Lecture Notes in Functional Analysis

Lecture Notes in Functional Analysis Please report any mistakes, misprints or comments to aulikowska@mimuw.edu.pl LECTURE 1 Banach spaces 1.1. Introducton to Banach Spaces Definition 1.1. Let X be a K vector space. A functional p X [0, + ) is called a seminorm, if (a) p(λx) = λ p(x), λ K, x X, (b) p(x + y) p(x) + p(y), x, y X. Definition 1.2. Let p be a seminorm such that p(x) = 0 x = 0. Then, p is a norm (denoted by ). Definition 1.3. A pair (X, ) is called a normed linear space. Lemma 1.4. Each normed space (X, ) is a metric space (X, d) with a metric given by d(x, y) = x y. Definition 1.5. A sequence {x n } n N in a normed space (X, ) is called a Cauchy sequence, if ε > 0 N(ε) N n, m N(ε) x n x m ε. Definition 1.6. A sequence {x n } n N converges to x (which is denoted by lim n + x n = x), if ε > 0 N(ε) N n N(ε) x n x < ε. Definition 1.7. If every Cauchy sequence {x n } n N converges in X, then (X, ) is called a complete space. Definition 1.8. A normed linear space (X, ) which is complete is called a Banach space. 1

2 LECTURE NOTES, FUNCTIONAL ANALYSIS Lemma 1.9. Let (X, ) be a Banach space and U be a closed linear subspace of X. Then, (U, ) is a Banach space as well. 1.2. Examples of Banach spaces Example 1. Let B(T ) be a space consisting of all bounded maps x T K. For each x B(T ) we set x = sup x(t). t T Then, (B(T ), ) is a Banach space. To prove the assertion we need to show that (a) is a norm, (b) each Cauchy sequence converges to an element from B(T ). Concerning claim (a), let λ K and x B(T ). Then, (1.10) λx = sup t T Let t o T and x, y B(T ). Then, x(t o ) + y(t o ) x(t o ) + y(t o ) sup t T λ x(t) = λ sup x(t) = λ x. t T x(t) + sup y(t) = x + y. t T The right hand side of the inequality above is independent on t. Therefore, taking supremum of both sides of the inequality over t T yields (1.11) x + y x + y. Finally, let x be such that x = 0, which is equivalent to sup t T x(t) = 0. This implies that x(t) = 0 for each t. Thus, x = 0. Now, we shall prove the statement (b). Let {x n } be a Cauchy sequence. Then, for every ε > 0 there exists N = N(ε) such that for all n, m N it holds that x n x m < ε. In particular, x n (t) x m (t) < ε, t T. Thus, for any t T the sequence {x n (t)} n N converges to some x(t), due to the completeness of K (real and complex numbers are complete spaces). Define a candidate for a limit of the sequence {x n } n N, that is, x T K as x(t) = lim n + x n(t). If follows from the statement above that there exists N o = N o (ε, t) such that (1.12) x n (t) x(t) < ε, n N o. Without loss of generality we can assume that N o (ε, t) N(ε) for each t T. Then, for n N it holds that x n (t) x(t) x n (t) x No(ε,t)(t) + x No(ε,t)(t) x(t) x n x No(ε,t) + ε < 2ε,

LECTURE 1. BANACH SPACES 3 where we used the fact that {x n } n N is a Cauchy sequence and (1.12). Moreover, for each t T and N = N(ε) we have x(t) x N (t) + x N (t) x(t) x N + 2ε, which implies x x N + 2ε and so forth x B(T ). Example 2. Let T be a metric space and C b (T ) be a space of bounded continuous functions on T. Then, (C b, ) is a Banach space. To prove the assertion, it is sufficient to show that C b (T ) is a closed subspace of B(T ) (due to the Lemma 1.9), that is, to show that every sequence in C b (T ) which converges in B(T ) converges to a point from C b (T ). Let {x n } n N be a sequence of bounded, continuous functions convergent to x B(T ). We need to show that x is a continuous function. For any ε > 0 there exists N = N(ε) N, such that x N x < ε/3, since the sequence is convergent. Now, let t o T. By the continuity of x N, there exists δ = δ(ε, t o ) > 0 such that d(t, t o ) < δ x N (t) x N (t o ) < ε/3. Therefore, for all t such that d(t, t o ) < δ it holds that which ends the proof. x(t) x(t o ) x(t) x N (t) + x N (t) x N (t o ) + x N (t o ) x(t o ) 2 x x N + x N (t) x N (t o ) 2 ε/3 + ε/3 = ε, Example 3. A space of continuous functions vanishing at infinity C o (R n ) = {f C b (R n ) with a norm is a Banach space. Example 4. The following spaces lim f(t) = 0} t + c o = {{t n } n N t n K, lim n + t n = 0}, c = {{t n } n N t n K, lim n + t n exists} with a norm are Banach spaces. Remark 1. B(N) is often denoted by l. 1.3. l p spaces Definition 1.13. Let 1 p < +. We define l p = {x l + n=1 x n p + < + } and x p = p x n p. Lemma 1.14. (Hölder inequality for sequences). Let 1 p, q + be such that 1 p + 1 q = 1. Then, for x l p and y l q it holds that n=1

