X is a Banach space, j N xj =: x X exists (see digression of series in Banach ∈ P ∈ spaces). T is continuous = T x = j N T xj. Using ( ) and the continuity of , ⇒ ∈ ∗∗ k · k n X y T x = lim y T xj = 0, k − k n − −→∞ j=1 thus y = T x and x < 2, i.e. y T (BX ). k k ∈ 2 4.13 Corollary (Inverse mapping theorem). Let X,Y be Banach spaces and T BL(X,Y ) ∈ a bijection. Then T 1 BL(Y,X). − ∈ 1 Proof. T is open by Thm. 4.12, thus T − is continuous.

74 4.14 Deﬁnition. Let X,Y be normed spaces and T : X dom T Y a linear operator. ⊇ −→ (a) Graph of T : n o (T ) := (x, T x) X Y : x dom(T ) . G ∈ × ∈

We equip X Y with the norm (x, y) X Y := x X + y Y . Recall: X,Y complete × k k × k k k k = X Y complete. ⇒ × (b) T closed iﬀ (T ) closed in X Y . G × 4.15 Remark.

(a) T is closed if and only if the following implication holds:

n n (x ) dom T with x −→∞ x T and T x −→∞ y Y n n ⊆ n −−−−→ ∈ n −−−−→ ∈ = x dom T and y = T x. ⇒ ∈

(b) Compare (a) with the deﬁnition of T continuous, where convergence of (T xn)n must be shown. Here, it is given!

4.16 Theorem (Closed graph theorem). Let X,Y be Banach spaces and

T : X dom T Y ⊇ −→ a closed linear operator. Then dom T is closed (in X) if and only if T is bounded. n Proof. “ ” Let (x ) dom T with x →∞ x X. Then (x ) is Cauchy in X. ⇐ n n ⊆ n −−−→ ∈ n n By hypothesis: T is bounded, thus (T xn)n is Cauchy in Y . But Y is complete, i.e. n there exists y Y such that T xn →∞ y and since T is closed limn xn dom T ∈ −−−→ →∞ ∈ (and T x = y).

“ ” Deﬁne the projection ⇒ (T ) dom(T ) Px : G −→ . (x, T x) x 7−→ – (T ) is closed in X Y , therefore (T ) is itself a Banach space. G × G – dom(T ) is closed in X (by hypothesis), therefore dom(T ) is itself a Banach space.

– Px is a bijection – P is bounded because let z = (x, T x) (T ), then x ∈ G

Pxz X = x X x X + T x Y = z X Y . k k k k ≤ k k k k k k × So P 1. k xk ≤ 1 By the inverse mapping theorem Px− : dom(T ) (T ), x (x, T x), is also 1 −→ G 1 7−→ bounded. = c < : Px− x X Y c x X . Since Px− x X Y = x X + T x Y , ⇒ ∃ ∞ k k × ≤ k k k k × k k k k = T x (c 1) x . ⇒ k kY ≤ − k kX So T is bounded.

75 4.17 Example. Let X = Y = Cc(R) with supremum norm. Consider

1 d C (R) Cc(R) T := : c −→ dx f f 0 7−→

Claim: T is closed. Let (fn)n N dom(T ) be a sequence with the properties: ∈ ⊆ n g Cc(R) such that fn g →∞ 0, •∃ ∈ k − k∞ −−−→ n h Cc(R) such that fn0 h →∞ 0, •∃ ∈ − ∞ −−−→ 1 i.e. (fn, f 0 ) (T ) and we need to show: (g, h) (T ) (equivalently g C (R) and n ⊆ G ∈ G ∈ c h = g0). By uniform convergence, we can exchange limits: Z x Z x Z x dt h(t) = dt lim fn0 (t) = lim dt fn0 (x) 0 0 n n 0 →∞ →∞ | {z } fn(x) fn(0) − = g(x) g(0) − where the last equality follows from pointwise convergence of fn to g. Z x = g(x) = g(0) + dt h(t). ⇒ 0

1 So by the fundamental theorem of calculus: g C (R) with g0 = h. Since g Cc(R) we ∈ ∈ have g C1(R). So T is closed. ∈ c 4.3 (Bi)-Dual spaces and weak topologies 4.18 Theorem. Let X be a normed space. Then for every x X ∈ f(x) x = sup | |. k k 0=f X∗ f 6 ∈ k k∗

Proof. Let S := sup0=f X∗ f(x) / f . 6 ∈ | | k k∗ Since f(x) f x we have S x . • | | ≤ k k∗ k k ≤ k k Corollary 4.9 gives the existence of f˜ X∗ such that f˜ = 1 and f˜(x) = x . So • ∈ k k∗ k k f˜(x) S | | = x . ≥ f˜ k k k k∗

4.19 Deﬁnition. Let X be a normed space. X∗∗ := (X∗)∗ is the bidual space of X. More generally, we introduce the n-fold dual space

n-fold (n−1)-fold z }| { z }| { ∗ X∗ · · · ∗ := X ∗ · · · ∗ for n N in a recursive way. ∈

76 4.20 Theorem. Let X be a normed space. The canonical embedding

X X∗∗ J : −→ x Jx 7−→ with

X∗ K Jx: −→ f f(x) 7−→ is well-deﬁned, linear and isometric. If J is surjective, X is called reﬂexive.

4.21 Remarks. Hilbert spaces are reflexive (Riesz!). • Every finite dimensional normed space is reflexive (use the dual basis!). • `p, Lp are reflexive for p ]1, [. • ∈ ∞ X reflexive = X complete. • ⇒ Proof of Theorem 4.20. J is well-defined ( Jx (X∗)∗ x X). Let α, β K and ⇐⇒ ∈ ∀ ∈ ∈ f, g X : ∈ ∗ (Jx)(αf + βg) = (αf + βg)(x) = αf(x) + βg(x) = α(Jx)(f) + β(Jx)(g).

So Jx is linear. J is bounded, because

(Jx)(f) = f(x) f x . | | | | ≤ k k∗ k k This implies (Jx)(f) | | x . f ≤ k k k k∗ So Jx x and Jx X∗∗. J is isometric, because k k∗∗ ≤ k k ∈ f(x) Jx = sup | | = x k k∗∗ 0=f X∗ f k k 6 ∈ k k∗ by Theorem 4.18. J is linear, because let α, β K, f X∗ and x, y X. Then ∈ ∈ ∈ J(αx + βy) (f) = f(αx + βy) = αf(x) + βf(y) = α(Jx)(f) + β(Jy)(f)

Since this is true for every f X , we have J(αx + βy) = αJ(x) + βJ(y). ∈ ∗ 4.22 Theorem. Let X be a Banach space. Then

X reﬂexive X∗ reﬂexive. ⇐⇒ Proof. “ ”( X ) = ((X ) ) = (X ) = X ⇒ ∗ ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗ “ ” See e.g. Theorem III.3.4 in Werner, Funktionalanalysis. ⇐

77 4.23 Theorem. Let X be a normed space. Then

X∗ separable = X separable. ⇒ 4.24 Remark. X = `1 shows that “ =” does not hold, since ` = (`1) which is not ⇐ ∞ ∼ ∗ separable.

