PMH3 - Functional Analysis. Andrew Tulloch

Transcription

1 PMH3 - Functional Analysis Andrew Tulloch

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3 Contents Chapter 1. Banach Spaces and Linear Operators 4 1. Banach Spaces 4 2. Linear Operators 6 Chapter 2. The Hahn-Banach theorem Zorn s Lemma 13 Chapter 3. An Introduction to Hilbert Spaces Hilbert Spaces Projections Orthonormal Systems The Stone-Weierstrass Theorem 27 Chapter 4. Uniform Boundedness and the Open Mapping Theorem The Principle of Uniform Boundedness The Open Mapping Theorem 39 Chapter 5. Spectral Theory 44 Chapter 6. Compact Operators 53 Chapter 7. The Hilbert Space Decomposition The Fredholm Alternative 68 3

4 CHAPTER 1 Banach Spaces and Linear Operators 1. Banach Spaces Definition 1.1 (Norm). Let X be a vector space. A norm on X is a function : X R satisfying x 0 with equality if and only if x = 0. αx = α x. x + y x + y for all x, y X. We call the pair (X, ) a normed vector space. Theorem 1.2 (Reverse triangle inequality). Let X be a normed vector space. For any x, y X, we have x y x y Definition 1.3 (Complete space). Let X be a normed vector space. Then X is complete if every Cauchy sequence in X converges to some x X. Definition 1.4 (Banach space). A Banach space is a complete normed vector space. Proposition 1.5 (Convergence). Let (V, ) be a normed vector space. A sequence (x n ) in V converges to x V if given ϵ > 0, there exists N such that x x n < ϵ whenever n < N. Lemma 1.6. If x n x, then x n x R. Proof. x n x x x n 0. Proposition 1.7. Every convergent sequence is Cauchy. Definition 1.8 (Banach space). A complete, normed, vector space is called a Banach space Proposition 1.9. (K, ) is complete. Proposition (l p, p ) is a Banach space for all 1 p Proof. A general proof outline follows. Use completeness of R to find a candidate for the limit. Show this limit function is in V. 4

5 1. BANACH SPACES 5 Show that x n x in V. Let x (n) be a Cauchy sequence in l p. Since x (n) j Cauchy sequence in K. Hence, lim n x (n) j x (n) j x (n) x (m), we know that x (n) j is a := x j exists, and is our limit candidate. We now show that j=1 x j p <. We have Proposition (l([a, b]), ) is a Banach space Proposition If 1 p <, then (l([a, b]), p ) is not a Banach space. Proof. Consider a sequence of functions that is equal to one on [0, 1 2 ], zero on [ n, 1], and linear between. This is a Cauchy sequence that does not converge to a continuous function. We ve seen that (l([a, b]), p ) is not complete for 1 p <. Theorem 1.13 (Completion). Let (V, ) be a normed vector space over K. There exists a Banach space (V 1, 1 ) such that (V, ) is isometrically isomorphic to a dense subspace of (V 1, 1 ). Furthermore, the space (V 1, 1 ) is unique up to isometric isomorphisms. Proof. Rather straightforward - construct Cauchy sequences, append limits, quotient out (as different sequences may converge to the same limit). Definition (V 1, 1 ) is called the completion of (V, ). Definition 1.15 (Dense). If X is a topological space and Y X, then Y is dense in X if the closure of Y in X equals X, that is, Y = X. Alternatively, for each x X, there exists (y n ) in Y such that y n x. Definition 1.16 (Isomorphism of vector spaces). Two normed vector spaces (X, X) and (Y, Y ) are isometrically isomorphic if there is a vector space isomorphism Ψ : X Y such that Ψ(x) Y = x X x X Example Let l 0 = {(x i ) #{i, x i 0} < }. The completion of l 0, p is (l p, p ), because, l 0 is a subspace of l p, It is dense, since we can easily construct a sequence in l 0 converging to arbitrary x l p. Example 1.18 ( L p spaces). Let µ be the Lebesgue measure on R. Let L p ([a, b]) = {measurable f : [a, b] K b a f p dµ < } ( b 1/p. Let f p = dµ) a f p Since f p = 0 f = 0 a.e, we quotient out by the rule f g f g = 0 a.e., and then our space of equivalence classes forms a normed vector space, denoted L p ([a, b]).

6 2. LINEAR OPERATORS 6 Theorem 1.19 (Riesz-Fischer). (L p ([a, b]), p ) is the completion of (C[a, b], p ), and is a Banach space. Proof. Properties of the Lebesgue integral. Remark. Let X be any compact topological space, let C(X) = {f : X K f is continuous}, and let f = sup x X f(x). Then C(X, ) is Banach. Let X be any topological space. Then the set of all continuous and bounded functions with the supremum norm forms a Banach space. Let (S, A, µ) be a measure space. Then we can define the L p and L p analogously, and they are also Banach. 2. Linear Operators Definition 2.1 (Linear operators on normed vector spaces). Let X, Y be vector spaces over K. A linear operator is a function T : X Y such that T (x + y) = T (x) + T (y) T (αx) = αt (x) for all x, y, α. We write Hom(X, Y ) = {T : X Y T is linear} Definition 2.2. T : X Y is continuous at x X if for all ϵ > 0, there exists δ > 0 such that x y X < δ T x T y y < ϵ Definition 2.3. L(X, Y ) = {T : X Y T is linear and continuous} Remark. If dim(x) < then Hom(X, Y ) = L(X, Y ). This is not true if X has infinite dimension. Definition 2.4 (Bounded linear operator). Let T : X Y be linear, then T is bounded if T maps bounded sets in X to bounded sets in Y. That is: for each M > 0 there exists M > 0 such that x X M T x Y M Consider the space L(X, Y ), the set of all linear and continuous maps between two normed vector spaces X and Y. Theorem 2.5 (Fundamental theorem of linear operators). Let (X, X ) and Y, Y be normed vector spaces. Let T Hom(X, Y ), the set of all linear maps from X to Y. Then the following are all equivalent.

7 2. LINEAR OPERATORS 7 1) T is uniformly continuous 2) T is continuous 3) T is continuous at 0 4) T is bounded 5) There exists a constant c > 0 such that T x Y c x X x X Proof. 1) 2) 3) is clear. 3) 4). Since T is continuous at 0, given ϵ = 1 > 0, there exists δ such that T x T 0 1 whenever X 0 δ, ( ) i.e. that x δ T x 1. Let y X. The δy y δ, and so T δy y < 1. Hence, and so δ T y 1 y T y y δ for all y X. Thus, for all y M, we have T y M, where M = M δ, and so T is bounded. 4) 5). If T is bonded, given M = 1 > 0, there exists c 0 such that x 1 T x c. Then Hence, T x c x. 5) 1). If 5) holds, then So if ϵ is given, taking δ = ϵ c, we have T ( ) x c x T x T y = T (x y) c x y. T x T y c x y < c ϵ c = ϵ. Corollary. If T Hom(X, Y ), then T continuous T bounded T x c x for all x X. Definition 2.6 (Operator norm). The operator norm of T L(x, y), T is defined by any one of the following equivalent expressions. (a) T = inf{c > 0 T x < c x }. (b) T = sup x 0 T x x. (c) T = sup x 1 T x. (d) T = sup x =1 T x. Proposition 2.7. The operator norm is a norm on L(x, y).

