Lecture Course. Functional Analysis

Transcription

1 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!

2 Version of April 10, 2014

3 Contents 1 Topological and metric spaces Topological spaces: basics Limits and continuity Metric spaces Example: sequence spaces l p Compactness Example: spaces of continuous functions Baire s Theorem Banach and Hilbert spaces Vector spaces Banach spaces Linear operators Linear functionals and dual space Hilbert spaces Measures, integration and L p -spaces Measures Integration L p -spaces Decomposition of Measures The cornerstones of functional analysis Hahn-Banach theorem Three consequences of Baire s theorem (Bi)-Dual spaces and weak topologies Bounded operators Topologies on the space of bounded linear operators Adjoint operators The spectrum Compact operators Fredholm alternative for compact operators

4 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

5 1 Topological and metric spaces 1.1 Topological spaces: basics 1.1 Definition. Let X be a set. T P(X) is a topology iff (1), X T. (2) T is closed under arbitrary unions (i.e. if I is an arbitrary index set and for every α I let a set A α T be given. Then A α T holds.). (3) T is closed under finite intersections (i.e. if n N and A 1,..., A n T, then holds.). α I n A k T k=1 (X, T ) is called topological space (often just X) A P(X) is open iff A T. Let T 1, T 2 be topologies on X. T 1 is finer than T 2 iff T 1 T 2 and coarser than T 2 iff T 1 T Examples. (a) Indiscrete topology: T = {, X}. (b) Discrete topology: T = P(X). (c) Euclidean (or standard) topology on R n, n N: A R n is open iff x A ε > 0 such that B ε (x) A, where B ε (x) := {y R n : x y < ε} is the Euclidean ball of radius ε > 0 about x R n. Induced topology on subsets 1.3 Definition. Let (X, T ) be a topological space, A P(X) (not necessarily open!). T A P(A) is the relative topology on A iff 1.4 Remark. (a) T A is topology on A. T A := {B A : C T with B = C A}. (b) If A / T and B T A then it may happen that B / T. Example. Let X = R with standard topology, A = [0, 1]. Then B := [0, 1/2[ T A but B / T. 1.5 Definition. Let X be a topological space, A X, x X. (a) A is closed iff A c := X \ A T. 5

6 (b) U X (not necessarily open) is a neighbourhood of x iff A T such that x A and A U. (c) X is a Hausdorff space iff for all x, y X, x y, there exist neighbourhoods U x of x and U y of y such that U x U y =. (d) x is a limit point of A (or accumulation point) iff for all neighbourhoods U of x U A. 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 U A c. Boundary of A: A := {x X : x boundary point of A}. (g) Interior of A: Å := 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 Definition. Let (X, T ) be a topological space, B T a family of open sets. (a) B is a base for T iff T consists of unions of sets from B. (b) B is a subbase for T iff finite intersections of sets from B form a base. (c) N T is a neighbourhood base at x iff every N N is a neighbourhood of x and for every neighbourhood U of x there exists N N with N U. 1.8 Remark. (a) Let S P(X). Then there exists a topology T on X such that S is a subbase for T and T is the coarsest topology containing S. Jargon: T is generated by S. (b) Example: R n with standard topology. Let x R n. (i) {B 1/k (x) : k N} is a neighbourhood base at x. (ii) {B 1/k (q) : k N, q Q n } is a base for the standard topology (see the proof of Thm later). 6

7 1.9 Definition. Let J be an arbitrary index set. For every j J let (X j, T j ) be a topological space. T is the product topology on the Cartesian product space { X j := f : J } X j with f(j) X j j J j J iff T has the base { } A j : A j T j j J, A j X j for at most finitely many j s. j J 1.10 Remark. If J is finite, then the condition A j X j for at most finitely many j s can be dropped Definition. Let X be a topological space. (a) X is separable iff A X countable with A = X. (b) X is 1 st countable iff every x X has a countable neighbourhood base. (c) X is 2 nd 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 2 nd countable = X is 1 st countable and separable. Proof. Let B be a countable base for the topology. Let x X and N x := {B B : x B}. Then N x is a neighbourhood base and countable (hence 1 st countable). B B choose x B B and let A := {x B : B B}. We claim A is countable (trivial) and A = X. For all x X and neighbourhoods U of x there exists C T such that x C U. C is a union from sets in B, so there exists B B such that x B U. On the other hand x B B means x B U and x B A implies A U. 1.2 Limits and continuity 1.13 Definition. Let X be a topological space and (x n ) n N X be a sequence. (x n ) n converges to x X iff for every neighbourhood U of x there exists n 0 N and for all n n 0 : x n U. Notations: lim n x n = x or x n n x Remark. (a) convergence is harder for finer topologies. (b) X Hausdorff = limits are unique (see Problem T3) Definition. Let (X, T X ) and (Y, T Y ) be topological spaces and let f : X Y. (a) f is sequentially continuous iff x n n x (in X) implies f(x n ) n f(x) (in Y ). (b) f is continuous iff for every A T Y : f 1 (A) T X (f 1 (A) is the inverse image). (c) f is open iff A T X : f(a) = {y Y : x A : y = f(x)} T Y. 7

8 (d) f is a homeomorphism iff f is bijective, open and continuous (bijection compatible with topological structure) 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 n x in X. Let V Y be a neighbourhood of f(x). W.l.o.g. 1 assume 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 = n 0 N : n n 0 : x n U. But that also means n 0 N: n n 0 : f(x n ) V. If V is not open there exists an open subset V 0 V with x V 0 and one can repeat the same argument with V 0 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: N U = N U C. (1) Let {N k } k N be a countable neighbourhood base at x. Consider Ñk := k j=1 N j. Then Ñ k+1 Ñk k and (2) Ñ k is a neighbourhood of x k. (3) So {Ñk} k N is a neighbourhood base of x. (1) and (3) imply that k N x k Ñk U C l k : x l Ñk U C = x l l x. f(x k ) f(u) C V C = f(x k ) V k. But f(x) V so f(x k ) cannot converge to f(x). 1.3 Metric spaces Known notions and notations: Let X be a set. 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: Cauchy sequence, completeness 2 B ε (x) := {y X : d(x, y) < ε}. 1 without loss of generality 2 Completeness is not a topological notion! See Problem 4. 8