Proof of Thm. 4.23. Let A := fn X∗ : n N be dense (exists by hypothesis). For { ∈ ∈ } n N choose xn X such that xn = 1 and fn(xn) fn /2. Let ∈ ∈ k k | | ≥ k k∗ D := span xn : n N X K{ ∈ } ⊆ Claim: D is dense in X. (Note: D is not countable, but a K-linear subspace). Suppose D is not dense. Then there exists z X such that dist(z, D) > 0. By Corollary 4.10 there ∈ exists a linear functional f X with f = 0 and f(z) > 0. Now: ∈ ∗ |D A dense in X implies, there exists a sequence (f ) A such that • ∗ nk k ⊆ k fnk f →∞ 0. − ∗ −−−→ We have • 1 fnk f fnk (xnk ) f(xnk ) = fnk (xnk ) fnk . − ∗ ≥ | − | | | ≥ 2 ∗ k k →∞ →∞ Then fnk 0 = fnk 0 in X∗ = f = 0 . So D is dense. ∗ −−−→ ⇒ −−−→ ⇒ Deﬁne K˜ := Q if K = R, resp. K˜ := Q + iQ if K = C and set ˜ D := span˜ xn : n N . K{ ∈ } Thus, D˜ is countable and dense in X.

4.25 Definition. Let X be a K-vector space, let pα α I be a separating family of semi- { } ∈ norms on X, i.e. 0 = x X α I : p (x) > 0. ∀ 6 ∈ ∃ ∈ α For given x X, the collection ∈ n o U (x) := x + U : α I, r > 0 , α,r α,r ∈ where U := y X : p (y) < r defines the neighbourhood subbase14 of the locally α,r { ∈ α } convex topology associated with pα α I . { } ∈ n o 4.26 Remarks. (a) U (x) = y X : p (y x) < r α,r ∈ α − (b) p separating implies that the locally convex topology is Hausdorff. { α} (c) The elements of the neighbourhood subbase are convex sets, i.e. for y , y U and 1 2 ∈ α,r λ [0, 1] we have λy + (1 λ)y U , because ∈ 1 − 2 ∈ α,r p (λy + (1 λ)y ) p (λy ) + p ((1 λ)y ) α 1 − 2 ≤ α 1 α − 2 = λ pα(y1) +(1 λ) pα(y2) | {z } − | {z }

14i.e. ﬁnite intersections give a base.

78 4.27 Deﬁnition. Let X be a normed space. The weak topology on X is the locally convex topology on X induced by the family of seminorms x f(x) f X∗ . { 7→ | |} ∈ 4.28 Lemma. Let X be a normed space. Then the weak topology is (a) Hausdorﬀ,

(b) the coarsest topology on X such that every f X∗ is continuous with respect to . In particular, the weakT topology is coarser than∈ the norm ( strong) topology. T ≡ (c) identical to the norm topology, if dim X < , ∞ (d) such that (xn)n X converges to x X in the weak topology (or “weakly”) if and only if ⊂ ∈ lim f(xn) = f(x) f X∗. n →∞ ∀ ∈ w n Notation: x x or x →∞+ x. n −→ n −−− Proof. (a) By Remark 4.26 (b), it suﬃces to check that the family of seminorms is separating. This follows from Corollary 4.9: for every x X f X : f(x) = x . ∈ ∃ x ∈ ∗ k k (b) Show the implication: f X∗ = f weakly continuous: • ∈ S⇒ Let G K be open. Then G = γ G Brγ (γ). ⊆ ∈ 1 [ 1 = f − (G) = f − (B ). ⇒ rγ γ G ∈ Using that the union over γ G can be restricted to γ G ran(f) = ∈ ∈ ∩ ⇒ 1 [ f − (G) = x X : f(x) γ < r ∈ | − | γ γ G ran(f) ∈ ∩ [ [ = x X : f(x x ) < r , ∈ | − γ | γ γ G ran(f) xγ X:f(xγ )=γ ∈ ∩ ∈ | {z } Uf,rγ (xγ ) 1 where Uf,rγ (xγ) is weakly open by deﬁnition. Thus f − (G) is weakly open. Let f X∗ be continuous with respect to some topology on X. Since : K • ∈ T |·| → [0, [ is continuous (with respect to the standard topologies on K and [0, [) ∞ ∞ the mapping (X, ) [0, [ T −→ ∞ x f(x) 7−→ | | is continuous = U := y X : f(x) < r , and since is a topology, ⇒ f,r { ∈ | | } ∈ T T the weak topology is coarser than . T (c) See Problem 43.

(d)( xn)n converges to x in the weak topology K \ K N f1, . . . , fK X∗ r1, . . . , rK > 0 : xn Ufk,rk (x) for ﬁnally all n ⇐⇒ ∀ ∈ ∀ ∈ ∀ ∈ k=1

f X∗ r > 0 : xn Uf,r(x) for ﬁnally all n . ⇐⇒ ∀ ∈ ∀ | ∈ {z } f(xn) f(x)

79 4.29 Remarks. (a) Weak limits are unique by Lemma 4.28 (a).

(b) If dim X = , then the weak topology is not 1st countable (hence not metrisable) ∞ [for a proof, see Problem 44].

Lemma 4.28( b) (c) Strong convergence = weak convergence (but in general not vice ⇒ versa).

(d) In a Hilbert space X (by Lemma 4.28 (d) and Riesz):

w n x x y, x →∞ y, x y X n −→ ⇐⇒ h ni −−−→h i ∀ ∈ (e) In `1: w n x x x x →∞ 0 n −→ ⇐⇒ k n − k −−−→ [I. Schur, J. Reine Angew. Math. 151, 79–111 (1921); see also J. B. Conway, A course in functional analysis, 2nd ed., Springer, New York, 1990, Prop. 5.2]

p 4.30 Example. Let X = ` and p [1, ], let (en)n N be the canonical basis with k k ∈ ∞ ∈ en = (en)k N, en = δnk (for p = just the sequence itself). Then ∈ ∞ (e ) has no -convergent subsequence. • n n k·kp 1 If p = 1, then (en)n is not weakly convergent. For, (` )∗ ∼= `∞ and choosing f • n ↔ ( 1, 1, 1, 1, ) `∞ = f(en) = ( 1) n N = not convergent as − − ··· ∈ ⇒ − ∀ ∈ ⇒ n . [Also consistent with Remark 4.29 (e)] → ∞ If 1 < p , then (en)n is weakly convergent to 0. For, in the case 1 < p < we • p ≤ ∞ q ∞ q have (` )∗ = ` by Theorem 2.38 with 1 < q < = for f y = (yk)k ` : ∼n ∞ ⇒ ↔ ∈ f(e ) = y →∞ 0. n n −→ In the case p = use also Problems 23 & 24. ∞ 4.31 Lemma. Let X be a normed space and x w x. Then n −→ (a) (b) x lim inf xn sup xn < . n k k ≤ k k ≤ n N k k ∞ →∞ ∈ w n Proof. Inequality (b): let x x, so f(x ) →∞ f(x) for every f X . n −→ n −−−→ ∈ ∗

= f X∗ (ﬁxed) : sup f(xn) < , ⇒ ∀ ∈ n N | | {z } | ∞ ∈ (Jxn)(f) with the canonical embedding J : X X . Since X is Banach = (uniform bound- −→ ∗∗ ∗ ⇒ edness principle Thm. 4.12) sup Jxn < . n N k k∗∗ ∞ ∈ | {z } xn k k Cor. 4.9 Inequality (a): = fx X∗: fx = 1 and fx(x) = x . ⇒ ∃ ∈ k k∗ k k

= x = fx(x) = lim fx(x) lim fx xn lim inf xn . n n n ⇒ k k | | →∞ | | ≤ →∞ |k {zk}∗ k k ≤ →∞ k k =1

80 w 4.32 Theorem. Let X be a normed space. Then xn x if and only if the following two statements hold: −→

supn N xn < . • ∈ k k ∞ F X∗ with span(F ) is ( -) dense in X∗ such that for every f F : •∃ ⊆ k · k∗ ∈

lim f(xn) = f(x). n →∞ Proof. “ ” from Lemma 4.31 and by deﬁnition of weak convergence. ⇒ “ ” use an ε/3-argument. Let ε > 0 and g X∗. Let K := x + supn N xn /2 < ⇐ ∈ k k ∈ k k . ∞ ε – Since span(F ) is dense in X∗, there exists f span(F ): f g < 3K . ∈ k − k∗ – For f span(F ) there exists N N such that for every n N: ∈ ∈ ≥ ε f(x ) f(x) < . | n − | 3 (Note: convergence f(x ) f(x) holds not only for f F but also for n −→ ∈ f span(F ), because of ﬁnite linear combinations.) ∈ = ⇒

g(x) g(xn) g(x) f(x) + f(x) f(xn) + f(xn) g(xn) | − | ≤ | − | | − | | − | g f x + xn + f(x) f(xn) ≤| k −{z k}∗ |k k {zk k} | −{z }| < ε 2K < ε 3K ≤ 3 < ε.