8 2. LINEAR OPERATORS 8 Proof. The following are simple to verify. (a) T 0, with equality if and only if T = 0. (b) αt = α T. (c) S + T S + T. Example 2.8 (Calculating T ). To calculate T, try the following. 1) Make sensible calculations to find c such that for all x X. 2) Find x X such that T x = c x. T x c x Remark. Ignore!2, Q3(b), Q8 on the practice sheet, as we will be ignoring Hilbert space theory for the time being. Definition 2.9 (Algebraic dual). Let (X, ) be a normed vector space over K. The algebraic dual of X is Elements of X are called linear functionals. X = Hom(X, K) = {φ : X K φ is linear}. Definition 2.10 (Continuous dual (just dual)). The continuous dual (just dual) of X is Remark. X X if dim(x) =. X = L(X, K) = {φ : X K φ is linear and continuous}. Example Let ( ([a, b]), ) be the normed vector space of polynomials p : [a, b] K. (a) The functional D : ([0, 1]) K given by D(p) = p (1) is linear, but not continuous. (b) The functional I : ([0, 1]) K given by I(p) = 1 p(t) dt is linear and continuous. 0 Proof. (a) Linearity is clear. The p n (t) = t n for all t [0, 1]. Then D(p n ) = n p n. So D is not continuous, as continuity implies that there exists c such that T x c x. (b) Exercise: Show I = 1. Describing the continuous dual space X is one of the first things to do when trying to understand a normed vector space. It is generally pretty difficult to describe X. Proposition 2.12 (Dual of the l p space for (1 < p < )). Let 1 < p <. Let q be the dual of p, defined by 1 q + 1 p = 1. Then (lp ) is isometrically isomorphic to l q.

9 2. LINEAR OPERATORS 9 Remark (Observation before proof). Let 1 p <. Let e i = (0, 0,..., 1, 0,... ) where 1 is in the i-th place. 1) If x = (x i ) l p, then x = x i e i i=1 in the sense that the partial sums converge to x. 2) If φ : l p K is linear and continuous, then φ(x) = x i φ(e i ) i=1 Proof of observations. Let S n = n i=1 x ie i. Then Write φ(x) as x S n p p = (0, 0,..., x n+1, x n+2,... ) p p = x i p 0 i=n+1 as it is the tail of a convergent sum. φ(x) = φ( lim n S n) = lim (φ(s n)) n ( n ) = lim φ x i e i n = lim = n i=1 i=1 n x i φ(e i ) x i φ(e i ) i=1 (continuity) (linearity) Proof. Define a map θ by θ : l q (l p ) y φ y where φ y (x) = x i y i for all x l p. (1) φ y is linear, as φ y (x + x ) = φ y (x) + φ y (x ) (valid as sums converge absolutely.) (2) φ y is continuous, as φ y (x) = x i y i x i y i x p y q

10 2. LINEAR OPERATORS 10 by Hölder s inequality. From the fundamental theorem of linear operators, as φ y (x) x p y q, we have that φ y is continuous, and that φ y y q ( ) (3) θ is linear. (4) θ is injective, as θ(y) = θ(y ) φ y = φ y φ y (x) = φ y (x) x l p φ y (e i ) = φ y (e i ) i N y i = y i i N y = y (5) θ is surjective. Let φ (l p ). Let y = (φ(e 1 ),..., φ(e n ),... ) = (y 1,..., y n,... ). We now show y l q. Then Let x (n) l q be defined by by Observation 2) above. x (n) i = On the other hand, we know φ(x (n) ) = y i q y i if i n and y i 0 0 otherwise i=1 x (n) i φ(e i ) = φ(x (n) ) φ x (n) p ( ) 1/p = φ x (n) i p i=1 ( n ) 1/p = φ y i (q 1)p i=1 ( n ) 1/p = φ y i q Now, using ( ) and ( ), we have ( n n ) 1/p y i q φ y i q and so we must have and so y l q. i=1 i=1 i=1 n y i q ( ) i=1 as 1/p + 1/q = 1. ( ) y q φ ( )

11 2. LINEAR OPERATORS 11 We also have, by ( ), y q φ y (6) Finally, we show that θ is an isometry. By ( ) and ( ), we have θ(y) = φ y = y q as required. How big is X? When is X {0}? Examples suggest that X is big with a rich structure.

12 CHAPTER 2 The Hahn-Banach theorem The Hahn-Banach theorem is a cornerstone of functional analysis. It is all about extending linear functionals defined on a subspace to linear functionals on the whole space, while preserving certain properties of the original functional. Definition 0.13 (Seminorm). A let X be a vector space over K. A seminorm on X is a function p : X R such that (1) p(x + y) p(x) + p(y) x, y X (2) p(λx) = λ p(x) x X, λ K Theorem 0.14 (General Hahn-Banach). Let X be a vector space over K. Let p : X R be a seminorm on X. Let Y X be a subspace of X. If f : Y K is a linear functional such that f(y) p(y) y Y then there is an extension f : X K such that f is linear f(y) = f(y) f(x) p(x) y Y x X Remark. This is great. Y can be finite dimensional (and we know about linear functionals on finite dimensional spaces) If p(x) = x, then f(x) x x X and so f X Corollary. Let (X, ) be a normed vector space over K. For each y X, with y 0, there is φ X such that φ(y) = y and φ = 1 Proof. Fix y 0 in X. Let Y = {Ky} = {λy λ K}, a one-dimensional subspace. Define f : Y K, f(λy) = λ y. This is linear. Set p(x) = x. Then f(λy) = p(λy) 12

13 1. ZORN S LEMMA 13 and so by Hahn-Banach, there exists f : X K such that f is linear f(λy) = f(λy) λ K f(x) x x X Then we have f X and f = 1 as required. 1. Zorn s Lemma Theorem 1.1 (Axiom of Choice is equivalent to Zorn s Lemma). See handout for proof that A.C. Z.L. Definition 1.2 (Partially ordered set). A partially ordered set (poset) is a set A with a relation such that (1) a a for all a A, (2) If a b and b athen a = b, (3) If a b and b c, then a c Definition 1.3 (Totally ordered set). A totally ordered set is a poset (A, ) such that if a, b A then either a b or b a. Definition 1.4 (Chain). A chain in a poset (A, ) is a totally ordered subset of A. Definition 1.5 (Upper bound). Let (A, ) be a poset. An upper bound for B A is an element u A such that b u for all b B. Definition 1.6 (Maximal element). A maximal element of a poset (A, ) is an element m A such that m x implies x = m, that is, m x x = m Example 1.7. Let S be any set. Let P(S) be the power set of S (the set of all subsets of S). Define a b a b. Maximal element is S Theorem 1.8 (Zorn s Lemma). Let (A, ) be a poset. Suppose that every chain in A has an upper bound. Then A has (at least one) maximal element. Example 1.9 (Application - all vector spaces have a basis). Definition 1.10 (Linearly independent). Let X be a vector space over F. We call B X linearly independent if λ 1 x λ n x n = 0 λ 1 = = λ n = 0 for all finite {x 1,..., x n } B.