9 Figure 1: Triangle inequality For A X, x X define diam(a) := sup a,a A d(a, a ) diameter of A. dist(x, A) := inf a A d(x, a) distance of x to A Lemma. Let X be a metric space. Then X is a topological space, which is 1 st countable and Hausdorff w.r.t. 3 the metric topology, i.e. the one given by the base {B 1/k (x)} k N. x X Proof. {B 1/k (x)} k N is a neighbourhood base at x which is clearly 1 st countable. X is Hausdorff: Let x, y X be two distinct points. Then ε := d(x, y) > 0. For arbitrary points u B ε/2 (x) and v B ε/2 (y) the triangle inequation yields (see Figure 1): 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 B ε/2 (x) B ε/2 (y) = and X is Hausdorff Corollary. All topological notions are available in a metric space X. In particular for (x n ) n X, x X, A X: (a) (x n ) n converges to x ε > 0 n 0 N: n n 0 : x n B ε (x) lim n d(x n, x) = 0. (b) x A (y n ) n A: lim n y n = 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 Theorem. X a metric space. Then X separable implies X 2 nd countable. Proof. There is a countable and dense subset A X (A = X). We claim that {B 1/n (a)} n N is a base of the (metric) topology: Let C X open. B ε (x) C. Thus C = B ε(x) (x). x C a A x C ε(x) > 0 such that Denseness of A in X implies: x C a(x) A and n(x) N such that x B 1/n(x) (a(x)) B ε(x) (x) (see Figure 2). But then we can also write C = B 1/n(x) (a(x)). 3 with respect to x C 9

10 Figure 2: Finding n(x) NB. This also proves Remark 1.8 (b): {B 1/k (q)} k N in R n. q Q n is a base of the Euclidean topology 1.20 Lemma. Let X be a complete metric space and A X. Then: A closed A complete. Proof. See Analysis II Example. Consider the space of all continuous functions f : [0, 1] C: with two different metrics (a) Supremum metric C([0, 1]) := {f : [0, 1] C: f is continuous} d (f, g) := f g, where f := sup f(x) x [0,1] The metric space (C([0, 1]), d ) is complete and separable (see Thms and 1.42 later). (b) 1-metric: d 1 (f, g) := 1 0 dx f(x) g(x) defines a metric on C([0, 1]): (i) d 1 (f, g) 0 is obvious. Suppose f g. Then there exists at least one point x 0 [0, 1] with f(x 0 ) g(x 0 ) and we find ε > 0 so small that f(x 0 ) g(x 0 ) > ε. But f, g are continuous, so there exists δ > 0 such that for any x [0, 1] with x x 0 < δ we have f(x) f(x 0 ) < ε/3 and g(x) g(x 0 ) < ε/3. Combining these inequalities we obtain f(x) g(x) f(x 0 ) g(x 0 ) f(x) f(x 0 ) g(x 0 ) g(x) > ε/3 >ε <ε/3 <ε/3 for all x [0, 1] with x x 0 < δ. Thus d 1 (f, g) = 1 0 dx f(x) g(x) > ε/3 δ > dx 1 ]x0 δ,x 0 +δ[(x) f(x) g(x) >ε/3 10

11 Figure 3: f n (ii) The symmetry d 1 (f, g) = d 1 (g, f) is obvious. (iii) Triangle inequality: d 1 (f, h) = Thus, d 1 is indeed a metric. 1 0 dx f(x) h(x) d 1 (f, g) + d 1 (g, h). f(x) g(x) + g(x) h(x) (C([0, 1]), d 1 ) is separable because d 1 (f, g) d (f, g) and (C([0, 1]), d ) is separable. (C([0, 1]), d 1 ) is not complete: Consider f n from Figure 3. (f n ) n is a Cauchy sequence: Let n, m N, m n: But d 1 (f n, f m ) = dx f n (x) f m (x) = dx f n (x) 1 [1/2,1] (x) 1/2 1/2 1/n 1/2 1/2 1/n dx f n (x) f m (x) 1 n. 1 dx f n (x) 1 [1/2,1] (x) 1 n 1 and 1 [1/2,1] (x) / C([0, 1]). Suppose f C([0, 1]) with d 1 (f n, f) 0 as n. Then 1 dx f(x) 1 [1/2,1] (x) = 0 0 which cannot be true (consider a neighbourhood of x = 1/2) = f n does not converge in C([0, 1])! 1.22 Definition. Let (X, d X ), (Y, d Y ) be metric spaces and T : X Y. T is called an isometry iff x, x X : d X (x, x ) = d Y (T (x), T (x )). X, Y are isometric iff there exists a bijective isometry T : X Y Remark. Let T : X Y be an isometry. Then (a) T is injective and continuous, (b) X and the range of T, are isometric. ran T := T (X) = {y Y : x X with y = T (x)}, 11

12 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: (x n ) n (y n ) n : lim n d(x n, y n ) = 0 ( is clearly reflexive, symmetric and transitive). We look at the equivalence classes we obtain in this way, so consider X := { equivalence classes x of Cauchy sequences in X}. We write (x n ) n x for a representative (x n ) n of the equivalence class x. Define a metric on X: d( x, ỹ) := lim n d(x n, y n ) x, ỹ X, (x n ) n x, (y n ) n ỹ We now have to check that d... (1)...is well defined: (i) Existence of the limit: d(x n, y n ) d(x n, x m ) + d(x m, y m ) + d(y m, y n ) d(x n, y n ) d(x m, y m ) d(x n, x m ) + d(y m, y n ) And the same inequality holds with m and n exchanged. Thus we have: d(x n, y n ) d(x m, y m ) d(x n, x m ) + d(y m, y n ) ( ) =:α n =:α m Because of (x n ) n and (y n ) n being Cauchy sequences we have ε > 0 N N such that n, m N: d(x n, x m ) < ε/2 and d(y n, y m ) < ε/2 So α n α m < ε and (α n ) n R is a Cauchy sequence, thus convergent (because R is complete). (ii) Independence of representatives: Let (x n ) n (x n) n and (y n ) n (y n) n. We have to show lim d(x n, y n ) = lim n n d(x n, y n) Repeat derivation of ( ) with exchanging x m x n and y m y n and obtain d(x n, y n ) d(x n, y n) d(x n, x n) + d(y n, y n) n 0 by definition of 12