4.33 Theorem (Eberlein-Smulian)ˇ . Let X be a Banach space and A X. Then ⊆ A weakly compact A weakly sequentially compact. ⇐⇒ Proof. R. Whitley, Math. Ann. 172, 116–118 (1967).

4.34 Deﬁnition. Let X be a normed space. Then the weak* topology on X∗ is the locally convex topology on X∗ induced by the family of seminorms f f(x) x X . { 7→ | |} ∈ 4.35 Lemma. Let X be a normed space. Then the weak* topology is

(a) Hausdorﬀ,

(b) the coarsest topology on X∗ such that x X the map X∗ K, f f(x), is continuous, ∀ ∈ → 7→

(c) coarser than the weak topology on X∗, and the two coincide if and only if X is reﬂexive, (d) such that (f ) X converges to f X in the weak* topology if and only if n n ⊆ ∗ ∈ ∗ n f (x) →∞ f(x) x X. n −−−→ ∀ ∈ ∗ Notation: f w f. n −−→

81 Proof. (a) f f(x) x X is a separating family of seminorms since for f = 0 x X { 7−→ | |} ∈ 6 ∃ ∈ with f(x) = 0. 6 (b) Analogous to Lemma 4.28 (b).

The next lemma is the weak* analogue of Lemma 4.31 and Theorem 4.32. But notice that here X must be complete in order to apply the uniform boundedness principle in (a).

4.36 Lemma. Let X be a Banach space, f X , (f ) X . Then ∈ ∗ n n ⊆ ∗ ∗ (a) If f w f, then n −→

f ∗ lim inf fn ∗ sup fn ∗ < . X n X X k k ≤ k k ≤ n N k k ∞ →∞ ∈

∗ (b) f w f if and only if n −→

supn N fn X∗ < • ∈ k k ∞ A X with span(A) (norm-) dense in X such that x A •∃ ⊆ ∀ ∈ n f (x) →∞ f(x). n −−−→ Proof. Exercise.

4.37 Theorem (Banach-Alaoglu). Let X be a Banach space. Then the closed unit ball in X∗ n o ¯ B1∗ := f X∗ : f 1 ∈ k k∗ ≤ is compact in the weak* topology.

Proof. Deﬁne A := Ax, where Ax := z K : z x compact in K. The x X { ∈ | k ≤ k k} elements of A are of× the∈ form X K f : −→ x f(x) 7−→ with f(x) A , i.e. f(x) x . By Tychonoﬀ (Thm. 1.38), A is compact in the product- ∈ x | | ≤ k k space topology (the coarsest topology such that for all x X the map Π : A A , ∈ x −→ x f f(x), is continuous). Now 7−→ n o B¯ = f A : f is linear • 1∗ ∈ The restriction of the product-space topology to B¯ coincides with the weak* topol- • 1∗ ogy.

82 We will show B¯ closed in A (= B¯ compact, because A is compact). For x, y X, 1∗ ⇒ 1∗ ∈ α, β K, let ∈ n o Kx,y,α,β := f A : f(αx + βy) αf(x) βf(y) = 0 ∈ | −{z − } (Παx+βy αΠx βΠy)(f) − − 1 = Π αΠ βΠ − ( 0 ). αx+βy − x − y { }

But Παx+βy αΠx βΠy is continuous and 0 is closed in K, thus Kx,y,α,β is closed in − − { } A and therefore also ¯ \ B1∗ = Kx,y,α,β. x,y X α,β∈ K ∈ 4.38 Theorem (Helly, version 2 of Banach-Alaoglu). Let X be a separable Banach space. ¯ Then B1∗ is weak* sequentially compact.

Proof. Let xk : k N X be (countable) dense in X. We need to show that any { ¯∈ } ⊆ sequence (fn)n B∗ has a weak* convergent subsequence. Fix k N arbitrary. Consider ⊆ 1 ∈ the sequence (fn(xk))n K. This sequence is bounded, because ⊆

fn(x) fn xk xk . | | ≤ k| {zk}∗ k k ≤ k k 1 ≤

Thus, for all ﬁxed k,(fn(xk))n has a convergent subsequence.

Claim: There exists a common subsequence (mj)j N : k N,(fmj (xk))j is convergent. ⊆ ∀ (1)∈ Proof: Use Cantor’s diagonal sequence trick: There exists (nj )j N such that (f (1) (x1))j ⊆ nj (2) (1) converges. Then there exists (nj) (nj )j such that (f (2) (x2))j converges. Continuing ⊆ nj (l+1) (l) this procedure, there exists (nj )j (nj )j such that (f (l+1) (xl+1))j converges. ⊆ nj (j) The claim then holds with mj := nj .

Now, deﬁne g(x) := limj fmj (x) x span(xk : k N) =: dom(g). →∞ ∀ ∈ ∈ dom(g) is a dense subspace of X with respect to • k · k g : dom(g) K is linear • −→ g is bounded: •

g(x) = lim fmj (x) x | | j | | ≤ k k →∞ | {z } fmj x ≤k k∗ k k | {z≤1 }

Since K is complete, we can apply the bounded linear extension theorem 2.32: There exists ¯ g˜: X K such thatg ˜ dom(g) = g and g˜ = g 1. Thusg ˜ B1∗. Then by Lemma → ∗ | k k∗ k k ≤ ∈ 4.36 (b), f w g˜ as j . mj −−→ → ∞ 4.39 Theorem (version 3 of Banach-Alaoglu). Let X be Banach space. Then n o X reﬂexive B¯ := x X : x 1 is weakly compact. ⇐⇒ 1 ∈ k k ≤

83 Proof. ”= ”: Use that ”weak topology on X” = ”weak* topology on (X ) ”, because X ⇒ ∗ ∗ is reﬂexive and Thm. 4.37. ” =”: See e.g. Dunford/Schwartz: Linear Operators, vol. I, Interscience, 1966, Thm. ⇐ V.4.7.