4 LECTURE NOTES, FUNCTIONAL ANALYSIS (a) x y l 1, (b) x y 1 x p y q. Lemma 1.15. (Minkowski inequality for sequences). Let 1 p + and x, y l p. Then x + y p x p + y p. Example 5. Spaces (l p, p ) are Banach spaces for 1 p +. Space (l, ) coincides with (B(N), ), therefore we assume that p < +. Similarly as in the Example 1, to prove the assertion we need to show that (a) p is a norm, (b) each Cauchy sequence converges to an element from l p. Claim (a) is straightforward, when one uses the Minkowski inequality for the proof of the triangle inequality. For the proof of completness, consider a Cauchy sequence {x n } n N. Each element x n l p is a sequence given by x n = (x n 1, xn 2,... ). Note that lp l, what holds due to the following estimate + x p = p x k p p k=1 sup k N x k p = sup x k = x, k N for x = (x 1, x 2,... ). Thus, if we consider the sequence {x n } n N as a sequence of elements of l space, we conclude that there exists exactly one x l such that lim n + x n x = 0 (this follows from the completeness of (l, )). In particular, (1.16) lim n + xn k = x k, for each k N. We shall show that x = {x k } k N is an element of l p space and that {x n } n N converges to x in l p. For any ε > 0 there exists N = N(ε), such that for all n, m N it holds that x n x m p < ε. In particular, for every K N K p x n k xm k p x n x m p < ε. Using (1.16) and passing to the limit with x m k we obtain K p x n k x k p < ε. k=1 k=1 Since the estimate is valid for all K and the right hand side of the inequality is independent on K, it holds also that + p x n k x k p < ε. k=1

LECTURE 1. BANACH SPACES 5 Therefore x n x p < ε, which proves that x is a limit of the sequence in l p. Moreover, 1.4. Minkowski functional x p x x N p + x N p ε + x N p < +. Definition 1.17. Set A is called an absorbing set if for each x X there exists t K, such that t x A. Definition 1.18. Set A is called a balanced set if x A x A. Definition 1.19. Let A be a convex, absorbing and balanced set. A functional µ A X [0, + ) defined by x (1.20) µ A (x) = inf {t (0, + ) t A} is called Minkowski functional. Lemma 1.21. Minkowski functional generates a seminorm on X. If additionally A is bounded in each direction, that is, for each x X a set (A lin{x}) is a bounded set, then it is a norm. Proof. We shall concentrate on the essential part of the proof, that is, showing that µ A fulfills the triangle inequality. Fix ε > 0 and let t = µ A (x) + ε, s = µ A (y) + ε. Then, t 1 x, s 1 y A, what follows from the definition of the Minkowski functional. Set A is convex, therefore t t + s x t + s t + s y s = x + y t + s A, which implies that x + y µ A (x + y) = inf {z (0, + ) A} t + s = µ A (x) + µ A (y) + 2ε. z Due to a freedom in the choice of ε the assertion is proved. 1.4.1. Examples of normed spaces with a norm introduced by the Minkowski functional Let F = F (Ω) be a space of real valued, Lebesgue measurable functions on Ω. Example 6. Orlicz spaces L M (Ω). Let M be a non-negative convex function on [0, + ), such that Define a set M(0) = 0 and lim M (t) = +. t + (1.22) A = {f F Ω M ( f(x) ) dx 1}. Orlicz space L M (Ω) is the smallest linear space containing A. Minkowski functional µ A defines a norm on L M (Ω). It can be checked that

6 LECTURE NOTES, FUNCTIONAL ANALYSIS Example 7. Lebesgue spaces L p (Ω) = {f F Ω f(x) p dx < + }. The most important class of Orlicz spaces arises when we set M(x) = x p, where 1 < p < +. In this case we obtain Lebesgue spaces L p (Ω). Analogously as in the example above, A = {f F Ω f(x) p dx 1}. It turns out that Minkowski functional µ A is given by the following formula µ A (f) = p Ω f(x) p dx. Note that A is a convex, absorbing and balanced set, therefore µ A is a seminorm on L p (Ω). Moreover, if µ A (f) = 0, then Ω f(x) p dx = 0, which implies f = 0 a.e. Thus, µ A defines a norm on L p (Ω). Example 8. Generalized Lebesgue spaces L p( ) (Ω) = {f F Ω f(x) p(x) dx < + }. The next important class of Orlicz spaces is created when one sets M(x) = x p(x), where p(x) fulfills 1 < p 1 p(x) p 2 < + for some p 1, p 2. In this case we obtain generalized Lebesgue spaces L p( ) (Ω). Similarly as before, A = {f F Ω f(x) p(x) dx 1}. 1.5. L p spaces. Definition 1.23. Let 1 p < +. The space L p (Ω) consists of equivalence classes of Lebesgue measurable functions f Ω R such that Ω f(x) dx < +, where two measurable functions are equivalent if they are equal a.e. f L p (Ω) is defined by L p = p Ω f(x) dx. The L p norm of For p = + the definition is slightly different. We say that a function f is essentially bounded, if essup f = inf sup f(x) < +. N N =0 (Ω/N) The space L (Ω) consists of equivalence classes (two functions are equivalent if they are equal a.e.) of measurable, essentially bounded functions f Ω R with a norm L = essup f.