13 (2)...fulfills the axioms of a metric: d 0 d( x, ỹ) = 0 lim n d(x n, y n ) = 0 for some representatives (x n ) n (y n ) n x = ỹ Symmetry and triangle inequality are clear from the properties of d. (b) Define i : X X b b 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, ã) = 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. Let x X and ε > 0. Pick any representative (x n ) n x. (x n ) n being a Cauchy sequence implies N N n, m N: d(x n, x m ) < ε. Let b W defined by (x N, x N,...) b So W is dense. = d( b, x) = lim n d(x N, x n ) < ε (c) Let ( x (k) ) k N ) be a Cauchy sequence in X; W is dense in X: k N z (k) W such that d( x (k), z (k) ) < 1 k Let (z k, z k, z k,...) z (k) be the corresponding constant representative for each k. For every k, l N we get (due to the isometry property of i): d(z k, z l ) = d(i(z k ), i(z l )) d( z z (k) z (l) (k), x (k) ) <1/k + d( x (k), x (l) ) + d( x (l), z (l) ) <1/l Thus (z 1, z 2, z 3,...) X is a Cauchy sequence and hence a representative of some x X. We show now, that lim d( x (k), x) = 0. n Let ε > 0, then <1/k for all k large enough, as (z n ) n is Cauchy. d( x (k), x) d( x (k), z (k) ) + d( z (k), x) < ε =lim n d(z k,z n) 13

14 (d) Suppose, there exists ˆX with a dense subset V and X with a dense subset W and bijective isometries j, respectively i between X and V, respectively X and W. We show: X and ˆX are isometric. We know: V := j(x) and W := i(x) are isometric with isometry j i 1. According to the Problem T5, j i 1 can be uniquely extended to an isometry X ˆX. Thus we have the uniqueness up to isometries. ˆX j V X j i 1 i W X Figure 4: Uniqueness 1.4 Example: sequence spaces l p 1.25 Definition (l p -spaces). Set { ( l p := x = (x n ) n N : x n C n and x p := x n p) } 1/p < whenever p [1, [ and { l := ( p will be a norm for every p [1, ]). n N x = (x n ) n N : x n C n and x := sup n N } x n < 1.26 Lemma. For every p [1, ], d p (x, y) := x y p, x, y l p defines a metric on l p. Proof. All properties clear, except for triangle inequality, this follows from Lemma Lemma. Let p, q [1, ] be conjugated exponents, i.e. 1/p + 1/q = 1 (convention: 1/ = 0 ). Then (a) Dual pairing and Hölder inequality: x l p, y l q : x, y := n N x ny n is well defined and x, y x n y n x p y q. n N (b) Minkowski inequality: Let x, y l p. Then x + y p x p + y p. Proof. From corresponding inequalities on C N and subsequent limit N. For example, to prove (a), use N ( N ) 1/p ( N ) 1/q x n y n x n p y n q n=1 n=1 (see e.g. Forster, vol. 1) and perform limit Remark. Minkowski inequality does not remain true for p ]0, 1[. Therefore we consider l p -spaces only for p Theorem. (a) l p is separable p [1, [. n=1 14

15 (b) l is not separable. Proof. (a) We use the separability of C. For n N let M n := {(x 1,..., x n, 0,...) l p with x j Q + iq, j = 1,..., n}. So M n is countable and M := n N M n is also countable. Claim M = l p. Let y l p and ε > 0, then there exists N N: j=n+1 y j p < εp 2 and since Q + iq is dense in C there exists x M N such that N j=1 x j y j p < εp 2. This implies ( dp (y, x) ) p = x y p p < εp (b) See Problem Theorem. l p is complete for every p [1, ]. Proof. (a) Case p [1, [. Let (x (n) ) n N l p be a Cauchy sequence (x (n) = (x (n) 1, x(n) 2,...)). Let ε > 0. Then there exists N N such that for every n, m N and J 1, J 2 N: J 2 x (n) j j=j 1 x (m) j p < ε p ( ) Step 1: We have to show the existence of a candidate for the limit using the completeness of C. Let J 1 = J 2 = J. ( ) implies (x (n) J ) n C is a Cauchy sequence and since C is complete, there is some x J C such that lim n x (n) J = x J for all J N. So our candidate is x := (x 1, x 2,...). Step 2: Set J 1 = 1 in ( ) and use the Minkowski inequality in C J 2 : ( J2 ) 1/p x j p j=1 ( J2 ) 1/p ( x j x (n) J2 ) 1/p j p + x (n) j p j=1 ε + x (n) p J 2 N. j=1 } {{ } x (n) p < So we obtain x p ε + x (n) p < for every n N and therefore x l p. For every n N and m in ( ): J 2 j=1 x (n) j x j p < ε p J 2 N, so for sending J 2 in addition, we have So x (n) x in l p. x (n) x p < ε. d p(x (n),x) (b) Case p = : Replace J 2 j=j 1 by sup J1 j J 2 and p by. 15