14 1. ZORN S LEMMA 14 Definition 1.11 (Span). We say B X spans X if each x X can be written as x = λ 1 x λ n x n for some λ 1,..., λ n F and {x 1,..., x n } B. Definition 1.12 (Hamel basis). A Hamel basis is a linearly independent spanning set. Equivalently, B X is a Hamel basis if and only if each x X can be written in exactly one way as a finite linear combination of elements of B. Theorem Every vector space has a Hamel basis Proof. Let L = {linearly independent subsets}, with subset ordering. Let C be a chain in L. Let u = a C a. Then (1) u L, (2) u is an upper bound for C. So Zorn s Lemma says that L has a maximal element b. Then b is a Hamel basis. b is linearly independent. If Span(b) X, there exists X X\Span(b), and b = b {x} L is linearly independent, contradicting maximality of b. Remark. If X, ) is Banach, every Hamel basis is uncountable. Theorem 1.14 (Hahn-Banach theorem over R). Let X be a real linear space and let p(x) be a seminorm on X. Let M be a real linear subspace of X and f 0 a real-valued linear functional defined on M. Let f 0 satisfy f 0 (x) p(x) on M. Then there exists a real valued linear functional F defined on X such that (i) F is an extension of f 0, that is, F (x) = f 0 (x) for all x M, and (ii) F (x) p(x) on X. Proof. We first show that f 0 can be extended if M has codimension one. Let x 0 X\M and assume that span(m {x 0 }) = X. As x 0 / M be can write x X uniquely in the form x = m + αx 0 for α R. Then for every c R, the map f c Hom(X, R) given by f c (m + αx) = f 0 (m) + cα is well defined, and f c (m) = f 0 (m) for all m M. We now show that we can choose c R such that f c (x) p(x) for all x X. Equivalently, we must show f 0 (m) + cα p(m + αx 0 )

15 1. ZORN S LEMMA 15 for all m M and α R. By positive homogeneity of p and linearity of f we have f 0 (m/α) + c p(x 0 + m/α) α > 0 f 0 ( m/α) c p( x 0 m/α) α < 0 Hence we need to choose c such that c p(x 0 + m) f 0 (m) c p( x 0 + m) + f 0 (m). This is possible if p( x 0 + m 1 ) + f 0 (m 1 ) p(x 0 + m 2 ) f 0 (m 2 ) for all m 1, m 2 M. By subadditivity of p we can verify this condition since f 0 (m 1 + m 2 ) p(m 1 m 2 ) = p(m 1 x 0 + m 2 x 0 ) p(m 1 x 0 ) + p(m 2 + x 0 ) for all m 1, m 2 M. Hence c can be chosen as required. Hence D(F ) = X, and the theorem is proven. Theorem 1.15 (Hahn-Banach over C). Suppose that c is a seminorm on a complex vector space X and let M sub a subspace of X. If f 0 Hom(M, C) is such that f 0 (x) p(x) for all x M, then there exists an extension f Hom(X, C) such that f M = f 0 and f(x) p(x) for all x X. Proof. Split f 0 into real and imaginary parts f 0 (x) = g 0 (x) + ih 0 (x). By linearity of f 0 we have 0 = if 0 (x) f 0 (ix) = ig 0 (x) h 0 (x) g 0 (ix) ih 0 (ix) = (g 0 (ix) + h 0 (x)) + i(g 0 (x) h 0 (ix)) and so h 0 (x) = g 0 (ix). Therefore, f 0 (x) = g 0 (x) ig 0 (ix) for all x M. We now consider X as a vector space over R, X R. Now considering M R as a subspace of X R. GSince g 0 Hom(M R, R) and g 0 (x) f 0 (x) p(x) and so by the real Hahn- Banach, there exists g Hom(X R, R) such that g MR = g 0 and g(x) p(x) for all x X R. Now set F (x) = g(x) ig(ix) for all x X R. Then by showing f(ix) = if(x), we have that f is linear. We now show f(x) p(x). For a fixed x X choose λ C such that λf(x) = f(x). Then since f(x) R and by definition of f, we have f(x) = λf(x) = f(λx) = g(λx) p(λx) = λp(x) = p(x)

16 1. ZORN S LEMMA 16 as required.

17 CHAPTER 3 An Introduction to Hilbert Spaces 1. Hilbert Spaces Definition 1.1 (Inner product). Let X be a vector space over K. An inner product is a function, : X X K such that (1) x + y, z = x, z + y, z (2) αx, z = α x, z (3) x, y = y, x (4) x, x 0 with equality if and only if x = 0 We then have and x, y + z = x, y + x, z x, αz = α x, z Definition 1.2 (Inner product space). Let (X,, ) be an inner product space. Defining x = x, x turns X into a normed vector space. To prove the triangle inequality, we use the Cauchy-Swartz theorem. Theorem 1.3 (Cauchy-Schwarz). In an inner product space (X,, ), we have x, y x y x, y X Proof. 0 x λy, x λy = x, x x, λy λy, x + λy, λy = x 2 λ x, y λ y, x + λ 2 y 2 = x 2 2Re(λ y, x ) + λ 2 y 2 17

18 1. HILBERT SPACES 18 Set λ = x,y y 2. Then as required. Corollary. 0 x 2 2Re( = x 2 x, y 2 y 2 x, y 2 x, y 2 y 2 ) + y 2 x + y x + y Definition 1.4 (Hilbert space). If (X,, ) is complete with respect to then it is called a Hilbert space. Example 1.5. (a) l 2, where x, y = i=1 x iy i. Cauchy-Schwarz then says x i y i x i 2 y i 2 i=1 (b) L 2 ([a, b]), where f, g = b f(x)g(x) dx. a Cauchy-Swartz then says b a i=1 i=1 f(x)g(x) dx... Definition 1.6 (Orthogonality). Let (X,, ) be inner product spaces. Then x, y X are orthogonal if x, y = 0 where x, y 0. Theorem 1.7. Let x i,..., x n be pairwise orthogonal elements in (X,, ). Then n n x i 2 = x i 2 Theorem 1.8 (Parallelogram identity). In (X,, ) we have i=1 i=1 x + y 2 + x y 2 = 2( x 2 + y 2 ) ( ) for all x, y X. Remark. If (X, ) is a normed vector space which satisfies parallelogram identity then X is an inner product space with inner products defined by the polarisation equation ( 1 4 x + y 2 x y 2) K = R x, y = ( 1 x + y 2 x y 2 + i x + iy 2 i x iy 2) K = C 4

19 2. PROJECTIONS Projections Definition 2.1 (Projection). Let X be a vector space over K. A subset M of X is convex if for any x, y M, then tx + (1 t)y M t [0, 1] Theorem 2.2 (Projection). Let (H,,, ) be a Hilbert space. Let M H be closed and convex. Let x H. THen there exists a unique point m x M which is closest to x, i.e. x m x = inf x m = d m M Proof. For each k 1 choose m k M such that d 2 x m k 2 d k Each m k exists as d is defined as the infimum over all m. Then m k m l 2 = (m k x) (m k x) 2 = 2 m k x m l x 2 m k + m l 2x 2 2d l + 2d2 + 2 k 4 m k + m l 2 x 2 and as m k /2 + m l /2 M, we have m k+m l 2 x 2 d 2. Then m k m l 2 2( 1 k + 1 l ) Thus (m k ) is Cauchy. So m k m x M as H is complete and M is closed. We then have and so now we show that m x is unique. x m x = d Suppose that there exists m x M with x m x = d. Then by the above inequality, we have from above. m x m x 2 = 2 m x x m x x 2 4 m x m x 2 x 2 0 Definition 2.3 (Projection operator). Let (H,,, ) be a Hilbert space. Let M H be closed and convex. Define P M : H H by P M (x) = m x from above. This is the projection of H onto M.

20 2. PROJECTIONS 20 Definition 2.4 (Orthogonal decomposition). If S H, let S = {x H x, y = 0 y S. We call S the orthogonal component. Theorem 2.5 (From previous lecture). If M H, then the projection of H onto M is P m : H H x m x where m x M is the unique element with x m x = inf m M x m. Lemma 2.6. Let M H be closed subspace. Then x P M x M for all x H. Proof. Let m M. We need to show x P M x, m = 0. This is clear if m = 0. Without loss of generality, assuming m 0, we can assume m = 1. Then write x P M x = x (P M x + x P M x, m m) + x P M x, m m. Let the bracketed term be m. Then x m x P M x, m m because x m, x P M x, m m = x P M x, m x m, m = C x P M x x P M x, m m, m = C( x P M x, m x P M x, m m ) = 0. So x P M x 2 = x m 2 + x P M x, m 2. So x P M x 2 x P M x 2 + x P M x, m 2 by definition of P M x. Thus, x P M x, m = 0 and thus x P M x M. Theorem 2.7. The following theorem is the key fundamental result. Let (H,,, ) be a Hilbert space. Let M be a closed subspace of H. Then H = M M. That is, each x H can be written in exactly one way as x = m + m with m M, m M. Proof. Existence - Let x = P m x + (x P M x). Uniqueness - Let x = x 1 + x 1, x = x 2 + x 2 with x 1, x 2 M, x 1, x 2 M. Then x 1 x 2 = x 2 x 1 M Then x 1 x 2, x 1 x n = 0 x 1 = x 2.