4.40 Example. Compactness of B¯ in diﬀerent spaces. Here p, q ]1, [ are H¨older 1 ∈ ∞ conjugate.

weak weak* seq. weak seq. weak*

1 ` (∼= c0∗) no yes no yes p q ` (∼= (` )∗) yes yes yes yes 1 `∞ (∼= (` )∗) no yes no yes L1 (not a dual!) no — no —

p q L (∼= (L )∗) yes yes yes yes 1 L∞ (∼= (L )∗) no yes no yes

84 5 Bounded operators

5.1 Topologies on the space of bounded linear operators 5.1 Deﬁnition. Let X,Y be normed spaces.

(a) uniform (operator) topology on BL(X,Y ) := norm topology with respect to X Y . k · k → (b) strong (operator) topology on BL(X,Y ) := locally convex topology induced by the seminorms T T x Y x X . { 7−→ k k } ∈ (c) weak (operator) topology on BL(X,Y ) := locally convex topology induced by the seminorms T `(T x) x X . { 7−→ | |}` ∈Y ∗ ∈ 5.2 Remark. (a) The families of seminorms in Definition 5.1 (b) and (c) are separating (check!) = Hausdorff topologies. ⇒ (b) Strong and weak operator topologies are not first countable if dim X = . ∞ 5.3 Lemma. (a) The strong topology is the coarsest topology on BL(X,Y ) such that all maps

BL(X,Y ) Y Mx : −→ , x X T T x ∈ 7−→ are continuous. The weak topology is the coarsest topology on BL(X,Y ) such that all maps

BL(X,Y ) K M : −→ , x X, ` Y ∗ `,x T `(T x) ∈ ∈ 7−→ are continuous.

(b)

(c) Let T,Tn BL(X,Y ), n N. Then ∈ ∈ s (Tn)n converges to T in the strong operator topology, in symbols, Tn T • (“strongly”) −−→ lim Tnx = T x x X. n ⇐⇒ →∞ ∀ ∈ w (Tn)n converges to T in the weak operator topology, in symbols, Tn T • (“weakly”) −−→

lim `(Tnx) = `(T x) x X ` Y ∗ n ⇐⇒ →∞ ∀ ∈ ∀ ∈ Proof. (a) analogous to the proof of Lemma 4.28 (b).

85 (b) Upper branch: follows from M BL(X,Y ) ∗ x X ` Y , because `,x ∈ ∀ ∈ ∀ ∈ ∗

M`,xT M`,xT = `(T x) ` Y ∗ T x Y = | | ` Y ∗ x T = 0 | | | | ≤ k k |k {zk } ⇒ T ≤ k k k k ∀ 6 T x k k ≤k kk k lower branch:

M : BL(X,Y ) Y is linear x X. Also bounded: • x → ∀ ∈

MxT MxT Y = T x Y T x = sup k k x , k k k k ≤ k kk k ⇒ 0=T BL(X,Y ) T ≤ k k 6 ∈ k k

i.e. Mx is continuous for all x. So from (a): strong operator topology is coarser than the uniform topology. M = ` M = continuity of M implies continuity of M , and the strong • `,x ◦ x ⇒ x `,x operator topology must be ﬁner than the weak operator topology by (a).

5.4 Lemma. Let be a Hilbert space and (T ) BL( ). H n n ⊆ H (a) If (T x) Cauchy in x then there exists T BL( ) such that T s T . n n H ∀ ∈ H ∈ H n −→ (b) If ( y, Tnx )n is Cauchy in K for all x, y , then there exists T BL( ) such that h w i ∈ H ∈ H T T . n −−→ Proof. (a) Deﬁne

T : H −→ H . x lim Tnx n 7−→ →∞ The map T is

well deﬁned (limit exists!) • linear • bounded: We have sup T x < for all x (due to convergence). Thus by • n k n k ∞ ∈ H the uniform boundedness principle: sup T < = n k nk ∞ ⇒

T = sup T x sup Tn < . k k x : x =1 k k ≤ n N k k ∞ ∈H k k | {z } ∈ lim Tnx n →∞ |k {z k} Tn x ≤ k k|{z} k k =1

Hence T BL( ) and T s T . ∈ H n −−→ (b) Deﬁne the form K Q: H × H −→ (x, y) lim Tnx, y n 7−→ →∞ h i Q is well-deﬁned (i.e. the limit exists) • Q is sesquilinear •

86 Q is bounded, i.e. for every x, y : • ∈ H Q(x, y) sup Tn x y | | ≤ n N k k k k k k |∈ {z } =:S< ∞ where S < will be shown below. ∞ By Problem T13 (a corollary of the Riesz representation) there exists a unique T w ∈ BL( ) such that Q(x, y) = T x, y which means T T . H h i n −−→ Proof of S < : Let x, y be ﬁxed. Then we know: ∞ ∈ H

sup Tnx, y < n N | h {z i} | ∞ ∈ =:`n(y)

with ` . By the uniform boundedness principle ?? and the Riesz-representation n ∈ H∗ Theorem we infer that

sup `n = sup Tnx < x , n N k k∗ n N k k ∞ ∀ ∈ H ∈ ∈ and again by the uniform boundedness principle ?? we have

sup Tn < . n N k k ∞ ∈ 5.5 Examples. Let (T ) BL(`2). n ⊆ 2 (a) For x = (x1, x2,...) ` and n N let ∈ ∈ 1 1 1 T x := x , x ,... = x. n n 1 n 2 n

n Then T = 1 →∞ 0, i.e. uniform convergence to zero (operator). k nk n −−−→ (b) Let Tnx := 0,..., 0, xn+1, xn+2,... . | {z } n times

T s 0, because • n −−→ 2 X∞ 2 n Tnx = xj →∞ 0. k k | | −−−→x `2 j=n+1 ∈ (T ) does not converge uniformly to 0, because • n n T e = e = T 1. n n+1 n+1 ⇒ k nk ≥ (c) Let Tnx := 0,..., 0, x1, x2,... | {z } n times “n-times iterated right shift”.

87 T w 0, because for y `2 by the Cauchy-Schwarz inequality • n −−→ ∈ v ∞ u ∞ X uX 2 y, Tnx = yn+jxj x t yn+j . | h i | ≤ k k · | | j=1 j=1

| n→∞{z } 0 −−−→y∈`2 (T ) does not converge strongly to 0, because • n n T x = x x `2. k n k k k ∀ ∈ 5.2 Adjoint operators 5.6 Missing. For the amusement of the reader. 5.7 Deﬁnition. (a) Let X,Y be normed spaces and T BL(X,Y ). Then ∈ Y ∗ X∗ T × : −→ ` T ` 7−→ × with (T ×`)(x) := `(T x) is a linear operator, the adjoint of T . (b) Let , be Hilbert spaces and T BL( , ). Then H1 H2 ∈ H1 H2 2 1 T ∗ : H −→ H y T y 7−→ ∗ where (by Riesz!) T y is the unique z such that for every x ∗ ∈ H1 ∈ H1

y, T x 2 = `(x) 1∗ h iH ∈ H = z, x =: T ∗y, x 1 1 h iH H is a linear operator, the (Hilbert) adjoint of T . 5.8 Remarks. (a) In the Hilbert space case we have that 1 T ∗ = T ×A− A1 2 with : is antilinear, isometric and bijective (Corollary 2.58). Aj Hj∗ −→ Hj (b) T ∗ is linear, because

T ∗(αy + βz), x = αy + βz, T x 1 2 H h iH

= α y, T x 2 +β z, T x 2 h| {z iH} |h {z iH} T ∗y,x T ∗z,x h iH1 h iH1

= αT ∗y + βT ∗z, x , 1 H and since this is true for every x we have linearity (analogous for T ). ∈ H1 × 1 5.9 Examples. (a) Let X = Y = ` and T x := (0, x1, x2,...), the right-shift operator. Then T BL(`1). T :(`1) (`1) (we know f (`1) acts as f(x) = P ξ x ∈ × ∗ −→ ∗ ∈ ∗ n N n n for every x `1 and some unique ξ ` by Riesz). So ∈ ∈ ∈ ∞ X X (T ×f)(x) = f(T x) = ξn(T x)n = ξn+1xn n N n N ∈ ∈ i.e. T × corresponds to (ξ2, ξ3, ξ4,...), which is a left shift.