LECTURE 1. BANACH SPACES 7 Remark 2. The reason to regard functions that are equal a.e. as equivalent is so that f L p = 0 implies that f = 0 and thus L p is a norm. For example, the characteristic function of the rational numbers Q is equivalent to 0 in L p (R), for 1 p +. Lemma 1.24. (Hölder inequality for integrals). Let 1 p + and 1 p + 1 g f L p (Ω), g L q (Ω). Then, f g L 1 (Ω) and f g L 1 f L p g L q. = 1. Let Theorem 1.25. Orlicz spaces L M (Ω), Lebesgue spaces L p (Ω) and generalized Lebesgue spaces L p( ) (Ω) are Banach spaces. We prove the theorem only for the Lebesgue spaces. In the proof we shall use the following lemma. Lemma 1.26. For any normed space (X, ) the following conditions are equivalent (a) (X, ) is a complete space. (b) If {x n } n N is a sequence in X, such that + n=1 x n < +, then there exists x X such that N lim x n x = 0. N + n=1 The condition (b) simply states that any absolutely convergent series is convergent. Proof of Lemma 1.26. (a) (b). The implication follows from the fact that S N = N n=1 x n is a Cauchy sequence. (b) (a). Let {x n } n N be a Cauchy sequence. For each k N there exists N k, such that We choose a subsequence {x nk } k N such that x m x n < 2 k, n, m N k. x nk+1 x nk < 2 k, k N and denote y 1 = x n1, y k = (x nk+1 x nk ) for k > 1. Therefore + i=1 y i < +. From assumptions it follows that there exists y X, such that N lim N + n=1 y n y = lim N + x n N+1 y = 0. Therefore, the subsequence {x nk } k N converges in X. A Cauchy sequence, which has a convergent subsequence, converges as well, which ends the proof. Proof of Theorem 1.25 for the Lebesgue spaces, p < +. Checking that L p is a norm, when one uses Minkowski inequality, is straightforward. Note that we have proved the assertion in an alternative way for 1 < p < + in the Example 7. For the proof of completeness we shall use claim (b) from Lemma 1.9. Let {f n } n N be

8 LECTURE NOTES, FUNCTIONAL ANALYSIS a sequence in L p (Ω) such that M = + n=1 f n L p < +. We need to construct a function f L p (Ω), such that lim N + N n=1 f n f L p = 0. Define ĝ n, ĝ Ω R as following ĝ n (x) = n i=1 From Minkowski inequality we obtain ĝ n L p = n i=1 f i (x) and ĝ(x) = f i L p n i=1 f i L p + n=1 + n=1 f n (x). f n L p = M < +, By construction, ĝ n converges monotonically to ĝ. Therefore, from the monotone convergence theorem and the inequality above it follows that (ĝ(x)) p dx = Ω lim (ĝ n(x)) p dx = lim (ĝ n (x)) p dx < M p. Ω n + n + Ω which implies that ĝ L p (Ω) and in particular ĝ is finite a.e. From the latter fact we conclude that f(x) = + n=1 f n (x) is finite a.e. and f L p (Ω) with f L p ĝ L p. Note that 0 f(x) n i=1 p f i (x) = + i=n+1 Thus, by the Lebesgue dominated convergence theorem lim n + Ω p p f i (x) ( f i (x) ) (ĝ(x)) p < M p. i=n+1 f(x) which ends the proof due to the Lemma 1.26. n i=1 f i (x) p dx = 0,

LECTURE 2 Linear Operators Definition 2.1. Let (E, E ) and (F, F ) be normed spaces. A space consisting of linear, bounded operators A E F is denoted as α(e, F ). A bounded operator is understood here as the operator, which maps bounded sets onto bounded sets. Remark 3. α(e, F ) is a normed space with a norm given by A α(e,f ) = A(x) sup F. x E,x 0 x E Remark 4. For any A α(e, F ) the following equalities hold (2.2) A α(e,f ) = sup A(x) F = sup A(x) F. x x E 1 x x E =1 Theorem 2.3. Let (E, E ) and (F, F ) be normed spaces and T E F be a linear operator. Then, the following conditions are equivalent: (a) T is continuous, (b) T is continuous at 0, (c) there exists M such that for each x E it holds that T x F M x E, (d) T is uniformly continuous. Proof of Theorem 2.3. (c) (d). Condition (c) implies that T is Lipschitz continuous and thus, also uniformly continuous. (d) (a) (b). The proof is trivial. (b) (c). We shall prove this implication by contradiction. Assume that T is continuous at 0 and (c) does not hold, which implies that there exists a sequence {x n } n N such that T x n F > n x n E. 9

10 LECTURE NOTES, FUNCTIONAL ANALYSIS Define y n = x n n x n E. Then, y n E = 1/n and lim n + y n 0 E = 0. Moreover, T y n F = T x n F n x n E > 1. Therefore, lim n + y n 0 E = 0 and T y n T (0) F > 1, which is a contradiction due to the fact that T is continuous. Lemma 2.4. In an infinite dimensional Banach space there exist unbounded operators, which are defined everywhere. Lemma 2.5. Let D E be a dense subset of a normed space (E, E ) and (F, F ) be a Banach space. Then, for each T α(d, F ) there exists a unique ˆT such that ˆT E F, ˆT D = T, and ˆT α(e,f ) = T α(d,f ). Proof of Lemma 2.5. For each x E there exists a sequence {x n } n N D, such that lim n + x n x E = 0. Define an operator ˆT x = lim n + T x n. The operator ˆT is well defined, since {T x n } n N is a Cauchy sequence and F is a Banach space, thus the limit exists. Note that ˆT is unique (it does not depend on the choice of the sequence {x n } n N ). Indeed, if there exist sequences {x n } n N and {x n} n N such that lim x n = lim n + k + x k = x, then lim n,k + x n x k = 0 and from the boundedness of T, E Moreover, ˆT is bounded since It also holds that lim T x n T x k n,k + F M lim x n x k n,k + E = 0. ˆT x F = lim T x n n + F lim M x n n + E sup M x n E < C. n N ˆT α(e,f ) = sup ˆT (x) F = sup ˆT (x) F T (x) = sup F = T x E, x 0 x E x D,x 0 x D x D,x 0 x α(d,f ). D