21 2. PROJECTIONS 21 Thus x 1 = x 2. Corollary. Let M H be a closed subspace. Then we have (a) P M L(H, H). (b) P M 1. (c) ImP m = M, Ker P M = M. (d) PM 2 = P M. (e) P M = I P M. Proof. (c), (d), (e) exercises. (a). Let x, y H. Write x = x 1 + x 1 and y = y 1 + y1 with x 1, y 1 M and x 1, y1 M. Then x = y = (x 1 + y 1 ) + (x 1 + y1 ) and so P M (x + y) = x 1 + y 1 and similarly P M (αx) = αp M x. We also have x 2 = P M x + (x P M x) 2 = P M x 2 + x P M x 2 P M x 2 and so P M 1. If y H is fixed, then the map φ y : H K x x, y is in H. Linearity is clear, and continuity is proven by Cauchy-Swartz, φ y (x) = x, y y x. So φ y y. Since φ y (y) = y 2, we then have φ y = y. Theorem 2.8 (Riesz Representation Theorem). Let H be a Hilbert space. The map θ : H H y φ y is a conjugate linear bijection, and φ y = y.

22 2. PROJECTIONS 22 Proof. Conjugate linearity is clear. Injectivity φ y = φ y φ y (x) = φ y (x) x so x, y = x, y = 0 y y, y y = 0 and so y = y. Surjectivity Let φ H. We now find y H with φ = φ y. If φ = 0, take y = 0. Suppose φ 0. Then Ker φ H. But Ker φ is a closed subspace of H. So H = (Ker φ) (Ker φ). Hence (Ker φ) {0}. Pick z (Ker φ), z 0. For each x H, the element Note that φ(z) 0 since z / Ker φ. Then and so and so letting y = φ(z) z 2 z, we have φ = φ y. x φ(x) φ(z) z Ker φ 0 = x φ(x) z, z φ(z) = x, z φ(x φ(z) z 2 φ(x) = x, φ(z) z x H, z 2 Example 2.9. From Hahn-Banach given y H there exists φ H such that φ = 1 and φ(y) = y. We can be very constructive in the Hilbert case, and let φ(x) = x, y y Example All continuous linear functionals on L 2 ([a, b]) are of the form for some g L 2 ([a, b]). φ(f) = b a f(x)g(x) dx

23 3. ORTHONORMAL SYSTEMS 23 Example 2.11 (Adjoint operators). Let H 1, H 2 be Hilbert spaces. Let T L(H 1, H 2 ). The adjoint of T is T L(H 2, H 1 ) given by for all x H 1, y H 2 Exercise Check all of the above. T x, y 2 = x, T y 1 Exercise Prove T = T t where T t is the transpose. 3. Orthonormal Systems Definition 3.1 (Orthonormal system). As subset S H is an orthonormal system (orthonormal) if e, e = δ e,e e, e S Definition 3.2 (Complete orthonormal system or Hilbert basis). An orthonormal system S is complete or a Hilbert basis if span S = H Remark. By Gram-Schmidt and Zorn s Lemma, every Hilbert space has a complete orthonormal system. Example 3.3. (1) l 2. Then is orthonormal and is complete. S = {e i i 1} (2) L 2 C ([0, 2π]). Then S = { 1 2π eint n Z} is orthonormal and is complete. Completeness follows from Stone-Weierstrass theorem. (3) L 2 R ([0, 2π]). Then S = { 1 2π, 1 π cos nt, 1 π sin nt n 1} is orthonormal and is complete, again by Stone-Weierstrass. We want to look at series e S..., which is tricky if S is not countable. Lemma 3.4. If {e k k 0} is orthonormal, then a l e k converges in H if and only if k=0 a k 2 k=0

24 3. ORTHONORMAL SYSTEMS 24 converges in K. If either series converges, then Note. If x n x, y n y, then 2 a k e k = a k 2 k=0 k=0 x n, y n x, y Proof. If k=0 a ke k converges to x, then x, x = lim n n a k e k, a k e k n = lim k=0 n k=0 n a k 2 Conversely, if k=0 a k 2 converges, then writing x n = n k=0 a ke k, we have m x m x n 2 = a k e k 2 = = k=n+1 m k=n+1 m k=n+1 k=0 a k e k 2 by Pythagoras a k 2 0 and so (x n ) is Cauchy, and hence converges by completeness of H. Lemma 3.5. Let {e 1,..., e n } be orthonormal. Then n x, e k 2 x 2 for each x H. k=1 Proof. Let y = n k=1 x, e k e k. Let z = x y. We claim that z y. We have x, y = x y, y = x, y y 2 = n x, e k x, e k k=1 = 0. n x, e k 2 k=1

25 3. ORTHONORMAL SYSTEMS 25 So x 2 = y + z 2 = y 2 + z 2 Pythagoras n y 2 = x, e k 2 k=1 We want to write expressions like e S x, e e. Corollary. Let x H and S orthonormal. Then {e S x, e = 0} is countable. Proof. {e S x, e = 0} = {e S x, e > 1 k k 1 From the lemma, #{e S x, e > 1 k } k2 x 2 For if this number were greater than k 2 x 2, then the LHS in Lemma is greater than 1 k 2 k 2 x 2. Therefore: Corollary (Bessel s Inequality). If S is orthonormal, then x, e 2 x 2 e S for all x H Proof. e S x, e 2 is a sum of countably many positive terms, and so order is not important. We want to write e S x, e e. This sum is over a countable set, but is the order important? Theorem 3.6. Let S be orthonormal. Let M = span S. Then P M x = x, e e e S where the sum can be taken in any order. Proof. Fix x H. Choose an enumeration {e k k 0} = {e S x, e = 0}.

26 3. ORTHONORMAL SYSTEMS 26 By Bessel s inequality, we have x, e k 2 x 2 k=0 and so the LHS converges. By Lemma 3.4, we know y = x, e k e k M converges in H. k=0 Write x = y+(x y) = M +M. We claim (x y) M. Then P M x = y from characterisation of projection operator. Let e S. Then x y, e = lim n x n x, e k e k, e k=0 = lim ( x, e n x, e k e k, e ) n = x, e k= x, e k e k, e. k=0 If e {e S x, e 0}, then e = e j for some j, and so x y, e = x, e j x, e j = 0 If x, e = 0, then e e j for all j, and so e j, e = 0, and so Thus x y (span S). x y, e = 0 0 = 0. Exercise 3.7. Show that x y (span S) = M. Recall that if {x } is a countable orthonormal system in a Hilbert space H. Then a k e k < a k 2 < k=1 k=1 and a k e k 2 = a k 2 ( ) We also had the following. k=1 k=1