16 1.5 Compactness 1.31 Definition. Let X be a topological space and A X. (a) A is compact iff for every open cover A α I B α with an index set I and B α X open for every α I there exists a finite open subcover, i.e. N N and α 1,..., α N I with A N n=1 B α n ( Heine-Borel-property ). (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 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 Theorem. Let X be a topological space. (a) If X is 1 st countable, then X compact implies X sequentially compact. (b) If X is 2 nd countable, then X compact is equivalent to X sequentially compact Theorem (Lindelöf). Let X be a 2 nd countable topological space and let X = α I A α be an open cover. Then there exists (α n ) n N I such that X = n N A α n, i.e. a countable subcover. Proof. Let {B k } k N be a countable base of the topology. Set Since every A α is a union of B k s, we have K := {k N : α I with B k A α }. k K B k = α I So for every k K pick α k I such that B k A αk. Then we get X ( ) = B k A αk ( ) A α = X k K α I So k K A α k = X. k K A αk A α = X (*) Proof of Theorem (a) By contradiction: Assume X is compact but there exists a sequence (x n ) n N X without convergent subsequence. Claim: x X there exists a neighbourhood U(x) such that x n U(x) for at most finitely many n. Proof of the claim: Consider a countable neighbourhood base {U i } i N at x and let V m := m i=1 U i for m N. Suppose the claim was false. Then for every m N there exist in finitely many k N such that x k V m. This means there exists a subsequence (k m ) m N increasing such that x km V m for every m N. So, for every m, m N, m m: x km V m. This implies lim m x km = x, i.e. convergent subsequence. Now X = y X U(y) = n i=1 U(y i) for some y 1,..., y n X (because X is compact). The claim implies that U(y i ) contains at most finitely many sequence members x k U(y i ), so there are at most finitely many k such that x k X. 16

17 (b) = follows from (a) and Lemma We now show = by contradiction: Assume every sequence has a convergent subsequence, but there exists an open cover without finite subcover. Since X is 2 nd countable there exists a countable subcover X = j N C j of this cover by Lindelöf. For every n N pick a point ( n ) x n X \ C j (always possible n N because there does not exist a finite subcover). Let (x n ) n X be a sequence in X. By hypothesis, it has a convergent subsequence x nk x X. There exists N N such that x C N, so C N is a neighbourhood of x. x nk being convergent means x nk C N for finally all k, but for every k such that n k N we have by definition of x n that x nk / C N Theorem. Let X be a compact topological space and A X. Then (a) A closed = A compact j=1 (b) If X is also Hausdorff, then A compact = A closed. k Proof. (a) Let A α I U α be an open cover. Since A is closed A c is open and ( ) X = A c U α is an open cover. Since X is compact, there exist α 1,..., α n I such that ( n ) X = A c U αi and A n i=1 U α i is a finite subcover. (b) See Problem Warning. Bounded and closed do not imply compact in general! Example: l p, p [1, ] and α I B 1 (0) = {x l p : x p 1} = {x l p : x p < 1} = B 1 (0) is bounded and closed but consider e (n) := (..., 0, 1, 0,...) l p (with 1 at the nth position) n N. It fulfills { d p (e (n), e (m) ) = e (n) e (m) 2 1/p p < p = n, m N, n m, 1 p = so there exists no convergent subsequence and B 1 (0) is not sequentially compact. Hence, by Theorem 1.33 (a), B 1 (0) is not compact Theorem. Let X be a metric space. Then i=1 X is compact X is 2 nd countable (a) (c) (b) X is sequentially compact X is separable. 17

18 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 2 nd countable and we can apply Theorem 1.33 (b). (b) For any topological space X 2 nd countability implies separability (Theorem 1.12). If X is a separable metric space, then it is also 2 nd countable (Theorem 1.19). (c) We show that sequential compactness implies separability by constructing a countable set M with M = X. Fix n N and use the following algorithm to define points x (n) k : Choose an arbitrary x (n) 1 X If R (n) 1 := X \ B 1/n (x (n) 1 ), pick x(n) 2 R (n) 1, otherwise stop. Suppose x (n) 1,..., x(n) k are chosen. If pick x (n) k+1 R(n) k ( R (n) k k := X \, otherwise stop. j=1 B 1/n (x (n) 1 ) ), Claim: This algorithm stops after finitely many steps. True, because for k l we have d(x (n) k, x(n) l ) > 1/n. Thus, if the algorithm did not stop after finitely many steps we would have an infinite sequence (x (n) l ) l N X without a convergent subsequence which is a contradiction to X being sequentially compact. The claim implies the existence of K n N such that X = K n j=1 B 1/n ( x (n) j Set M n := {x (n) j : j = 1,..., K n } and M := k N M n. M is countable and M = X Theorem (Tychonoff). Let J be an index set and X α a compact topological space for all α J. Then { X α = f : J } X α such that f(α) X α α J α J is compact (in the product topology). Proof. See any textbook on topology, e.g. Kelley, v. Querenburg. ) 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) Theorem. Let X, Y be topological spaces, X compact, f C(X, Y ). Then (a) f(x) is compact. 18

19 (b) If in addition X, Y are Hausdorff and f a bijection, then f is a homeomorphism. (c) If in addition X, Y are metric spaces, then f is uniformly continuous iff ε > 0 δ δ ε : x X : f(b δ (x)) B ε (f(x)) (Note that B δ (x) and B ε (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. (a) Let f(x) α J F α be openly covered. Then X f 1 ( α J ) F α = f 1 (F α ). α J Because f is continuous f 1 (F α ) is open α J and we have an open cover of X. But X is compact which by definitions means that we find N N and α 1,..., α n such that X N n=1 f 1 (F αn ) and, thus, f(x) N n=1 F α n. 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 (well defined by Theorem 4.40 (a)) Theorem. C(X, K) is complete. d (f, g) := sup f(x) g(x) = f g. x X Proof. Follows from completeness of the bounded continuous functions C b (X, K), see Problem 5, and C(X, K) = C b (X, K), which follows from compactness of X and Theorem 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 Here, we only show = : For m, n N define G mn := Compactness of X implies that, { f C(X, K) : f(b 1/m (x)) B 1/n (f(x)) } x X 19