LECTURE 2. LINEAR OPERATORS 11 Theorem 2.6. Let (F, F ) be a Banach space. Then, (α(e, F ), α(e,f ) ) is a Banach space as well. Proof of Theorem 2.6. We need to prove that a normed space (α(e, F ), α(e,f ) ) is complete. Let {A n } n N be a Cauchy sequence in α(e, F ), that is, A n x A m x (2.7) ε > 0 N = N(ε) n, m N A n A m α(e,f ) = sup F < ε, x E, x 0 x E which implies that {A n x} n N is a Cauchy sequence in F for each fixed x E as well. Since F is a Banach space, this sequence is convergent. Thus, an operator A E F given by Ax = lim n + A nx is well defined. It is also linear and bounded. The latter claim holds due to the following Ax F x E A n x = lim F n + x E A n x sup F n N x E < +. Now, we want to prove that lim n + A n A α(e,f ) = 0. Let x be such that x E = 1. Condition (2.7) implies then ε > 0 N = N(ε) n, m > N A n x A m x F < ε. If we let m +, then A n x Ax F < ε. This holds for each x such that x E = 1, therefore sup A n x Ax F < ε, x x E =1 which ends the proof due to the equality (2.2). Lemma 2.8. Let X, Y, Z be normed spaces and S α(x, Y ), T α(y, Z). Then, T S α(x,z) T α(y,z) S α(x,y ). Proof of Lemma 2.8. Let x X and x 0. Then, (T S)(x) Z = T (Sx) Z T α(y,z) Sx Y T α(y,z) S α(x,y ) x X. Thus, Taking supremum over x X ends the proof. (T S)(x) Z x X T α(y,z) S α(x,y ).

12 LECTURE NOTES, FUNCTIONAL ANALYSIS Example 9. Examples of linear operators T X Y : (a) Identity : X = Y, T = Id. The norm T α(x,y ) of T is equal to 1. (b) A linear map between finite dimensional spaces : X = R m, Y = R n and A R n R m is a matrix A = {a ij }, 1 i n, 1 j m, T (x) = A x. (c) Differentiation : X = C 1 ([0, 1]), Y = C([0, 1]), T (f) = f. (d) Let g L q ([0, 1], µ) and p, q be such that 1 p + 1 q = 1. X = Lp ([0, 1], µ), Y = L 1 ([0, 1], µ), T g (f) = 0 1 fg dµ. (e) Fredholm integral operator : X = Y = C([0, 1]), where k C([0, 1] [0, 1]). (T (f))(y) = 0 1 k(x, y)f(x)dx, (f) An operator defined analogously as in the example (e), but with X = Y = L 2 ([0, 1], µ) and k L 2 ([0, 1] [0, 1], µ). Remark 5. (to example (e)) Note that continuity of T (f) follows directly from the uniform continuity of k. Indeed, from a definition of uniform continuity ε > 0 δ > 0 such that ( y y < δ sup k(x, y) k(x, y ) < ε). x [0,1] Therefore, it also holds that Note that (T (f))(y) (T (f))(y ) T f = sup y [0,1] 0 1 0 1 k(x, y) k(x, y ) f(x) dx ε f. k(x, y)f(x)dx f sup k(, y) f k. y [0,1] Definition 2.9. An operator T α(x, Y ) is called an isomorphism, if there exists an inverse operator T 1 α(y, X). Moreover, if T α(x,y ) = T 1 α(x,y ) = 1, then T is called an isometry and X, Y are said to be isometrically isomorphic. Two isometrically isomorphic normed spaces share the same structure, so they are usually identified with each other.

LECTURE 2. LINEAR OPERATORS 13 Theorem 2.10. Let (X, X ) be a normed space and T α(x, X) = α(x). Then, the following implication holds + n=0 where T n = T T. n times T n converges in α(x) (Id T ) 1 exists and (Id T ) 1 = Proof of Theorem 2.10. + n=0 T n, Define S m = m n=0 T n. Then, (Id T )S m = S m (Id T ) and (Id T )S m = Id S m T S m = m n=0 T n m+1 n=1 T n = Id T m+1. Since + n=0 T n converges in α(x), we have that lim n + T n x = 0 for each x X. Therefore, passing to the limit in the equalities above yields Id(x) = lim (Id T m+1 + )x = lim (Id T )S mx = (Id T ) lim S mx = (Id T )( T n )x. m + m + m + Similar argument proves that Id(x) = ( + n=0 T n )(Id T )x. Therefore, (Id T ) 1 x = + n=0 T n x, x X. Remark 6. If (X, ) X is a Banach space it is sufficient to assume that T α(x) < 1. In this case (2.11) (Id T ) 1 1 α(x). 1 T α(x) Indeed, note that + n=0 T n α(x) + n=0 T n α(x) < +, since T α(x) < 1. Since X is a Banach space, then the convergence of the series + n=0 T n follows from Lemma 1.26. The inequality (2.11) follows from a formula on a sum of a geometric series. Exercise 1. Find a solution x C([0, 1]) to the following equation x(s) 0 1 k(s, t)x(t) dt = y(s), where k C([0, 1] [0, 1]) and y C([0, 1]) are given. n=0