27 4. THE STONE-WEIERSTRASS THEOREM 27 Theorem 3.8. Let S be orthonormal in H. Let M = span S. Then P M x = e S x, e e x H where the sum has only countable many terms and convergence is unconditional. Theorem 3.9. Let S be orthonormal in H. Then following are equivalent. (a) S is a complete orthonormal system (span S = H). (b) x = e S x, e e for all x (Fourier series). (c) x 2 = e S x, e 2 for all x (Parseval s formula). Proof. (a) (b). If M = span S = H, then P M x = x = x, e e e S by Theorem 3.8. (b) (c). By the infinite Pythagoras theorem ( ). (c) (a). Let M = span S. Suppose that z M. Then z = 0 + z M + M. Hence 0 = P M z 2 = z, e e 2 = z, e 2 = z 2 e S e S which implies z = 0, so M = H, and so S is complete. Remark. Consider L 2 ([0, 2π]), and let S = {e n n Z}. Then we can write f = n Z c n e n where c n = f, e n = 1 2π 2π f(t)e int dt. 0 We do not claim that convergence is pointwise, what we have proven is convergence is in L 2, f c n e n 2 0 n N as N. This is not the same as pointwise or uniform convergence ( ). 4. The Stone-Weierstrass Theorem This is a useful tool to show an orthonormal system is complete. In fact, this theorem is about uniformly approximating elements of C(X), where X is a compact Hausdorff space. it is a generalisation of the Weierstrass approximation theorem. Theorem 4.1 (Weierstrass approximation theorem). Let f C([a, b]) and let ϵ > 0 be given. Then there exists a polynomial p(x) such that f(x) p(x) < x [a, b],

28 4. THE STONE-WEIERSTRASS THEOREM 28 that is, f p < ϵ. Corollary. This implies the following important results: Continuous functions can be uniformly approximated by polynomials. P([a, b]), the space of polynomials on [a, b], is dense in C([a, b]). P([a, b]) = C([a, b]). We now prove Stone s 1930 s generalisation. First some setup: Let X be a compact Hausdorff space throughout. We then know that C(X) is a vector space. It also has sensible vector multiplication, (fg)(x) = f(x)g(x). Thus C(X) is a unital, commutative, associative ring. As we have f(λg) = λ(fg) then C(X) is a unital, commutative, associative algebra over K. Definition 4.2 (Subalgebra). A subalgebra of C(X) is a subset A which is closed under scalar multiplication, vector addition, and vector multiplication. A is unital if it contains the constant function f(x) = 1. Example 4.3. P([a, b]) is a subalgebra of C([a, b]). When is A dense in C(X)? Theorem 4.4 (Stone-Weierstrass theorem). Let X be a compact Hausdorff space, and let A be a subalgebra of C(X). If (1) A is unital, (2) f A f A, where f (x) = f(x), (3) A separates points of X. Then A = C(X). Definition 4.5. A separates points of X if, given x y, there is a function f A with f(x) f(y). Corollary. (a) P([a, b]) is dense in C([a, b]), as f(x) = x separates points. (b) Trigonometric polynomials are dense in {f C([0, 2π]) f(0) = f(2π)}. (c) Trigonometric polynomials are dense in L 2 ([0, 2π]), and S = {e n n Z} is complete.

29 4. THE STONE-WEIERSTRASS THEOREM 29 Setup Lemma 4.6. The function f(t) = t can be uniformly approximated by polynomials on [ 1, 1] Proof. The binomial theorem says We then have (1 + x) 1/2 = n=0 t = t 2 = 1 + (t 2 1) = ( 1 ) 2 x n x [ 1, 1] n n=0 ( 1 ) 2 (t 2 1) n t [ 2, 2] n Now let p N (t) = sumn=0( N 1 ) 2 n (t 2 1) n, and ( 1 ) t p N (t) = 2 (t 2 1) n n n=n+1 and so t p n 0 as N on [ 1, 1]. n=n+1 1 ) ( 2 n Theorem 4.7 (Stone-Weierstrass theorem). Let X be a compact Hausdorff space, and let A be a subalgebra of C(X). If (1) A is unital, (2) f A f A, where f (x) = f(x), (3) A separates points of X. Then A = C(X). Proof. We first prove for C R (X). Lemma 4.8. Let A be a unital subalgebra of C R (X). Then (a) f A, (b) min(f 1,..., f n ), max(f 1,..., f n ) A for all f, f 1,..., f n A C R (X). Proof. (a) Replace f by f f so we can assume that f = 1. From the previous lemma, we know for each n 1 there is a polynomial p n : [ 1, 1] R such that t p n (t) < 1 n for all t [ 1, 1]. Since f(x) f = 1 for all x X, we have f p n (f) 1 n But p n (f) is a finite linear combination of 1, f, f 2, f 3,... and so in in A, as A is unital. Thus f A.

30 4. THE STONE-WEIERSTRASS THEOREM 30 (b) Use the formulas and induction. max(f, g) = f + g f g f + g f g, min(f, g) = A 2 2 Proof of Stone-Weierstrass for C R (X). Let f C R (X) and let ϵ > 0 be given. We need to find g A such that f(z) g(z) < ϵ z X Step 0. We can assume that A is closed. Exercise 4.9. Why? Step 1. Let x, y X be fixed. Proposition There exists f xy A with f xy (x) = f(x), f xy (x) = f(y) Proof. If x = y then trivial (take f xy (z) = f(x)1(z)). If x y, since A separates points, there is h A with h(x) h(y). Then take f xy = ah + b1 A we can invert the coefficient matrix to find our coefficients a and b. Step 2. Let x X be fixed. Proposition There exists f x A such that f x (x) = f(x). f x (z) < f(z) + ϵ Proof. For each y X,let O y = {z X f xy (z) < f(z) + ϵ} where f xy is the function from Step 1. These are all open sets (why?) and thus X = y X since y O y. By compactness of X, we have m X = Letting f x = min(f xy1,..., f xyn ). Then i=1 O y O yi

31 4. THE STONE-WEIERSTRASS THEOREM 31 Since f xyi (x) = f(x) for all i, f x (x) = f(x) If z X, then z O yi for some i, and so f x (z) f xyi (z) < f(z) + ϵ as required. Step 3. Proposition There exists a function g A such that f(z) g(z) < ϵ for all z X. Proof. For each x X, let U x = {z X f x (z) > f(x) ϵ} where f x is from Step 2. These sets U i are open and since x U x, for an open cover, we can write X = n U x = U xj. x X j=1 Define g = max(f x1,..., f xn ). If z X, g(z) = f xi (z) for some i, which is less than f(z) + ϵ from Step 2. If z U xj for some j = 1,..., n, then g(z) f xj (z) > f(x) ϵ. Exercise Where did we use the Hausdorff property? We now prove for C C (X). Let A R = {f A f is real valued}. THen A R is an R-subalgebra of C R (X). It is unital, as 1 A and it is real valued. We now show A R separates points. If x y, there is f A such that f(x) f(y). Write f = u + iv with u, v real valued. Either u(x) u(y) or v(x) v(y), and so A R separates points. Hence A R is dense in C R.

32 4. THE STONE-WEIERSTRASS THEOREM 32 Now, let f C C (X). Then write f = u + iv. Then u, v C R (X). Then given ϵ > 0, there exists u 1, v 1 A R such that u u 1 ϵ 2, v v 1 ϵ 2 Writing f 1 = u 1 + iv 1 A, we have f f 1 (u u 1 ) + i(v v 1 ) u u 1 + v v 1 < ϵ and thus A is dense in C C (X). Corollary. Polynomials are dense in C([a, b]). Proof. A = P([a, b]) is an algebra, is unital, is closed under complex conjugation, and separates points. Thus, A is dense in C([a, b]). Definition 4.14 (Trigonometric polynomials). A trigonometric polynomail is an expression c n e int n Z with finitely many c n 0. So these are polynomials in s = e it and s 1 = s = e it. Corollary. The space A of all trigonometric polynomials is dense in C(Π), where Π = {z C z = 1} Proof. A is a sub-algebra of C(Π), it is unital, closed under complex conjugation, c n e int = c n e int n Z n Z and separates points. T is a compact Hausdorff space, and thus Stone-Weierstrass states that A is dense in C(Π). Corollary. The orthonomal system S = { 1 2π e int n Z} is complete in L 2 ([0, 2π]). Proof. span S = A is a space of trigonometric polynomials, which is dense in C(Π). Define Φ : C p ([0, 2π])] C(Π) f f where C p ([0, 2π]) = {f C([0, 2π]) f(0 = f(2π))}. Then Φ is an isometric isomorphism, and therefore functions of the form f(t) = c n e int is dense in C p ([0, 2π]).