20 ... 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 δ (x)) B 1/n (f(x)). Then we can find m N such that 1/m δ and get that f G mn. This shows C(X, K) = m N G mn n N. (1)... for any m N we can find p(m) N and a 1,..., a p(m) X such that X can be written a union of open balls of radius 1/m: X = p(m) j=1 B 1/m (a j ). (2) Now, K is separable, i.e. there exists a countable set {κ ν K: ν N} that is dense in K. For fixed m N and ϕ: {1,..., p(m)} N (or equivalently ϕ N p(m) ) define G (ϕ) mn := {f G mn : f(a k ) κ ϕ(k) < 1/n k = 1,..., p(m)}. We only want to consider those ϕ with G (ϕ) mn, so we set Φ mn := {ϕ N p(m) : G (ϕ) mn }. Φ mn is not empty (see below). For ϕ Φ mn pick some g ϕ G (ϕ) mn. Now define L mn := {g ϕ : ϕ Φ mn } and L := m,n N L mn is countable because Φ mn N p(m). 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: n N m N such that f G mn. There also exists ϕ f Φ mn such that f G (ϕ f ) mn : We only have to choose ϕ f (k) such that f(a k ) κ ϕf (k) < 1/n. But {κ ν : ν N} is dense in K, so this is obviously possible. For each x X choose k x {1,..., p(m)} such that x B 1/m (a kx ) (this is possible because of (2)). Now we consider g ϕf (x) L mn L. Using that f and g ϕf both lie in G (ϕ f ) mn L mn we can estimate the upper bound of the distance f(x) g ϕf (x) : f(x) g ϕf (x) f(x) f(a kx ) + <1/n by definition of G mn + κ ϕf (k x) g ϕf (a kx ) <1/n by definition of G ϕ f mn f(a kx ) κ ϕf (k x) <1/n by definition of G ϕ f mn + + g ϕf (a kx ) g ϕf (x) < <1/n by definition of G mn 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: 4 n 20

21 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) 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, 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 Corollary. Let K R d 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 B 0 be the K-Algebra generated by the monomials K K x = (x 1,..., x d ) x n α n N 0, α {1,..., d} Set B := B 0. Then Stone-Weierstraß gives us B = C(K, K). Now we show separability: If K = R, let B Q 0 be the Q-Algebra generated by the monomials, if K = C, let BQ 0 be the (Q + iq)-algebra. Then B Q 0 is countable Since K is compact, B Q 0 = B 0 = C(K, K) Definition. Let X be a metric space (not necessarily compact). Let F be a family of continuous functions f : X K. F is equicontinuous iff for every ε > 0 and x X there exists a δ > 0 such that for every f F: f(b δ (x)) B ε (f(x)) F is uniformly equicontinuous iff for every ε > 0 there exists δ > 0 such that for every x X and f F: f(b δ (x)) B ε (f(x)) 4 This property is called unital 21

22 1.47 Remark. (a) If X is even compact, then (by Problem T7): equicontinuous uniformly equicontinuous. (b) Examples X = [0, 1]: X = [0, 1]: {x cos(x/n) : n N} is uniformly continuous {x x 1/n : n N} is not equicontinuous. X = ]0, [: {x arctan(nx) : n N} is equicontinuous, (individually) uniformly continuous, but not uniformly equicontinuous Theorem. Let X be a metric space (not necessarily compact) and (f n ) n N C(X, K) an equicontinuous sequence. Suppose there exists a dense subset D X such that for every x D the limit lim n f n (x) exists. Then lim n f n (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 n (B δ (x)) B ε (f n (x)). (1) Since D = X, there is some y D B δ (x). At y, the sequence (f n (y)) n converges and is therefore a Cauchy sequence, so there exists N N such that n, m N So we have f n (y) f m (y) < ε. (2) f n (x) f m (x) f n (x) f n (y) + f n (y) f m (y) + f m (y) f m (x) <ε by (1) <ε by (2) <ε by (1) < 3ε So lim n f n (x) =: f(x) exists for every x X. Secondly, we show the continuity of f. (1) is equivalent to δ > 0 n N x X with d(x, x ) < δ = f n (x) f n (x ) < ε, so for n there exists δ > 0, such that for every x X with d(x, x ) < δ we have f(x) f(x ) < ε Lemma. In addition to the assumptions in Theorem 1.48 suppose that X is compact. Then we have even lim n d (f n, 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: (see Remark 1.47) and there is l N and a 1,..., a l X: We have f n (B δ (x)) B ε (f(x)) (1) X = f n (x) f(x) f n (x) f n (a jx ) l B δ (a j ). (2) j=1 + f n (a jx ) f(a jx ) =:T 1 =:T 2 where a jx is the centre of some Ball B δ such that x B δ (a jx ) f(a jx ) f(x) T 3

23 T 1 < ε for every n N by (1) (and the definition of a jx ) T 2 < ε for all n sufficiently large (there are only finitely many j s and pointwise convergence) T 3 < ε by taking n in (1) (compare the end of the proof of Thm. 1.48). So finally we have ε > 0 N N n N x X : f n (x) f(x) < 3ε Theorem (Arzelà-Ascoli). Let X be a compact metric space and (f n ) n N C(X, K) an equicontinuous and pointwise bounded sequence, i.e. sup f n (x) < x X. n N Then there exists a uniformly convergent subsequence (f nj ) j N. Equivalently, every equicontinuous and pointwise bounded subset F C(X, K) is relatively compact. Proof. Since C(X, K) is a metric space the equivalence of the 2 statements follows from Theorem We prove the sequence version : Since X is compact, due to Theorem 1.37, X is also separable, so there exists a dense subset {a l X : l N} X. Pointwise boundedness gives sup f n (a l ) < l N, n N so by the Bolzano-Weierstraß theorem for every l there exists a subsequence (n (l) j ) j N N such that lim (a l ) exists. W.l.o.g. assume (n (l+1) j f j n (l) j ) j (n (l) j ) j. Define n j := n (j) j (diagonal sequence trick), then (n j ) j l (n (l) j ) j and lim j f nj (a l ) exists for every l N. Since the set of the a l s is dense, Lemma 1.49 gives the claim Remark. Both assumptions, compactness and equicontinuity are essential for Theorem 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 Remark. Let X be a metric space and A 1, A 2 X be open and dense. Then A 1 A 2 is also open and dense. Completeness gives more: 1.53 Theorem (Baire). Let X be a complete metric space and n N let A n X be open and dense. Then n N A n is dense in X Remark. (a) Openness of n N A n is false in general. (b) Completeness is essential for the theorem: Consider Q with the metric inherited from R. Let {q n Q : n N} be an enumeration of Q. Define A n := Q \ {q n }. This is open and dense in Q but n N A n =. 23