88 (b) For = = and ϕ, ψ let P := ψ ϕ, , i.e. P x = ψ ϕ, x x . H1 H2 H ∈ H h ·i h i ∀ ∈ H Then P = ϕ ψ, because x, y ∗ h ·i ∀ ∈ H y, P x = y, ψ ϕ, x h i h i h i = ψ, y ϕ, x = P ∗y, x . h i 5.10 Theorem. Let X,Y be normed spaces and T BL(X,Y ). Then ∈

T × Y ∗ X∗ = T X Y k k → k k → Proof. By deﬁnition and Theorem 4.18 we have

T X Y = sup T x k k → x X k k x∈=1 k k = sup sup `(T x) x X ` Y ∗ | | x∈=1 ` ∈ ∗ =1 k k k kY 15 = sup sup (T ×`)(x) ` Y ∗ x X | | ∈ ∈ ` Y ∗ =1 x =1 k k |k k {z } T ×` k kX∗ = T × Y ∗ X∗ . k k →

5.11 Corollary. Let be Hilbert spaces and . Then 1, 2 T BL( 1, 2) T ∗ 2 1 = . H H ∈ H H k kH →H T 1 2 k kH →H Proof. This rests on

1 T = T − • ∗ A1 ×A2 Theorem 5.10 • , j = 1, 2, are isometric and bijective. •A j 5.12 Theorem. Let be a Hilbert space and T,S BL( ). Then H ∈ H (a) The map BL( ) BL( ) : H −→ H ∗ T T ∗ 7−→ is antilinear, isometric and bijective.

(b) (TS)∗ = S∗T ∗

(c) (T ∗)∗ = T . (d) If T has an inverse T 1 BL( ), then (T ) 1 = (T 1) . In particular (T ) 1 − ∈ H ∗ − − ∗ ∗ − ∈ BL( ). H 15interchange of suprema:

sup sup Aab ≤ sup sup Aab˜ ≤ sup sup Aa˜˜b =⇒ sup sup Aab = sup sup Aab. a b |{z} b a˜ |{z} a˜ ˜b a b b a ≤supa˜ Aab˜ ≤sup˜b Aa˜˜b

89 (e) TT = T 2. k ∗k k k Proof. (a) Antilinear by deﬁnition of T ∗, isometric by Cor. 5.10, surjective by part (c) of this theorem.

(b) See tutorial sheet.

1 1 (d) TT − = 1 = T − T together with (b) gives

1 1 T ∗(T − )∗ = 1 = 1∗ = (T − )∗T ∗.

(e) On the one hand, we have TT T T = T 2 by Corollary 5.11. On the other k ∗k ≤ k k k ∗k k k hand

y, T T x h ∗ i TT ∗ = sup 0=x,y x y 6 ∈H k k k k 2 T ∗x sup k 2k ≥ 0=x x 6 ∈H k k 2 = T ∗ = T 2 . k k 5.13 Definition. Let be a Hilbert space and T BL( ). H ∈ H T is unitary iff T is bijective and T 1 = T (i.e. TT = 1 = T T ). • − ∗ ∗ ∗ T is self-adjoint iff T = T . • ∗ T is normal iff TT = T T . • ∗ ∗ 5.14 Remarks. (a) Self-adjoint or unitary = normal. ⇒ (b) If T is unitary, then T x, T y = x, y = T x, T y x, y . h i h i h ∗ ∗ i ∀ ∈ H (c) If T is self-adjoint, then

x, T y = T x, y x, y . ( ) h i h i ∀ ∈ H ∗ Setting x = y gives x, T x R for every x . h i ∈ ∈ H [In Problem 41, the property ( ) was called symmetry; the notions symmetric and ∗ self-adjoint agree for bounded linear operators]

(d) T normal = T x = T x for every x . In particular ker(T ) = ker(T ). ⇒ k k k ∗ k ∈ H ∗ 5.15 Examples. (a) T = z1, z K. Then T ∗ = z1. So T is self-adjoint if and only if ∈ z R. ∈ (b) Let = L2([0, 1]), k C([0, 1]2) and H ∈ Z (T f)(x) := dy k(x, y)f(y).

If k(x, y) = k(x, y) x, y = T is self-adjoint. ∀ ∈ H ⇒

90 5.3 The spectrum In this subsection: X is a Banach space.

5.16 Deﬁnition. Let T BL(X). ∈ resolvent set (of T ): • n o ρ(T ) := z C : T z 1 bijective ∈ − ·

resolvent (Green function) of T : • 1 R := (T z)− z − – if z ρ(T ), then R BL(X) (in particular this inverse exists), ∈ z ∈ – need not exist for z ρ(T ). 6∈ spectrum of T : • spec(T ) := σ(T ) := C ρ(T ) \ if there exists 0 = x X such that T x = λx for some λ C then λ is an eigenvalue • 6 ∈ ∈ of T and x the corresponding eigenvector.

point spectrum (of T ): • n o n o spec (T ) := σp(T ) := z C : T z not injective = eigenvalues of T p ∈ −

continuous spectrum (of T ): • n o spec (T ) := σc(T ) := z C : T z injective and ran(T z) = X, but dense c ∈ − − 6

residual spectrum (of T ) • n o spec (T ) := σr(T ) := z C : T z injective and ran(T z) not dense r ∈ − −

[specr(T ) = ∅ for most T of interest] 5.17 Lemma. Let T BL(X). Then ∈ (a) C = σ(T ) ˙ ρ(T ) • ∪ σ(T ) = σ (T ) ˙ σ (T ) ˙ σ (T ) • p ∪ c ∪ r (b) If dim X < , then σc(T ) = ∅ = σr(T ). ∞ Proof. (a) clear by deﬁnition. (b) follows from linear algebra: T z injective = dim ker(T z) = 0 = dim ran(T − ⇒ − ⇒ − z) = dim(X) = ran(T z) = X. ⇒ − 5.18 Deﬁnition. Let D C be a region and ⊆ D X x: −→ . z x(z) 7−→

91 x is strongly differentiable (strongly analytic) in z D iff • 0 ∈ x(z + h) x(z ) lim 0 − 0 C h 0 h 3 → exists in X. x is weakly differentiable (weakly analytic) in z D iff z `(x(z)) is differentiable • 0 ∈ 7−→ in C in z0 for all ` X∗. ∈ x is strongly (weakly) analytic in D iff x is strongly (weakly) differentiable in z for • 0 all z D. 0 ∈ 5.19 Remark. (a) X-valued strongly analytic functions have analogous properties to C- valued analytic functions (e.g. a power-series expansion converging w.r.t. , etc.) k · k Moral: Replace (absolute value in C) by .16 | · | k · k (b) strongly analytic weakly analytic [see e.g. Reed/Simon Thm. VI.4 for ” =”] ⇐⇒ ⇐ 5.20 Theorem. Let T BL(X). Then ρ(T ) is open in C and the map ∈ ρ(T ) BL(X) −→ z R 7−→ z is strongly analytic. For λ, µ ρ(T ) the first resolvent identity holds ∈ R R = (λ µ)R R , (1) λ − µ − λ µ in particular RλRµ = RµRλ (they commute!). The proof uses: 5.21 Lemma (Neumann series). Let T BL(X) with T < 1. Then ∈ k k

1 1 X∞ j (1 T )− BL(X) and (1 T )− = T . − ∈ − j=0 Proof. From T j T j, T < 1 and the digression on series in Banach spaces (see the k k ≤ k k k k P j proof of Lemma 2.49) = S := ∞ T BL(X). Now, for N N consider ⇒ j=0 ∈ ∈ N N X X (1 T ) T j = 1 T N+1 = T j(1 T ). − − − j=0 j=0

N N N N Note that P T j →∞ S and T N+1 →∞ 0 (since T N+1 T N+1 →∞ 0). j=0 −−−−→ −−−−→ k k ≤ k k −−−−→ Thus (1 T )S = 1 = S(1 T ), i.e. S = (1 T ) 1. − − − − 5.22 Corollary. Let T BL(X). Then σ(T ) z C : z T . ∈ ⊆ { ∈ | | ≤ k k} Proof. For z = 0, consider 6 1 1 1 T − R = (T z)− = 1 . z − z z − For z > T we have T/z < 1 and get R BL(X) by Lemma 5.21. Thus z ρ(T ) = | | k k k k z ∈ ∈ C σ(T ). \ 16For a discussion of Banach-space valued functions of a complex variable, see e.g. Hille, Phillips, Func- tional analysis and semigroups, AMS, 1957; or Dunford, Schwartz, Linear operators, Vol. 1, Sect. III.14.