LECTURE 3 Dual Spaces Definition 3.1. Let (X, X ) be a normed space. Space α(x, K) consisting of linear, bounded functionals on X is called a dual space to X. It is denoted by X or X. Remark 7. Dual space X is a Banach space even if X is not a Banach space. Theorem 3.2. Let 1 p, q < + be such that 1 p + 1 q = 1. Then, an operator T l q (l p ), (T x)(y) = + n=1 x n y n, where x = (x 1, x 2,... ) l q, y = (y 1, y 2,... ) l p, is an isometric isomorphism. Proof of Theorem 3.2. from Hölder inequality Linearity of T is straightforward. T is bounded, which follows (3.3) (T x)(y) x q y p T x (l p ) x q. To show that T is an isomorphism we need to prove that it is injective and surjective. Indeed, T x = 0 x n = (T x)(e n ) = 0, n N, where e n = (0,..., 0, 1, 0,... ), that is, the n-th coordinate is the only non-zero element of e n. Thus, ker(t ) = {0}, which proves that T is an injection. For the proof of surjectivity we have to show that y (l p ) x l q such that T (x) = y, i.e., T (x)(y) = y (y) y l p. It is sufficient to prove that the equality holds for y = e n, n N. Indeed, assume that T (x)(e n ) = y (e n ) for each n N. Since T (x) and y are linear, then the equality holds for all y {lin{e n } n N}. However, the operators are also continuous and therefore the equality holds for {lin{e n } n N} p = l p 15

16 LECTURE NOTES, FUNCTIONAL ANALYSIS as well. Now, fix y (l p ) and define s n = y (e n ), x = {s n } n N. We prove that x l q. To this end, let {t n } n N be a sequence given by t n = { Therefore, for each N N it holds that and Therefore, N n=1 s n q = N ( s n q ) n=1 N n=1 N n=1 t n p = t n s n = s n q s n for s n 0, 0 for s n = 0. N n=1 N n=1 N s n p(q 1) = y (l p ) ( t n p ) 1 q n=1 N n=1 s n q N t n y (e n ) = y ( t n e n ) 1 p n=1 N = y (l p ) ( s n q ) n=1 y (l p ) x q y (l p ) x l q, since the first inequality holds for all N N. It is straightforward now, that T (x)(e n ) = s n = y (e n ). The fact that x q y (l p ) together with (3.3) imply that T is an isometry. Theorem 3.4. Let 1 p < + and q be such that 1 p + 1 q = 1 and (Ω, Σ, µ) be a measure space, where µ is a σ-finite measure. Then, is an isometric isomorphism. T L q (Ω, µ) (L p (Ω, µ)), (T g)(f) = Ω fg dµ Proof of Theorem 3.4. T is clearly a linear operator. It follows from Hölder inequality that T (g)(f) fg L 1 f L p g L q T g (L p ) g L q. Thus, the norm of the operator T is at most 1. In particular, the function f defined as below f = ḡ g ( g q p ), for 1 < p < +, g L q (where ḡ is a complex conjugate of g) is such that f L p = 1 and (T g)(f) = fg dµ = g L q, which implies that T g (L p ) g L q and thus T is an isometry. For p = 1 we set f = ḡ g and use similar arguments. 1 p.

LECTURE 3. DUAL SPACES 17 Injectivity of T is straightforward. Indeed, assume that T (g) = 0, which implies that Ω fg dµ = 0, f L p (Ω, µ). Thus, g = 0 µ a.e. and g = 0 in L q (Ω, µ). Now, we need to show that the operator is surjective, that is, y (L p ) g L q such that T (g) = y, i.e., (T g)(f) = Ω fg dµ = y (f), f L p. Assume that µ(ω) < + and define ν Σ K, ν(e) = y (χ E ). Since µ is finite, χ E L p (Ω, µ) and thus ν is well defined. It is clear that ν is additive and in the case p < + it is also σ-additive (why?). Thus, ν is a signed (or complex) measure. From the construction it follows that ν is absolutely continuous with respect to µ. Indeed, µ(e) = 0 χ E = 0 µ a.e. χ E L p (Ω, µ). Since χ E = 0 in L p (Ω, µ), it holds that ν(e) = y (χ E ) = y (0) = 0. By Radon-Nikodym theorem, there exists g L 1 (Ω, µ) such that y (χ E ) = ν(e) = E g dµ = Ω χ E g dµ, E Σ. Characteristic functions are dense in (L, L ) (and in (L p, L p) ), which implies that (3.5) y (f) = Ω fg dµ f L (Ω, µ). Now, it remains to prove that g L q (Ω, µ). Cases for q < + and q = + are considered separately. Assume that q < + and define a µ-measurable function f(x) = { g(x) q g(x) if g(x) 0, 0 if g(x) = 0. Then, g(x) q = (fg)(x) = f(x) p. Define a µ-measurable set E n = {x g(x) n}. Then, χ En f L (Ω, µ) and thus we can plug it into (3.5). Ω (χ En f)g dµ = y (χ En f) y (L p ) χ En f L p 1 = y (L p ) ( f p p dµ) En = y (L p ) ( En 1 g q p dµ), On the other hand Ω (χ En f)g dµ = En g q dµ.