24 Figure 5: Constructing the sequence (x n ). Proof. Define D := n N A n. Let x 0 X be arbitrary and fix and ε > 0. To prove the denseness of D in X we have to show that D B ε (x 0 ). We do this by constructing a sequence (x n ) that converges against x D B ε (x 0 ) (see Figure 5). A 1 dense implies that A 1 B ε (x 0 ) is non-empty and we can pick x 1 A 1 B ε (x 0 ) and ε 1 > 0 with ε 1 ]0, ε/2[ such that B ε1 (x 1 ) A 1 B ε (x 0 ). Proceeding analogously with A 2 and B ε1 (x 1 ) we pick x 2 A 2 B ε1 (x 1 ) and ε 2 ]0, ε 1 /2[ such that B ε2 (x 2 ) A 2 B ε1 (x 1 ) A 1 A 2 B ε (x 0 ). We proceed in the same fashion and get two sequences: (a) (ε n ) n with 0 < ε n < 2 n ε n N, (b) (x n ) n X with B εn+1 (x n+1 ) A n+1 B εn (x n ) A 1 A n B ε (x 0 ) In particular we have: N N n N : x n B εn (x N ). ( ) That is, (x n ) n is Cauchy and since X is complete, (x n ) converges to some x X. But because of ( ) we have x B εn (x N ) N N and (b) yields x n N B ε n+1 (x n+1 ) D B ε(x0 ). Thus, D B ε (x 0 ) Definition. Let X be a topological space and A X. 1. A is a G δ (-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 1 st category) iff A is a countable union of nowhere dense sets. 4. A is non-meagre (or of 2 nd category) iff A is not meagre. 24

25 1.56 Example. Q is meager in R (just consider the countable union Q = q Q {q}) 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) A n X meagre n N = n N A n meagre. Proof. (b) and (c) are clear by the very definitions. Statement (a) follows from the equivalence B has no interior points B c dense. To show =, consider x X. If x B then x B C. If x B it is by hypothesis no interior point of B and thus must be a limit point of B C. 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) A n X open and dense n N = n N A n dense. (ii) A n X is a dense G δ n N = n N A n is a dense G δ. (iii) A X, A and A is open = A non-meagre. (iv) A X meagre = A C dense 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. Proof of Lemma (i) (ii) by definition of G δ. (i) (iii) Let A X be open. Suppose A is meagre, that is A = n N A n with A n nowhere dense n N. According to 1.57 (a), this is equivalent to (A n ) C dense and open n N and because of implication (i), the intersection n N (A n) C =: B is dense as well. But then B C = n N A n A has no interior points. Thus A has no interior points either and this is clearly a contradiction to A being open. (iii) (iv) Problem 13. (iv) (i) Problem

26 2 Banach and Hilbert spaces 2.1 Vector spaces General assumption: X {0} is a K-vector space, where K = R or C. 2.1 Definition. Let M X. M is linearly independent iff all non-empty finite subsets F M are linearly independent, i.e. the following implication holds: f F α f K f = 0 = α f = 0 f F. 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 c c := {x = (x j ) j N : x j C j N and x j 0 for only finitely many j s}. The index c stands for compact support. Let e m := (..., 0, 1, 0,... ) with a 1 at the n-th position. Claim: B := {e n : n N} is a Hamel basis for c c. Remark. Even though l 1 is separable there exists no countable hamel basis for l Theorem. Every vector space X {0} has a Hamel basis. Proof. Uses Zorn s lemma. See later. 2.5 Corollary. X has infinite dimension iff for every n N there exists M n X such that M n = n and M n is linearly independent. Proof. Existence of a Hamel basis B with B =. 2.6 Example. Infinite dimensional vector spaces: c c, l p, C(X) (where X R d open). 26

27 2.2 Banach spaces 2.7 Definition. Let X be a vector space. A mapping X [0, [, x x is a norm iff (1) x > 0 0 x X, (2) λx = λ x λ K x X, (3) x + y x + y x, y X (X, ) is called a normed space. If only (2) and (3) hold, is called a seminorm. 2.8 Remark. Let X be a normed space. Then d(x, y) := x y is a metric on X. Thus all topological notions and results from the theory of metric spaces are available. 2.9 Example. l 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) x X R n, C n are normed spaces for n N (Euclidian norm or p-norm) Lemma. Let X be a normed space. Then the mapping X x x is continuous. Addition and multiplication are continuous as well: Let x k Let α k k x and y k k α in K and x k k y. Then x k + y k k x. Then α k x k k x + y. k αx. k Proof. Let (x k ) k X with x k x. This is equivalent to x k x x 0 (Corollary 1.18). To show that is sequentially continuous, we have to prove that x k x k 0. This follows from the following inequalities: x k x x k x x k + x k = x x k + x x k = lim sup x k x lim inf x k k k The following sets are a base of the metric topology of X: {x + B 1/k (0) : x X, k N} = {B 1/k (x)} = {y X : x y 1/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: x k + y k x y x k x k 0 + y k y k 0 k 0. 27