92 Proof of Thm. 5.20. Equation (1) follows from • (T λ)(R R )(T µ) = 1 (T λ)R (T µ) = − λ − µ − − − µ − = T µ (T λ) = λ µ − − − −

and by multiplication by Rλ from the left and by Rµ from the right. Commutativity: Exchange λ µ in (1) and equate. • ←→ Openness of ρ(T ): Let λ ρ(T )(= ∅ by Cor. 5.22) and z C such that z λ < • ∈ 6 ∈ | − | 1 . Then Rλ k k T z = T λ (z λ) = (T λ) 1 (z λ)Rλ . − − − − − − | −{z } =:V

By deﬁnition V < 1 and therefore by Lemma 5.21:(1 V ) 1 BL(X). Thus k k − − ∈ 1 1 R = (T z)− = (1 V )− R BL(X)( ) z − − λ ∈ ∗ and z ρ(T ), i.e. ρ(T ) is open. ∈ From ( ) and Lemma 5.21: • ∗

X∞ j X∞ j j+1 Rz = V Rλ = (z λ) Rλ for z B1/ Rλ (λ). − ∈ k k j=0 j=0

Thus B1/ Rλ (λ) z Rz is strongly diﬀerentiable, i.e. Rz is analytic in B1/ Rλ (λ). k k 3 7→ k k But since λ ρ(T ) was arbitrary this shows the claim. ∈ 5.23 Lemma. Let T BL(X). Then σ(T ) = ∅. ∈ 6 Proof. Let z > T . Then (see the proof of Cor. 5.22) | | k k 1 j 1 1 T − 1 X∞ T R = (T z)− = 1 = . z − z z − −z zj j=0

This expansion shows that

lim Rz = 0. ( ) z k k ∗ | |→∞ Assume σ(T ) = ∅. Then ρ(T ) = C and by Thm. 5.20,

C BL(X) −→ z R 7−→ z is entire (i.e. analytic on all of C). Also C z Rz is bounded. By Liouville’s 3 7−→ k k theorem17 z R is constant and by ( ) this constant has to be 0, i.e. R = 0 for all 7−→ z ∗ z z C. ∈ 17 For C-valued functions, see e.g. J. B. Conway, Functions of one complex variable, 2nd ed., Springer, New York, 1978, Thm. IV.3.4.

93 5.24 Deﬁnition. For T BL(X) ∈ Spectral radius of T : r(T ) := sup λ . ⇐⇒ λ σ(T ) | | ∈ 5.25 Theorem. Let T BL(X). Then ∈ (a) r(T ) = lim T n 1/n = inf T n 1/n, in particular, r(T ) T . n k k n N k k ≤ k k →∞ ∈ (b) If, in addition, X is a Hilbert space and T is normal, then lim T n 1/n = T = n k k k k ⇒ r(T ) = T . →∞ k k Proof. (a) We show ﬁrst lim T n 1/n = inf T n 1/n. • n k k n N k k n →∞ ∈ W.l.o.g. assume T = 0 n N (otherwise the claim is clear). Set an := n 6 ∀ ∈ ln T = an1+n2 an1 + an2 n1, n2 N. Fix m N arbitrary. k k ⇒ ≤ ∀ ∈ ∈ Write n N, n m as n = qm + r with q N and r 0, . . . , m 1 . Then ∈ ≥ ∈ ∈ { − } a qa + a a a n m r m + r n ≤ qm + r ≤ m qm + r

q an am =→∞ lim sup m N ⇒ n n ≤ m ∀ ∈ →∞ a a = lim sup n inf m . ⇒ n n ≤ m N m →∞ ∈ an am But (trivially) lim inf inf . Taken together, limn an/n exists and n n ≥ m N m →∞ →∞ ∈ lim ean/n = inf ean/n. n n N →∞ ∈ Thm. 5.20= z Rz is strongly analytic in D := C z C : z r(T ) • ⇒ 7−→ \{ ∈ | | ≤ } and therefore has a Laurent series expansion about 018

X j Rz = z Aj,Aj BL(X) for j Z, ∈ ∈ j Z ∈ which is norm-convergent in BL(X) for all z D. For z C : z > T D, ∈ { ∈ | | k k} ⊆ the expansion

1 X∞ T j R = (1) z −z zj j=0 holds, see the proof of Lemma 5.23. Thus, by uniqueness of the Laurent series, j 1 Aj = 0 for j N0 and A j = T − for j N. That is, the series (1) is norm ∈ − − ∈ convergent z D. ∀ ∈ On the other hand: Given any ε > 0, the series (1) is not norm-convergent on C z C : z r(T ) ε , because then we would get convergence for some z \{ ∈ | | ≤ − } ∈ spec(T ). Thus 1 is the radius of convergence of the series ξ P ξjT j 1, r(T ) 7→ − j N − and Hadamard’s root criterion gives ∈ r(T ) = lim sup T j 1/j = lim T j 1/j, j k k j k k →∞ →∞ as we have already shown.

18 For C-valued functions, see e.g. J. B. Conway, Functions of one complex variable, 2nd ed., Springer, New York, 1978, Thm. V.1.11.

94 (b) Using the property TT = T 2 from Thm. 5.12 (e): k ∗k k k

2 2 2 2 T normal 2 5.12(e) 4 T = T (T )∗ = (TT ∗)(TT ∗)∗ = TT ∗ = T , k k k k k k k k k k i.e. T 2 = T 2. By induction on k N: k k k k ∈ k k T 2 = T 2 . (2) k k k k Thus k k (2) lim T n 1/n = lim T 2 1/2 = T . n k k k k k k k →∞ →∞ 5.4 Compact operators Throughout we assume that X and Y are Banach spaces and is a Hilbert space. H 5.26 Deﬁnition. An operator T BL(X,Y ) is compact iﬀ for every bounded subset ∈ A X, the closure T (A) Y is compact, i.e. T (A) is relatively compact in Y . ⊆ ⊆ 5.27 Remark. T is compact if and only if for every bounded sequence (x ) X, the n n ⊆ sequence (T xn)n has a convergent subsequence (in metric spaces, compactness is equivalent to sequential compactness)

5.28 Examples. (a) For k C([0, 1]2) let T : L2([0, 1]) L2([0, 1]), ∈ −→ Z x (T f)(x) := dy k(x, y)f(y). 0 Then, T is compact (exercise, use Arzel`a-Ascolitheorem 1.50).

(b) Finite-rank operators. T BL(X,Y ) is of ﬁnite rank iﬀ dim ran(T ) < , i.e. J N ∈ ∞ ∃ ∈ such that J X T x = αj(x)fj, j=1 where the f Y are linearly independent and α X for j = 1,...,J. T is compact j ∈ j ∈ ∗ because Corollary 4.10 yields the existence of ` , . . . , ` Y such that 1 J ∈ ∗

`k(fj) = δkj.