18 LECTURE NOTES, FUNCTIONAL ANALYSIS Combining the both expressions we obtain 1 ( g q q dµ) y (L p ). En The inequality above holds also when the integral is taken over the whole Ω. Indeed, sup g q dµ = lim n N E n n N g q dµ = En g q dµ, Ω due to the monotone convergence theorem. Thus, Now, consider the case q = +. Define 1 ( g q q dµ) = g L q y (L p ). Ω E = {x g(x) > y (L 1 ) } and f(x) = χ E (x) g(x) g(x). It is clear that f L (Ω, µ) and f L 1 = µ(e). If µ(e) > 0, then µ(e) y (L 1 ) < E g dµ = Ω fg dµ = y (f) y (L 1 ) f L 1 and µ(e) < f L 1, which is a contradiction. Definition 3.6. Let Σ be a σ-algebra of subsets of Ω and K = {N, C}. ν Σ K is called a signed (or complex) finite measure, if (a) ν( ) = 0, (b) ν ( + i=1 A i ) = + i=1 ν(a i ), if A i A j = for i j. Theorem 3.7. Let Σ be a σ-algebra of subsets of Ω, µ be a finite measure and ν be a signed (or complex) measure on Σ. Moreover, let ν be absolutely continuous with respect to µ, that is, µ(e) = 0 ν(e) = 0. Then, there exists g L 1 (Ω, µ) such that ν(e) = E g dµ, E Σ. 3.1. Extensions of functionals Definition 3.8. Let X be a vector space. Functional p X R is called sublinear, if (a) p(λx) = λp(x), λ 0, x X, (b) p(x + y) p(x) + p(y), x, y X. Remark 8. Any seminorm is a sublinear functional.

LECTURE 3. DUAL SPACES 19 Theorem 3.9. (Hahn-Banach) Let X be a vector space and U be its subspace. If p X R is a sublinear functional, l U R is a linear functional such that l(x) p(x), x U, then there exists an extension L X R, such that L U = l and L(x) p(x), x X. In the proof we shall use the following lemma. Lemma 3.10. (Zorn lemma) Let (A, ) be a non-empty partially ordered set, such that every non-empty totally ordered subset has an upper bound in A. Then, the set A contains at least one maximal element. A totally ordered set is the set with partial order under which every pair of elements is comparable. Proof of Theorem 3.9. Assume that dim(x/u) = 1. With this assumption our task is to extend the functional l to X = U Rx o, where x o X/U. Note, that in this case any x X can be written as x = u + λx o, where u U and λ R. Define (3.11) L r (x) = l(u) + λr. We claim that L r is a linear extension of l, such that L r (x) p(x). If λ = 0, we have that L r (x) = l(u) + λr = l(u) p(u). Hence, we can assume that λ 0. Let λ > 0. Then, inequality L r (x) p(x) is equivalent to the following Similarly, for λ < 0 we obtain λr p(u + λx o ) l(u) r p ( u λ + x o) l ( u λ ) r inf v U (p(v + x o) l(v)). λr p(u + λx o ) l(u) r p ( u λ x o) l ( u λ ) r sup(l(w) p(w x o )). w U Therefore, there exists r R such that L r p, if which is equivalent to l(w) p(w x o ) p(v + x o ) l(v), l(v) + l(w) p(v + x o ) + p(w x o ), v, w U. But l(v)+l(w) = l(v+w) p(v+w) p(v+x o )+p(w x o ), since v+w U and p is sublinear.

20 LECTURE NOTES, FUNCTIONAL ANALYSIS Now, consider the collection A = {(V, L) V X is a linear subspace, L V R is a linear extension of l We define a partial order on A by setting (V 1, L 1 ) (V 2, L 2 ) V 1 V 2 and V 2 Z1 = V 1. Now, we check that every chain (i.e., totally ordered set) {(V i, L i ) i I} such that L(x) p(x), x V }. in A has an upper bound. Let V = i I V i. Thus, V is a subset of X and V contains each V i. Let define L V R in the following way. If x V, then there exists i I such that x V i and we set L(x) = V i (x). The definition does not depend on the choice of i because of the chain definition. Moreover, it implies that L is linear and (V i, L i ) ( V, L) for every i I. In other words this is an upper bound for the chain. Zorn s lemma guarantees the existence of a maximal element (V, L) A. If V X, then it follows from the first part of the proof, that we can extend the functional and thus, it is a contradiction due to the fact that (V, L) is maximal. Theorem 3.12. Let X be a C-vector space. Then, (a) If l X R is a R-linear functional, that is, then, l(λ 1 x 1 + λ 2 x 2 ) = λ 1 l(x 1 ) + λ 2 (x 2 ), λ 1, λ 2 R, x X, h(x) = l(x) il(ix) is a C-linear functional such that Re h = l. (b) If h X C is a C-linear functional, then l = Re h is a R-linear functional. (c) If p is a seminorm and h X C is a C-linear functional, then for all x X the following equivalence holds h(x) p(x) Re h(x) p(x). (d) If (X, X ) is a normed space and h X C is a C-linear continuous functional, then h α(x,c) = Re h α(x,r). Theorem 3.13. (Hahn-Banach theorem for C) Let X be a C-vector space and U be its subspace. If p X R is a sublinear functional, l U C is a linear functional such that Re l(x) p(x), x U, then there exists an extension L X C, such that L U = l and Re L(x) p(x), x X.