28 Multiplication is continuous: α k x k αx = α k x k αx + α k x α k x = = α k (x k x) + x(α k α) α k (x k x) + (α k α)x α k bounded in k x k x + α k α x k 0. k k Definition. A complete normed space is called a Banach space Examples. All spaces in Example 2.9 are Banach spaces. Consider C([0, 1]) with L 1 norm: f 1 := 1 0 f(x) dx is not a Banach space. This was already discussed in Example 1.21 (b) Theorem. Every normed space X can be completed, so that X is isometric to a dense linear subspace W of the Banach space ˆX, which is unique up to isometric isomorphisms. Proof. Analogous to the proof of Theorem Note that the isometry is even a linear bijection (thus a isomorphism) in this case (see Section 2.3) Definition. Let X be a normed space. A set {e n X : n N} is a (Schauder) basis in X iff for all x X there exists a sequence (x n ) n K N such that lim N N x x n e n = 0. Notation: x = n N x ne n infinite linear combinations, convergent series. n= Example. Let p [1, [. Then (e n ) n N with e n := (0,..., 1, 0,... ) (where the 1 is at the nth position) is a (Schauder) basis of l p : For x = (x n ) n N l p we have N x p x n e n = n=1 Note that this construction fails for l! p n=n Lemma. Let X be a normed space. Then x n p N 0. X has a (Schauder) basis = X is separable. Proof. Let K 0 := Q for K = R, respectively K 0 := Q + iq for K = C. Define N A N := x n e n : x n K 0. n=1 Then the union A := N N A N is dense in X and countable Remark. The implication = in Lemma 2.16 does not hold (Enflo, 1973) Lemma. Let X be a Banach space, A X a (linear) subspace. Then A is closed A is complete. 28

29 Proof. See Analysis II, like proof of Lemma 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 1,..., e n } in F. (F, ) is a normed space and isometric to (K n, ) with n α := α j e j α = (α 1,..., α n ) K n. j=1 K n is complete with respect to the Euclidean norm. But all norms in finite-dimensional spaces are equivalent 5. Thus (K n, ) is complete and because of the isometry, (F, ) 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 λ X \ U such that x λ = 1 and x λ u λ u U. Proof. Since U is closed, we have d := dist(x, U) = inf u U d(x, u) > 0 for all x X \ U (see Problem 9(c)!). Since λ < 1 there exists u λ U such that d x u λ d λ, hence γ := 1 x u λ λ d. Define x λ := γ(x u λ ) X \ U. The first property x λ = γ x u λ = 1 holds by definition of γ and x λ u = γ(x u λ ) u = γx (u + γu λ ) = = γ x u λ + u γd λ u U. γ U Warning 1.36 illustrates the more general 2.21 Theorem. Let X be a normed space. Then B 1 (0) = {x X : x 1} compact dim X < Proof. ( ) Let dim X <. Proof of Theorem 2.19: X is isometric to (K n, ). The statement then follows by Heine-Borel and the equivalence of norms. ( ) Suppose that dim X =. We show that this implies that B 1 (0) is not sequentially compact by constructing a sequence (x n ) n in B 1 (0) without a convergent subsequence: Pick an arbitrary x 1 X with x 1 = 1. Let U 1 := span{x 1 } be the subspace spanned by x 1. U 1 is closed in X. Riesz Lemma, applied with λ = 1/2 shows the existence of x 2 X \ U 1 such that x 2 = 1 and x 2 x 1 1/2. Let U 2 := span{x 1, x 2 }. If we continue this procedure we get a sequence (x n ) n N B 1 (0) that satisfies x n x m 1/2 for all n m. (x n ) n clearly has no convergent subsequence. Thus B 1 (0) is not compact. 5 This is a theorem from linear algebra: For norms and on K n we can find constants c, c ]0, [ such that c c. 29

30 2.3 Linear operators 2.22 Definition. Let X, Y be vector spaces (over the same field), X 0 X a subspace and T : X 0 Y. T is linear (a linear operator) iff T (αx + βy) = αt (x) + βt (y) α, β K, x, y X 0. Notation: T x := T (x). T is a (vector space) homomorphism. dom(t ) := X 0 is the domain of T. ran T := T (X 0 ) is the range of T. ker(t ) := {x X 0 : T x = 0} is the kernel of T Examples. (a) The identity operator 1 := 1 X : X X, x x is obviously a linear operator. (b) Consider X = Y = C([0, 1]). Then the differentiation operator T : X 0 = C 1 ([0, 1]) C([0, 1]) with f f, so (T f)(x) = f (x) and this is a linear operator. x (T f)(x) := f(t)dt has dom T = X and is called the anti-derivative. 0 (T f)(x) := xf(x) is the multiplication by x Lemma. Let T be a linear operator. Then (a) ran(t ) and ker(t ) are vector subspaces (b) dim ran(t ) dim dom T (c) ker(t ) = {0} there exists an inverse T 1 of T such that T 1 : ran(t ) dom(t ) and T 1 T = 1 dom(t ). Proof. Copy from linear algebra 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 T T 1 = 1, but only T T 1 = 1 ran(t ). In other words, ker(t ) = {0} does not imply T is bijective (only injective) Example (illustrating Remark 2.25). Let X = l and x = (x 1, x 2,...) X. Consider the shift operator defined by T x := (0, x 1, x 2,...), i.e. { 0 n = 1, (T x) n := x n 1 n 2. Then we have dom T = l and ran T = {y = (y 1, y 2,...) l : y 1 = 0}. Define (T 1 x) n := (x 2, x 3,...). Then dom T 1 = l and T 1 T = 1. But T T 1 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

31 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 dom(t ) Y T := 2.28 Examples. (cf. Example 2.23) T x sup x dom(t ) x = x 0 (a) 1: (X, ) (X, ) is bounded with 1 = 1. (b) Let X = C([0, 1]) with. sup x dom(t ) x =1 T x <. T = d dx is unbounded. For n 2 take f n : x sin(nx). Then f n = 1 and so that The anti-derivative fulfills T f n = sup n cos(nx) = n x [0,1] T f = sup T f n f n = n. x [0,1] for every f C([0, 1]) and f 0. Because of we obtain T 1. Moreover, T 1 = sup x [0,1] and since 1 = 1 we have T = 1. x 0 T f f 1 The multiplication operator by x has T = 1. x 0 f(t)dt 1 f dt = sup x = 1, x [0,1] 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 0 dom(t ) (c) There exists c ]0, [ such that T x c x for every x dom(t ). (d) T is bounded. Proof. (a) (b) obvious. 31