We have

αj(x) = `j(T x) `j T x | | | | ≤ k k k k

max `j T x . ≤ j J k k k k ∈

Assuming (xn)n X is bounded, the vector of coeﬃcients (αj(xn))j=1,...,J is bounded J ⊂ in K = Heine-Borel tells the existence of xnk such that for every j = 1,...,J we ⇒ k have αj(xnk ) →∞ βj with some βj K. Hence −−−→ ∈ J J X k X T x = α (x )f →∞ β f Y. nk j nk j −−−→ j j ∈ j=1 j=1

95 5.29 Theorem. Let T BL(X,Y ) be compact, (x ) X such that x w x X as ∈ n n ⊆ n −−→ ∈ n . Then (T x ) Y is strongly convergent. → ∞ n n ⊆ w Proof. Weak convergence xn x and Lemma 4.32 give supn N xn < . Let yn := −−→ ∈ k k ∞ T xn, n N and y := T x. Let ` Y ∗. Then, using the adjoint T × : Y ∗ X∗, ∈ ∈ −→ n `(y ) `(y) = `(T x ) `(T x) = (T ×`)(x x) →∞ 0. n − n − n − −→ w n →∞ Hence yn y. Suppose yn y. Then there is some ε > 0 and (ynk )k N such that −−→ −−−→s ∈ ynk y ε for every k N. But (xnk )k N is bounded. By the compactness of T , there ∈ − ≥ l ∈ w is a subsequence T xnk →∞ z Y , z = y. Since ynk y = z, this is a contradiction, l −−−→ ∈ 6 l −−−→l 6 n →∞ so y →∞ y. n −−−→ 5.30 Theorem. Let T BL(X,Y ). ∈

(a) Assume (Tn)n N BL(X,Y ), Tn compact for each n N and ∈ ⊆ ∈ n T T →∞ 0. k n − k −−−→ Then T is compact.

(b) T is compact if and only if T × is compact (Schauder’s Theorem). (c) Let Z be a Banach space and S BL(Y,Z). If S or T is compact, then ST BL(X,Z) is compact. ∈ ∈

Proof. (a) Let (xm)m X bounded, w.l.o.g. xm 1 for m N. Then Tnxm has ∈ k k ≤ ∈ a convergent subsequence as m with the limit y Y . Now by (Cantor’s) → ∞ n ∈ diagonal sequence argument, there is a common convergent subsequence (xmk )k N k ∈ 0 with Tnxmk →∞ yn, n N. Let ε > 0 and N N such that Tn Tn ε for −−−→ ∈ ∈ k − k ≤ every n, n N. Then 0 ≥

0 0 0 0 yn yn yn Tnxmk + Tnxmk Tn xmk + Tn xmk yn k − k ≤ | −{z } | −{z } | {z− } (1) Tn T 0 ε (2) ≤k − n k≤ < ε + ε + ε

because (1) and (2) tend to 0 as k . So (yn)n N is a Cauchy sequence, hence n → ∞ ∈ convergent, i.e. y →∞ y Y . n −−−→ ∈ k Claim: T xmk y →∞ 0. Indeed, let ε > 0. There is some n N such that − −−−→ ∈ T T ε/3 and y y ε/3. k − nk ≤ k n − k ≤ This leads to

T xmk y T xmk Tnxmk + Tnxmk yn + yn y − ≤ − − k − k ε/3 + Tnxmk yn +ε/3 ≤ | {z− } ε/3 ≤ for k big enough. So T is compact.

96 X (b)“ ” Let (`n)n K be a bounded sequence in Y ∗. Then K := T (B1 (0)) Y is a com- ⇒ ∈ ⊆ pact metric space. The sequence of restrictions f := ` C(K) is bounded n n|K ∈ and equicontinuous, because fn(y1) fn(y2) sup `k y1 y2 Y . | − | ≤ k N k k k − k | ∈ {z } < ∞ By the theorem of Arzel`a-Ascoli 1.50 there exists a uniformly convergent subse-

quence (fnk )k N. Hence ∈

T ×`n T ×`n = sup T ×`n x T ×`n x K k l ∗ k l − X x BX (0) | − | ∈ 1

= sup `nk (T x) `nl (T x) x BX (0) | − | ∈ 1 k,l →∞ = fnk fnl ,K 0, − ∞ −−−−→ X where the last equality is due to the fact that T (B1 (0)) is dense in K. So (T ` ) converges in X and T : Y X is compact. × nk k ∗ × ∗ −→ ∗ “ ” If T compact, then T : X Y is compact. So T J is compact, ⇐ × ×× ∗∗ −→ ∗∗ ×× X where J : X X is the canonical embedding (J x)(f) = f(x) for every X −→ ∗∗ X x X and f X . Now, for x X and ` Y we calculate ∈ ∈ ∗ ∈ ∈ ∗

(T ××JX (x))(`) = JX (x)(T ×`)

= (T ×`)(x) = `(T x)

= (JY (T x))(`)

This implies T ××JX = JY T being compact. Since Y is closed in Y ∗∗, T is compact aswell.

5.31 Theorem. Let be a separable Hilbert space. Then every compact T BL( ) is H ∈ H the uniform limit of a sequence of operators (T ) BL(X,Y ) of ﬁnite rank. n n ⊆

Proof. Let ϕj j N be an orthonormal basis and for n N let { } ∈ ⊆ H ∈

λn := sup T ψ ψ span(ϕ1,...,ϕn) k k ⊥ ψ =1 k k

(λn)n N is non-negative and decreasing, so there is some limit ∈

lim λn =: λ 0 n →∞ ≥

Claim: λ = 0: True, because for n N ψn span(ϕ1, . . . , ϕn) with ψn = 1 such ∈ ∃ ⊥ k k

97 that T ψ λ /2 λ/2. Now, for every y k nk ≥ n ≥ ∈ H 2 X∞ y, ψ 2 = y, ϕ ϕ , ψ | h ni | j j n j=1 !2 X∞ y, ϕ ϕ , ψ ≤ | j j n | j=1 X∞ ψ 2 y, ϕ 2 ≤ k nk | j | j=n+1

X∞ 2 n y, ϕ →∞ 0. ≤ | j | −−−→ j=n+1

w n So ψ 0 and by Theorem 5.29 T ψ →∞ 0. So λ = 0. Let n −−→ n −−−→ n X R := ϕ , T ϕ BL(X,Y ) n j · j ∈ j=1 with dim ran R n. Then, by Parseval’s Equality 2.49 n ≤ X∞ (T R )ψ = (T R ) ϕ , ψ ϕ − n − n h l i l l=1 n X∞ X = ϕ , ψ T ϕ ϕ , ψ T ϕ h l i l − h l i l l=1 l=1 X∞ = ϕ , ψ T ϕ h l i l l=n+1 X∞ = T ϕ , ψ ϕ h l i l l=n+1 | {z } span(ϕ1,...,ϕn) ⊥ and

X∞ ϕl, ψ ϕl ψ . h i ≤ k k l=n+1 So

X∞ n T Rn = sup (T Rn)ψ λn ϕl, ψ ϕl λn →∞ 0. k − k − ≤ h i ≤ −−−→ ψ l=n+1 ψ∈H=1 k k 5.5 Fredholm alternative for compact operators 5.32 Motivation.