33 4. THE STONE-WEIERSTRASS THEOREM 33 By the construction of the Lebesgue integral, simple functions n are dense in L 2 ([0, 2π]). i=1 a i 1 Ai Exercise Given f L 2 ([0, 2π]) and ϵ > 0, there exists g C p ([0, 2π]) such that f g 2 < ϵ. Thus A is dense in L 2 ([0, 2π]). Corollary. The following are separable (have a countable dense subset): (a) C([a, b]), (b) L p ([a, b]), 1 p < Proof. (a) We have P([a, b]) is dense in C([a, b]) and set P Q ([a, b]) with rational coefficients is dense in P([a, b]). Clearly, P Q ([a, b]) is countable, and thus is dense in C([a, b]). (b) Use the fact that C([a, b]) is dense in L p ([a, b]). Corollary. Let X be a compact metric space. Then C(X) is separable. Proof. As X is a compact metric space, then X is separable. Exercise Why? by Let {x n n 1} be a countable dense subset of X. For each n 1 and m 1 define f n,m (x) = f n,m : X K inf d(x, z) z / B(x n, 1 m ) We then claim f n,m is continuous. Now, let A be the space of all K-linear combinations of f k 1 n 1,m 1,..., f k l n l,m l, k 1,..., k l N. ( ) This is a sub-algebra of C(X), as A is unital, closed under conjugation, and separates points - if z 1, z 2 X with z 1 z 2, Choose n, m such that z 1 B(x n, 1 m ), z n / B(x n, 1 m ). Thus the sub-algebra A is dense by Stone-Weierstrass. The subset of Q-linear combinations of ( ) is countable and dense. Lemma If X is compact metric space then X is separable. Proof. For each m 1, X = B(x; 1 m ) x X

34 4. THE STONE-WEIERSTRASS THEOREM 34 has a finite subcover N m X = n=1 B(x m,n 1 m ) and thus the subset of all {x m,n } is a countably dense subset. Corollary. pi 2 6 = 1 n 2. Proof. S = { 1 2π e int n Z} is complete, and so Parseval s formula holds, f 2 2 = n Z n=1 f, e n 2. Apply to f(x) = x. A common strategy is to prove for polynomials, and then Stone-Weierstrass proves it for continuous functions. Corollary. If f C([a, b] [c, d]) then b d a c f(x, y) dydx = d b c a f(x, y) dxdy Proof. By direct calculation, the result is true for two-variable polynomials. Let f C([a, b] [c, d]) and ϵ > 0 be given. By Stone-Weierstrass, the space of polynomials in 2 variables is dense in C([a, b] [c, d]) and so there exists a polynomial p(x, y) with f(x, y) p(x, y) < The result then follows by direct calculation. ϵ (b a)(d c).

35 CHAPTER 4 Uniform Boundedness and the Open Mapping Theorem The following is at the core of two of the cornerstone theorems of functional analysis - the uniform boundedness principle and the open mapping theorem. 1. The Principle of Uniform Boundedness Theorem 1.1 (Baire s theorem). Let X be a complete metric space. If U 1, U 2,... are open dense subsets of X, then U = is dense in X. Proof. Let x X and ϵ > 0 be given. We need to show that B(x, ϵ) U. Lemma 1.2. There exists sequences (x n ) in X and (ϵ n ) in R + with the property that (a) x 1 = x, ϵ 1 = ϵ. (b) ϵ n 0 (c) B(x n+1, ϵ n+1 ) B(x n, ϵ n ) U n for all n 1. Proof. Let x 1,..., x n and ϵ 1,..., ϵ n be chosen. By density of U n, B(x n, ϵ n ) U n. n=1 U n Choose x n+1 B(x n, ϵ n ) U n. Choose ϵ n+1 > 0 such that B(x n+1, ϵ n+1) B(x n, ϵ n ) U n (openness). We have ϵ n+1 ϵ n. Choose 0 ϵ n+1 min( ϵ n+1 2, 1 n+1 ), then we have B(x n+1, ϵ n+1 ) B(x n+1, ϵ n+1) B(x n, ϵ n ) U n and ϵ n+1 < ϵ n with ϵ n+1 < 1 n+1. Given the lemma, the theorem follows. If m n, then by (c), B ( x m, ϵ m ) B(x n, ϵ n ) U n ( ) 35

36 1. THE PRINCIPLE OF UNIFORM BOUNDEDNESS 36 In particular, x m B(x n, ϵ n ). Thus, d(x n, x m ) < ϵ n for all m n. Thus (x n ) is Cauchy, and so x n ζ in X by completeness. By ( ), we then have d(x n, ζ) ϵ n for all n 1. So ζ B(x n, ϵ n ). So by (c), ζ B(x n+1, ϵ n+1 ) B(x n, ϵ n ) U n. Thus ζ B(x, ϵ) and thus ζ U = n=1 U n. The following corollary is often used Corollary. Let X be a complete metric space. Int(C n ) for some n. If C 1, C 2,... are closed with X = n=1 then Proof. If Int(C n ) = for all n then U n = X\C n are open and dense. So by Baire s theorem, n=1u n is sense, and in particular, n = 1 U n. We have X = C n = (X\U n ) n=1 n=1 = X\( U n ) X, n=1 a contradiction. There are three cornerstone theorems. Hahn-Banach, Uniform Boundedness, Open Mapping. Theorem 1.3 (Uniform boundedness). Let X, Y be Banach spaces. Let T α, α A, a family of continuous linear operators T α : X Y. Then if sup T α x < α A for each fixed x X, then sup T α < α A Remark. Rather amazing - you get a global bound from pointwise bounds. Proof. For each n 1, let X n = {x X T α x n α A} These are closed (T α is continuous) and X = n=1 X n

37 1. THE PRINCIPLE OF UNIFORM BOUNDEDNESS 37 by the hypothesis. By the corollary to Baire s theorem, we know there exists n 0 1 with Int(X n0 ). Choose x 0 Int(X n0 ), and let r > 0 such that B(x 0, r) Int(X n0 ). If z 1 then x 0 + rz B(x 0, r). So x 0 + rz X n0, and T α (x 0 + rz) n 0 α A, but a b a + b, so T α (rz) T α (x 0 ) T α (x 0 + rz) n 0. So r T α z n 0 + n 0, and For a general x X, T α z 2n 0 r z 1, α A T α x = T α ( x x ) x 2n 0 r x and thus T α 2n0 r, which implies sup T α < α A Recall, the Fourier series of f L 2 ([ π, π]) is f, e k e k k Z where e k (t) = eikt 2π. This converges to f in the L 2 norm. Exercise 1.4. If f is 2π-periodic and continuous, does the Fourier series converge pointwise? There are explicit (complicated) examples, but the easiest existence is using the uniform boundedness principle. Proposition 1.5. There is a 2π periodic continuous function whose Fourier series does not converge at 0. Proof. Let C p ([ π, π]) = {f C([ π, π]) f( π) = f(π)}. This is a Banach space with. If f C p, let f n = k n f, e k e k. Remark. We can now define, for each n 1, a linear operator T n : C p K by T n (f) = f n (0).