32 (b) (c) Suppose T is continuous at x 0. Claim: It follows continuity at 0. This is true because let (x n ) n dom(t ) be a sequence which converges to 0. Then (x n + x 0 ) n n x 0, and by linearity and continuity in x 0 we obtain n T x n + T x 0 = T (x n + x 0 ) n T x 0. So T x n 0. From this we know that for ε = 1 there is a δ > 0 such that whenever x δ for x dom(t ) we have T x 1. For general 0 x dom(t ), let x := δ, so that x δ. Then we have x x i.e. c = 1/δ. (c) (d) We know T x c x and so (d) (a) Let (x n ) n dom(t ) with x n δ x T x = T x 1 = T x δ 1 x, T x sup x dom(t ) x c <. x 0 n x then T x n T x = T (x n x) T x n x n Lemma. Let T be linear and dim dom(t ) <. Then T is bounded. Proof. See linear algebra 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 : U Y, x T U x := T x Extension of T to W : T : W Y with T x = T x x dom(t ), i.e. T dom(t ) = T 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. 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 n ) n dom(t ) with x n x in X. Since (x n ) n is a Cauchy sequence, (T x n ) n is a Cauchy sequence in Y because T x n T x m = T (x n x m ) T x n x m. n Using that Y is a Banach space, there exists y Y such that T x n y in Y. Define T x := y (then T dom(t ) = T ). We have to check several things: 32

33 m Well-definedness, i.e. independence of approximating sequence: Let x m x in X. Then we have T x n T x m T x n x m. In the limit n, m we get y lim m T x m = 0. Linearity: Let x n n x, x n n x. Then T (αx + α x) = lim n T (αx n + α x n) = lim n αt x n + lim n α T x n = α T x + α T x. Norm: as T is an extension, we have T T, because T = On the other hand, since is continuous sup T x x dom( T ) x sup T x x dom(t ) x = T. x 0 x 0 T x = lim T x n n x n dom(t ) = lim T x n lim T x n n n = T lim x n n = T x. So T x x T for every x dom(t ) and therefore T T. So T = T. Uniqueness: The fact that we defined T x := y is necessary to ensure the continuity in x 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 ). Proof. First of all, BL(X, Y ) is a K-vector space with zero element 0: X Y x 0. For T 1, T 2 BL(X, Y ) and α K define T 1 + T 2 and αt 1 by for every x X. X Y is a norm on BL(X, Y ). (T 1 + T 2 )x := T 1 x + T 2 x (αt 1 )x := αt 1 x T X Y 0 and if T X Y = 0 then T x = 0 for every x X, so T x = 0 for every x X and so T = 0. 33

34 We have αt X Y = (αt )x αt x T x sup = sup = sup α = α T. 0 x X x 0 x X x 0 x X x We have (T 1 + T 2 )x = T 1 x + T 2 x T 1 x + T 2 x. Then ( T1 x T x + T ) 2x T 1 + T 2. x sup 0 x X Let Y be complete. We show completeness of BL(X, Y ). Let (T k ) k N BL(X, Y ) be a Cauchy sequence. For every x X and k, l N: T k x T l x T k T l x. ( ) This implies that (T k x) k N Y is a Cauchy sequence for every x X and since Y is complete, there is some limit This defines a map lim T kx =: T x Y. k T : X Y x T x := lim k T kx. T is linear, because let α, α K and x, x X. Then T (αx + α x ) = lim k T k(αx + α x ) = lim k (αt kx + α T x ) = αt x + α T x. We show T is bounded and that it is the norm limit of the T k s. According to ( ) we have that for every ε > 0 there exists an N N such that k, l N and x X: T k x T l x ε x. So applying the limit l gives T k x T x ε x and so T k T ε. This gives two things: 1. T is bounded because T k BL(X, Y ) and (T k T ) BL(X, Y ) and since BL(X, Y ) is a vector space: 2. T k k T in X Y. T = T k (T k T ) BL(X, Y ). 34

35 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. The dual space of X is X := BL(X, K) 6. Notations for the norm on X : X K =: X =: =: Corollary (from Theorem 2.34). (X, X ) is a Banach space (no matter whether X is complete) Examples. Let X = C([a, b]) (where a < b R) equipped with. (a) For f X let I(f) := b a dx f(x) This is clearly a linear functional I : X K with I(f) b a dx f(x) f f (b a) so I X (b a) and the function f 1 yields equalities above and so I X = b a. (b) For f X and t [a, b] let δ t (f) := f(t). This gives a linear functional δ t : X K which is called the Dirac-δ functional with δ t (f) = f(t) f and equality again by considering f 1. So δ t X = 1. NB. Pay attention that the boundedness of δ t is heavily dependent on the norm on X. For example it s not bounded, if X is equipped with the 1-norm 1 = 1 0 dx. Notation. Let X, Y be metric spaces. We write X = Y iff X is isometrically isomorphic to Y Theorem. Let p [1, [ and 1 p + 1 q = 1. Then (lp ) = l q. Proof. Case 1 < p < : Let x = (x n ) n N l p, f (l p ). We work with the canonical (Schauder) basis {e n } n N for l p (introduced in Example 2.15). Therefore x can be written as a p -convergent series x = n N x ne n with x n K. Since f is continuous and linear: f(x) = n N f(x n e n ) = n N x n f(e n ). ( ) For N N (fixed), define x := ( x n ) n N l p by f(e n) q f(e x n := n) if n N f(e n ) 0, 0 otherwise. 6 Note: 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