Let M be a N N-matrix for N N and let z C. Then • × ∈ ∈ either: MΨ = zΨ has a solution 0 = Ψ CN , 6 ∈ or: (M z) 1 exists. − −

98 Let = L2([0, 2]) and M BL(L2([0, 2])) be deﬁned by • H ∈ (MΨ)(x) := xΨ(x) for a.e. x [0, 2] Ψ L2([0, 2]). ∈ ∀ ∈ – MΨ = Ψ has no solutions 0 = Ψ L2 because xΨ(x) = Ψ(x) for Lebesgue 6 ∈ almost every x [0, 2] = Ψ(x) = 0 for almost every x [0, 2]. ∈ ⇒ ∈ – (M 1) is not surjective, because − (M 1)Ψ (x) = (x 1)Ψ(x) ∗ − − The mapping ∗ 1/3 ϕ: x x 1 − 7−→ | − | is an element of L2([0, 2]) but ϕ ran(M 1): Suppose ϕ ran(M 1). 6∈ − ∈ − Then there exists Ψ L2([0, 2]) such that (M 1)Ψ = ϕ and thus Ψ(x) = 1 1/3 ∈ 2 − x 1 x 1 − which is not an element of L ([0, 2]). − | − | For compact operators the above either/or statement still holds. • In the following is a separable Hilbert space. H 5.33 Theorem (Analytic Fredholm theorem). Let D C be open and connected, let ⊆ f : D BL( ) be analytic and f(z) compact for all z D. Then −→ H ∈ either (A1): (f(z) 1) 1 does not exist for any z D. − − ∈ or (A2): there exists a discrete subset S D without accumulation point in D such that the mapping ⊆

D S BL( ) \ −→ H 1 z (f(z) 1)− 7−→ − is well-deﬁned and analytic. Moreover, z S iﬀ f(z)Ψ = Ψ has a solution 0 = Ψ . ∈ 6 ∈ H 5.34 Corollary. Let T BL( ) be compact and z C 0 . Then ∈ H ∈ \{ } either: T Ψ = zΨ has a solution 0 = Ψ 6 ∈ H or: (T z) 1 BL( ) exists. − − ∈ H Proof. Choose D := C 0 and f(z) := T . Alternative (A1) does not hold in Thm. 5.33 \{ } z because T 1 has an inverse in BL( ) z > T (Neumann series!) = Alternative z − H ∀| | k k ⇒ (A2) holds.

Proof of Thm. 5.33. It suﬃces to prove the following: For every z D exists a neighbour- 0 ∈ hood (z ) such that eiher (A1) or (A2) holds on (z ). [Indeed, color a neighbourhood N 0 N 0 (z ) red or blue, depending on which alternative holds. D is connected, i.e. if both colors N 0 appear then there existsz ˜ such that (˜z ) cannot be uniquely coloured. Contradiction.] 0 N 0 Fix z D. Since f is continuous there exists r > 0 such that f(z) f(z ) < 1 for 0 ∈ k − 0 k 2 all z B (z ) = D . By Thm. 5.31 (separability!) there exists a ﬁnite-rank operator ∈ r 0 r F BL( ) such that f(z ) F < 1 . Let N := rank(F ). By Lemma 5.21 (Neumann ∈ H k 0 − k 2 series)

D BL( ) r −→ H 1 z (1 f(z) + F )− (1) 7−→ −

99 is well deﬁned and analytic. Ex. 5.28 (b) and Riesz = Ψ ,..., Ψ linearly ⇒ ∃ 1 N ∈ H independent and ϕ , . . . , ϕ such that F = PN ϕ , Ψ . For n = 1,...,N deﬁne 1 N ∈ H k=1 h n ·i n the analytic -valued function H Dr γ : −→ H n z (1 f(z) + F ) 1 ∗ϕ . 7−→ − − n Then N 1 X g(z) := F (1 f(z) + F )− = γ (z), Ψ , (2) − n · n n=1 and

f(z) 1 = (g(z) 1)(1 f(z) + F ). (3) − − − For all z D we have the equivalences ∈ r f(z) 1 is not invertible g(z) 1 is not invertible − ⇐⇒ − ~ w w( ) [proof below] ∗ solution 0 = Ψ of f(z)Ψ = Ψ solution 0 = ϕ of g(z)ϕ = ϕ ∃ 6 ∈ H ⇐⇒ ∃ 6 ∈ H ~ w PN (2) w ϕ= n=1 βnΨn

N X β1, . . . ,βN C (not all = 0) : βn = γn(z), ψm βm ∃ ∈ h i m=1 ~ w w d(z) := det A(z) 1N N = 0, where A := (Anm)1 n,m N ,Anm(z) := γn(z), Ψm − × ≤ ≤ identity theorem But d : Dr C is analytic (because Anm is so m, n) = → ∀ ⇒ either S := z D : d(z) = 0 has no accumulation point in D or S = D = r { ∈ r } r r r ⇒ claim It remains to prove ( ), i.e. the problem is now reduced to the ﬁnite-rank case: “ ” ∗ ⇑ clear N 1 “ ” By contradiction. Assume d(z) = 0 = ξ C : A(z) N N β = ξ has ⇓ 6 ⇒ ∀ ∈ − × N a solution β C . Choose ξn := γn(z), ψ for ψ ﬁxed but arbitrary = ∈ ∈ H ⇒ ϕ˜ := ψ + PN β ψ solves g(z) 1 ϕ˜ = ψ, because − n=1 n n − N X g(z) 1 ϕ˜ = ψ g(z)ψ + g(z) 1 β ψ − − − m m m=1 N N (2) X X = ψ + ψ A (z) δ β γ (z), ψ n nm − nm m − n n=1 m=1 | {z } =0 = g(z) 1 is invertible, a contradcition. ⇒ −

100 5.35 Corollary (Riesz-Schauder). Let T BL( ) be compact. Then σ(T ) has no accu- ∈ H mulation point in C, except possibly at zero. Moreover, any 0 = λ σ(T ) is an eigenvalue 6 ∈ of ﬁnite multiplicity (i.e. the corresponding eigenspace is ﬁnite-dimensional).

Proof. Choose f(z) := T and D := C 0 in Thm. 5.33. Then C ( 0 S) ρ(T ), also z \{ } \ { } ∪ ⊆ S σ (T ) σ(T ) 0 S. Thus S 0 = σ (T ) 0 . Also zero is the only possible ⊆ p ⊆ ⊆ { } ∪ \{ } p \{ } accumulation point of S. Finite multiplicity: Suppose not (only possible for dim( ) = ). Then there exists an H ∞ inﬁnite ONB Ψn n N of the eigenspace corresponding to some eigenvalue λ = 0 and { } ∈ 6 T Ψn = λΨn for all n. Compactness of T implies that (T Ψn)n = (λΨn)n has a convergent subsequence, which is a contradiction since λ = 0. 6 5.36 Remark. (a) If dim = = 0 σ(T ). Proof : If not, then 0 ρ(T ), i.e. 1 1 H 1∞ ⇒ ∈ ∈ 1 T − BL( ) and = TT − . Let Ψn n N be an ONB. Then Ψn = T (T − Ψn) n. ∈ H { } ∈ ∀ Since T 1ψ is bounded and T is compact, Ψ has a convergent subsequence. { − n}n ⊆ H n Contradiction.

(b) Corollaries 5.34 and 5.35 have generalisations to Banach spaces and to K = R.

Outlook: Spectral theorem for compact self-adjoint operators

5.37 Theorem (Hilbert-Schmidt theorem). Let T BL( ) be self-adjoint and compact. ∈ H Then there exists an ONB Ψn n of and to each Ψn and eigenvalue λn R such that { } H ∈ T Ψ = λ Ψ for all n and T = P λ Ψ , Ψ (convergence in operator norm). n n n n n h n ·i n

101