38 1. THE PRINCIPLE OF UNIFORM BOUNDEDNESS 38 If f n (0) converges (as n ) for each f C p, then sup n 1 T n f = sup f n (0) < n 1 for all f C p, which by uniform boundedness implies We now show that ( ) is false. sup T n. n 1 ( ) We have f n (x) = k n = 1 π 2π π 1 π f(t)e ikt dte ikx 2π π f(t) k n e ik(x t) = 1 π f(t)d n (x t) dt 2π π where D n (t) = k n eikt is the Dirichlet Kernel. The Dirichlet kernel is real, and even, with D n (t) = sin(n )t sin t. 2 Note. T n is continuous, with norm T n = 1 π 2π π D n(t) dt. Proof. and so T n 1 π 2π π D n(t) dt. Going the other way, let π T n (f) 1 2π ( 1 2π π π π f(t) D n (t) dt dt ) D n (t) dt f 1 D n (t) 0 s t =. 1 D n (t) < 0 We have seen hat set functions can be approximated in L 1 -norm by continuous (periodic) functions. So if ϵ > 0 is given, there is a g C p such that 1 π (g(t) s(t))d n (t) dt 2π < ϵ. g can be chosen with g = 1. π

39 2. THE OPEN MAPPING THEOREM 39 So T n (g) 1 π D n (t) dt 2π < ϵ. π Thus 1 π D n (t) dt T n (g) < ϵ. 2π π So T n (g) g π D n (t) dt ϵ. 2π Since ϵ > 0 was arbitrary, π T n 1 π D n (t) dt. 2π π All that remains is to show that We have as n. T n = 1 2π T n = 1 π = 1 π 2 π = 2 π 2 π = 2 π π π 0 π 0 π π D n (t) dt D n (t) dt sin(n )t sin t 2 dt sin(n )t dt t 0 (n+ 1 2 )π 0 nπ n k=1 0 n k=1 1 kπ = 4 π 2 n k=1 sin v v sin v dv v kπ (k 1)π kπ (k 1)π 1 k dv sin v v dv sin v dv Thus there exists f C p such that the Fourier series of f diverges at x = The Open Mapping Theorem This theorem is tailor-made to deal with inverse operators.

40 2. THE OPEN MAPPING THEOREM 40 Definition 2.1 (Open mapping). Let X, Y be metric spaces. A function f : X Y is open if open sets in X are mapped to open sets in Y. Theorem 2.2 (Open mapping theorem). Let X, Y be Banach spaces. If T L(X, Y ) is surjective then T is open. Corollary (Bounded inverse theorem). Let X, Y be Banach spaces. If T L(X, Y ) is bijective, then T 1 L(Y, X). Proof. Let O X be open. theorem). Thus T 1 is continuous. Then (T 1 ) 1 (O) = T (O) is open (by the open mapping Corollary. Let (X, 1 ) and (X, 2 ) be Banach spaces. If x 1 C x 2 x X then 1 and 2 are equivalent. Proof. i : (X, 2 ) (X, 1 ) x x is linear, surjective and injective, and also continuous, as So the bounded inverse theorem gives i(x) = x 1 C x 2. i 1 : (X, 1 ) (X, 2 ) is continuous. Thus there exists A > 0 such that i 1 (x) 2 A x 1, which implies x 2 A x 1. So 1 A x 2 x 1 x X More generally, if T L(X, Y ) is bijective, then by the bounded inverse theorem, where c = 1 T 1, C = T. c x T x C x Lemma 2.3. Let X be a Banach space and Y a normed space. Then for T L(X, Y ), the following are equivalent. (a) T is open (b) There exists r > 0 such that B(0, r) T (B(0, 1))

41 2. THE OPEN MAPPING THEOREM 41 (c) There exists r > 0 such that B(0, r) T (B(0, 1)). Proof. (a) (b), (c). As B(0, 1) is open, the set T (B(0, 1)) is open in Y. Since 0 T (B(0, 1)) there exists > 0 such that the set B(0, r) T (B(0, 1)) T (B(0, 1)) T (B(0, 1)). (c) (b). Assume that there exists r > 0 such that B(0, r) T (B(0, 1)). We now show that B(0, r 2 ) T (B(0, 1)) which proves (b). Let y B(0, r 2 ). Then 2y B(0, r) and since B(0, r) T (B(0, 1)) there exists x 1 B(0, 1) such that 2y T x 1 r 2 Hence 4y 2T x 1 B(0, r) and by the same argument as before there exists x 2 B(0, 1) such that 4y 2T x 1 T x 2 r 2 Continuing this way we construct a sequence (x n ) B(0, 1) such that for all n. Dividing by 2 n we obtain 2 n y 2 n 1 T x 1 2T x n 1 T x n r 2 y n k=1 2 k T x k r 2 n+1 Hence y = k=1 2 k T x k. Since x k 1 for all k N we have that 2 k x k 2 k = 1 and so the series k=1 x = k=1 2 k x k k=1 converges absolutely in X as X is Banach and hence complete. We have also that x 1 and so x B(0, 1). Because T is continuous we have T x = lim n k=1 n 2 k T x k = y by construction of x. Hence y T (B(0, 1)) and (b) follows. (b) (a). By (b) and the linearity of T we have T (B(0, ϵ)) = ϵt (B(0, 1))

42 2. THE OPEN MAPPING THEOREM 42 for all ϵ > 0. Since the map x ϵx is a homeomorphism on Y the set T (B(0, ϵ)) is a neighbourhood of zero for all ϵ > 0. Now let U X be open and y T (U). As U is open there exists ϵ > 0 such that B(x, ϵ) = x + B(0, ϵ) U where y = T x. Since z x + z is a homeomorphism and T is linear we have T (B(x, ϵ)) = T x + T (B(0, ϵ)) = y + T (B(0, ϵ)) T (U). Hence T (B(x, ϵ)) is a neighbourhood of y in T (U). As y was arbitrary in T (U) it follows that T (U) is open. Lemma 2.4. Let X be a normed vector space and S X convex with S = S. non-empty interior, then S is a neighbourhood of zero. If S has a Proof. First note that S is convex. If x, y S and x n, y n S with x n, y n x, y then tx n + (1 ty n ) S for all n and t [0, 1]. Letting n we get tx + (1 t)y S for all t [0, 1] and so S is convex. We also easily have S = S. If S has a non-empty interior, there exists z S and ϵ > 0 such that B(z, ϵ) S. Therefore z ± h S whenever h < ϵ and since S = S we also have (z ± h) S. By the convexity of S we have y = 1 ((x + h) + ( x + h)) S 2 whenever h < ϵ. Hence B(0, ϵ) S, and so S is a neighbourhood of zero. Theorem 2.5 (Open mapping theorem). Suppose that X and Y are Banach spaces. If T L(X, Y ) is surjective, then T is open. Proof. As T is surjective we have Y = T (B(0, n)) n N with [T (B(0, n))] closed for all n N. Since Y is complete, by a corollary to Baire s theorem, there exists n N such that T (B(0, n)) has non-empty interior. Since the map x nx is a homeomorphism and T is linear, the set T (B(0, 1)) has non-empty interior as well. Now B(0, 1) is convex and B(0, 1) = B(0, 1). By linearity of T we have that T (B(0, 1)) = T (B(0, 1)) is convex as well. Since we know that T (B(0, 1)) has non-empty interior, the previous lemma implies that T (B(0, 1)) is a neighbourhood of zero, and thus there exists r > 0 such that B(0, r) T (B(0, 1)) and since X is Banach the previous lemma shows that T is open.