LECTURE 3. DUAL SPACES 21 Theorem 3.14. Let (X, X ) be a normed space and U be its subspace. Then, for every continuous linear functional u U K there exists a continuous linear funtional x X K, such that x U = u and x α(x,k) = u α(u,k). Proof of Theorem 3.14. (only for K = R) Define p X R as following p(x) = u α(u,k) x X. It is clear that p is a sublinear functional and u (x) u α(u,k) x X, for all x U. Thus, by Hahn-Banach theorem there exists x X R, such that x U = u and x (x) p(x), x X. Note that x ( x) p( x) = p(x), which implies that x (x) u α(u,k) x X, x X, which implies that x is continuous and x α(x,k) u α(u,k). For the proof of the inverse inequality notice that where u α(u,k) = sup u B u u (u) = sup u B u x (u) sup x B x x (x) = x α(x,k), B u = {u U u X 1} and B x = {x X x X 1}. Remark 9. Let X be a normed space. Then, for every x o X, x o 0 there exists x X such that x α(x,k) = 1 and x (x o ) = x o X. Indeed, let M = Kx o. Define x M R by One can easily check that x (λx o ) = λ x o X, λ K. x α(m,r) = 1 and x (x o ) = x o X. The existence of the extension x X K follows from the Hahn-Banach theorem. Remark 10. Let X be a normed space. Then, for every x 1, x 2 X such that x 1 x 2 there exists x X such that x (x 1 ) x (x 2 ). This is the immediate conequence of the previous remark. Remark 11. For any normed space (X, X ) it holds that Indeed, since x B x = {x X x X = sup x (x), x X. x B x x 1} we have that x (x) x x X x X, x B x.

22 LECTURE NOTES, FUNCTIONAL ANALYSIS On the other hand, it follows from the previous remark that there exists x such that x = 1 (and thus x B x ) and x X = x (x) x (x) sup x (x). x B x Lemma 3.15. Let (X, X ) be a normed space and V X be its convex and open subset such that 0 V. Then, there exists x X such that Re x (x) < 0, for each x V. Proof of Lemma 3.15. Moreover, for x o V define (only for K = R) Define sets A + B and A B as following A ± B = {a ± b a A, b B}. y o = x o and U = V {x o }. Set U is convex, open and y o U, 0 U. For the set U define a Minkowski functional p U. Note that p U is sublinear and p U (y o ) 1. Now, consider a linear functional y lin{y o } R, y (t y o ) = t p U (y o ). It holds that y (y) p U (y), for each y Y. Indeed, y (t y o ) 0 p U (t y o ), for t 0, y (t y o ) = p U (t y o ), for t > 0. Let x be an extension of y such that x p U. There exists ε > 0 such that B(0, ε) U such that x (x) = max{x (x), x ( x)} max{p U (x), p U ( x)} 1 ε x X. It is straightforward that x (y o ) = p U (x o ) 1 and for each x = (u y o ) V, where u U it holds that x (x) = x (u) x (y o ) p U (u) 1 < 0. Theorem 3.16. (Hahn-Banach theorem) Let X be a normed space, V 1, V 2 X be convex and V 1 be open. If V 1 V 2 =, then there exists x X such that Proof of Theorem 3.16. Re x (v 1 ) < Re x (v 2 ) v 1 V 1, v 2 V 2. Let define V = V 1 V 2 = (V 1 + ( V 2 )), which is in fact given by V = x V 2 (V 1 {x}). Note that V is open and since V 1 V 2 = it holds that 0 V. Therefore, by Lemma (3.15) there exists a functional x such that Re x (v 1 v 2 ) < 0 for all v 1 V 1 and v 2 V 2 and thus, Re x (v 1 ) < Re x (v 2 ).

LECTURE 3. DUAL SPACES 23 Theorem 3.17. (Hahn-Banach theorem) Let X be a normed space, V X be a closed and convex set and x V. Then, there exists a functional x X such that that is, there exists ε > 0 such that Re x (x) < inf{re x (v) v V }, Re x (x) < Re x (x) ε Re x (v). Proof of Theorem 3.17. Since V is closed, there exists an open ball with radius r, such that B(x, r) V. From Theorem 3.16 if follows that Therefore, and Re x (x + u) < Re x (v), u B(0, r), v V. Re x (x) + Re x (u) < Re x (v) Re x (x) + Re x X r Re x (v) Re x (x) + Re x X r inf{re x (v) v V }.

LECTURE 4 Weak convergence and reflexivity Let (X, X ) be a normed space. X is defined as a space of linear functionals given by i(x) X K (4.1) (i(x))(x ) = x (x). Remark 12. Operator i is a linear isometry (in general not surjective). Indeed, (i(x))(x ) = x (x) x x X i(x) x X. According to Remark 9, for every x X, x 0, there exists a functional x such that x (x) = x X and x = 1. Thus, x X = x (x) = i(x)(x ) i(x). Definition 4.2. A Banach space X is called reflexive if i defined as in (4.1) is surjective. We write then X X. Example 10. (a) Finite dimensional spaces are reflexive, (b) Banach spaces l p and L p for 1 < p < + are reflexive, (c) Banach spaces L 1, L, (C(K), ) are not reflexive. Definition 4.3. Let {x n } n N be a sequence in a space X. We say that {x n } n N converges weakly to some x and denote it as x n x, if lim n + x (x n ) = x(x) in K, x X. Definition 4.4. Let {y n } n N be a sequence in a space Y X. We say that {y n } n N converges weakly to some x and denote it as x n x, if lim y n(x) = y(x) in K, x X. n + 25