43 2. THE OPEN MAPPING THEOREM 43 Exercise 2.6. If X, Y are vector spaces, and if T : X Y is linear, then Γ(T ) is a subspace of X Y. Moreover, if X, Y are normed vectors paces, with (x, T x) Γ = x + T x. Theorem 2.7 (Closed Graph theorem). Let X, Y be Banach spaces, and T Hom(X, Y ). Then T L(X, Y ) if and only if Γ(T ) is closed in X Y. Proof. Suppose T L(X, Y ). If x n x in X, then (x n, T x n ) (x, T x) by continuity of T, and so Γ(T ) is closed. Conversely, suppose that Γ(T ) is closed in X Y. Define a norm Γ on X by x Γ = x + T x. Since Γ(T ) is closed, and since (X, ) is Banach, then (X, Γ ) is also a Banach space (exercise). Note that x x Γ. So by a corollary to the Open Mapping theorem, and Γ are equivalent. So there is c > 0 with x Γ c x x X. So x + T x c x, and so T x (c 1) x, and so T is continuous.

44 CHAPTER 5 Spectral Theory The eigenvalues of an n n matrix T over C are the λ C with det(λi T ) = 0 that is, λi T is not invertible. Remark. Showing existence of eigenvalues is equivalent to the fundamental theorem of algebra. Remark. We need our base field to be C to get reasonable spectral theory. Definition 0.8. Write L(X) = L(X, X). Definition 0.9. Let X be a Banach space over K, and let T L(X). Then the spectrum of T is σ(t ) = {λ K λi T is not invertible}. Remark. λi T is non invertible if either λi T is not injective, or λi T is not surjective. Remark. If dim(x) <, then X\Ker (T ) im(t ), and so T is injective if and only if T is surjective. This fails in the infinite dimensional case - consider the left and right shift operators on l 2. Definition 0.10 (Eigenvalue). λ K is an eigenvalue of T L(X) if there is x 0 with T x = λx, i.e. λ is an eigenvalue if and only if λi T is not injective. Theorem Let X {0} be a Banach space over C, and let T L(X). non-empty, compact (closed and bounded) subset of Then σ(t ) is a {λ C λ T } Example Let L, R : l 2 l 2 be the left and right shift operators. Then L = 1, and so σ(l) D(0, 1). If λ < 1, then L(λ, λ 2, λ 3,... ) = (λ 2, λ 3, λ 4,... ) = λ(λ, λ 2, λ 3,... ) and so λ is an eigenvalue. Thus D(0, 1) σ(l) D(0, 1). But σ(l) is closed, and so σ(l) = D(0, 1). Are the λ with λ = 1 eigenvalues? No - suppose λ = 1 and x 0 with Lx = λx. 44

45 5. SPECTRAL THEORY 45 Then L n (x) = λ n x. Thus, x n+1 = λ n x 1. Then x = (x 1, λ, x 1, λ 2 x 1,... ) which is not in l 2. Then R = 1, and so σ(r) D(0, 1). Note. LRx = L(0, x 1,... ) = (x 1, x 2,... ), so LR = I ( ) Remark. Unlike dim(x) <, ( ) does NOT say that R is invertible (RL = I). Consider the operator L(λI R) = λl I = λ(λ 1 I L). If 0 < λ < 1, then we know that λ 1 I L is invertible (as λ 1 / σ(l)). So if λi R were invertible, then L is invertible, which is false. Thus λ σ(r). Hence Since σ(r) is closed, σ(r) = D(0, 1). D(0, 1)\{0} σ(r) D(0, 1). Theorem Let X {0} be a Banach space over C. Let T L(X). Then σ(t ) is a nonempty, compact subset of {λ C λ T }. Lemma With above assumptions σ(t ) {λ C λ T }. Proof. We need to show that if λ > T then λi T is invertible. Technique: Geometric series. We guess We now verify this guess. Since (λi T ) 1 = k=0 1 λi T = T k λ k+1. k=0 T k λ k+1 T k <, λ k+1 k=0 the series S = k=0 T k converges in X. λ k+1 We now show that S is the inverse of λi T. As we are working in infinite dimensions, we ned to check left and right inverses. Let S n = n 1 k=1 T k S n (λi T ) =. Then λ k+1 ) T k λ k+1 ( n 1 k=0 = I T n λ n I (λi T )S n = I T n λ n I (λi T )

46 5. SPECTRAL THEORY 46 and so S(λI T ) = (λi T )S and so λi T is invertible. Exercise Show that if I T < 1 then T is invertible with inverse k=0 (I T )k Hint: Consider 1 T = 1 I (I T ). In particular, the ball B(I, 1) in L(X) consists of invertible elements. right. The following is used to show σ(t ) is closed and nonempty, it is also interesting in its own Proposition Let X be Banach over K. Let GL(X) = {T L(X) T invertible. Then (a) GL(X) is a group under composition of operators. (b) GL(X) is open in L(X). (c) The map φ : GL(X) GL(X) T T 1 is continuous. Proof. (a) The open mapping theorem tells us that if T GL(X) then T 1 L(X), and so T 1 GL(X). The rest is clear. (b) Let T 0 GL(X). We claim We have ( B T 0, 1 T 1 0 ) GL(X). (c) We have I T 1 0 T = T 1 0 (T 0 T ) T0 1 T 0 T < 1 as T B ( T 0, ) 1 T0 1 T 1 0 T 1 = T 1 (T T 0 )T 1 0 T 1 T T 0 T 1 0 ( )

47 5. SPECTRAL THEORY 47 1 If T T 0 2 T 1 0, then I T T0 1 = (T 0 T )T0 1 We then have T 0 T T T 0 T 1 = (T T0 1 ) 1 = (I T T0 1 ) k 2 k=0 k=0 I T T 1 0 k Hence T 1 = T 1 0 (T 0 T 1 ) T 1 0 T 0 T 1 2 T 1 0, and from ( ), we have T 1 0 T 1 2 T T T 0 and so T T 1 is continuous. Corollary. σ(t ) is closed. Proof. Let f : C L(X) λ λi T This is continuous, as f(λ) f(λ 0 ) = (λ λ 0 )I = λ λ 0 and σ(t ) = f 1 (L(X)\GL(X)) which is the inverse image of a closed set, and hence is closed. So σ(t ) is a compact subset of {λ C λ T }. Write ρ(t ) = C\σ(T ) (the resolvent set), and let R T = R : ρ(t ) L(X) with R T (λ) = (λi T ) 1. Theorem Let K = C and X {0} and T L(X). Then σ(t ).

48 5. SPECTRAL THEORY 48 Proof. We use Lioville s theorem - a bounded entire function must be constant. Let φ = L(X) (hence φ : L(X) C.) Let Lemma f φ is analytic on ρ(t ). f φ : ρ(t ) C λ φ(r(λ)) Proof. We show f φ is differentiable. Consider ( ) f φ (λ) f φ (λ 0 ) R(λ) R(λ0 ) = φ λ λ 0 λ λ 0 ( (λi T ) 1 (λ 0 I T ) 1 ) = φ λ λ 0 ( (λ0 I T ) 1 ((λ 0 λ)i)(λi T ) 1 ) = φ λ λ 0 = φ ( (λ 0 I T ) 1 (λi T ) 1) φ ( (λ 0 I T ) 2) as λ λ 0, where we use the fact that φ is continuous and T T 1 is continuous. So f φ is analytic on ρ(t ) for all φ L(X). Now suppose that σ(t ) =. Then f φ : C C is analytic. Lemma f φ is bounded. Proof. If λ > T, then f φ (λ) = φ ( (λi T ) 1) ( ) = φ T k λ k+1 k=0 T k φ φ = λ k+1 k=0 k=0 T k λ k+1 φ λ T 0 as λ. So f φ is bounded, entire, and thus f φ = c by Lioville s theorem. By the above, f φ (λ) = 0 for all λ. Hence φ(r(λ)) = 0 for all λ, φ. Thus from Hahn-Banach, R(λ) = 0 for all λ which is a contradiction, as the zero operator is not invertible if X {0}.