26 LECTURE NOTES, FUNCTIONAL ANALYSIS Theorem 4.5. (Sequential version of Banach-Alaoglu theorem) Let (X, X ) be a separable normed space and Y X. If {y n } n N is a sequence such that sup n N y n Y 1, then there exists a subsequence {y nk } k N and an element y Y such that y nk y. Proof of Theorem 4.5. If X is separable, then there exists a set L = {x j } l N X which is linearly independent and linearly dense in X. Since y n (x 1 ) is uniformly bounded with respect to n, then there exists a subsequence n 1 k such that y n 1(x 1) converges to some k element y 1 Y. Similarly, y n 1 k (x 2 ) is bounded and thus, there exists a subsequence n 2 k such that y n 2 k (x 2 ) converges to some y 2 Y. Iterating this procedure we obtain sequences y n m k y n m 1 y k n 1 k y n. By a diagonal argument (setting y nk = y n k k ) we chose a subsequence {y nk } k N such that lim y n k (x l ) = y l, l N. k + To complete the proof we need to show that y nk converges for all x X. To this end, fix ε > 0. Since L is linearly dense, there exist N = N(ε) and a sequence {(λ li, x li )} i N, λ li K, x li L, such that N(ε) x λ li x li < ε. i=1 X Define a linear functional y lin{x l x l L} K as N(ε) y( i=1 λ li x li ) = N(ε) λ li y l i. i=1 Since the assumptions of Hahn-Banach theorem are fulfilled, we can extend y on X preserving y Y 1. Note that for each ε > 0 it holds that lim y n k (x) y(x) 2ε + lim y nk ( k + k + N(ε) i=1 N(ε) λ li x li ) y( i=1 λ li x li ) = 2ε. Remark 13. If X is reflexive, then weak and weak convergence are equivalent. In such a case, Theorem 4.5 can be formulated in terms of the weak convergence as well. Remark 14. For a general predual space X there is a topological version of Theorem 4.5, which states that the unit ball in X is compact in the weak topology. Definition 4.6. (a) A subset M of a metric space X is nowhere dense, if int( M) =, (b) A set M is of the 1-st category, if M = n N M n, where M n are nowhere dense sets. Theorem 4.7. (Baire category) A non-empty, complete metric space is not of the 1-st category.

LECTURE 4. WEAK CONVERGENCE AND REFLEXIVITY 27 Theorem 4.8. Let X be a Banach space and Y be a normed space. Let T i α(x, Y ), where i I. If for each x X sup T i x Y < +, i I then sup T i α(x,y ) < +. i I Proof of Theorem 4.8. E n = {x X sup i I Let n N and define T i x Y n} = {x X T i x n}. i I From the assumptions it follows that sets E n are closed and X = n N E n. Therefore, by Theorem 4.7 at least one of the sets E n has a non-empty interior. Let n o N be such that inte no. Then, for each x o inte no there exists ε > 0 such that B(x o, ε) E no. Let x X be such that x X 1. Then, εx + x o B(x o, ε) E no. If so, then T i (εx + x o ) Y n o, i I. Therefore, for any x X such that x X 1 it holds that T i x Y = 1 ε T i(εx) Y = 1 ε T i(εx + x o ) T i x o Y 1 ε (n o + T i x o Y ). Since the right hand side of the inequality above does not depend on the choice of x, it also holds that T i α(x,y ) 1 ε (n o + T i x o Y ). Taking supremum over i I yields sup i I T i α(x,y ) 1 ε (n o + sup T i x o Y ) < +. i I

LECTURE 5 Open mapping and closed graph theorems Definition 5.1. We say that T X Y is open, if for any open set U X a set T (U) Y is open. Lemma 5.2. Let (X, X ), (Y, Y ) be normed spaces and T X Y be a linear operator. Then, the following conditions are equivalent. (a) T is open, (b) r > 0 ε > 0, such that B Y (0, ε) T (B X (0, r)), (c) ε > 0, such that B Y (0, ε) T (B X (0, 1)). Proof of Lemma 5.2. i) ii) and ii) iii) are trivial. We shall prove ii) i). Let U X be an open set and x U. It means that r > 0, such that x + B X (0, r) U Moreover, T x T (U) and due to the linearity of T T x + T (B X (0, r)) = T (x + B X (0, 1)) T (U). From the assumption we have that there exists ε > 0 such that B Y (0, ε) T (B X (0, r)). Therefore it holds that T x + B Y (0, ε) T (U) and thus T (U) is open. Theorem 5.3. Let (X, X ), (Y, Y ) be Banach spaces and T α(x, Y ) be a surjective operator. Then, T is open. To prove the theorem we need the following lemma. Lemma 5.4. Let (X, X ), (Y, Y ) be Banach spaces and T α(x, Y ) be a surjective operator. Then, there exists ε > 0 such that B Y (0, ε) T (B X (0, 1)). Proof of Lemma 5.4. Since T is surjective it holds that Y = n N T (B X (0, n)). From linearity of T it follows that Y = n T (B X (0, 1)). n N 29

30 LECTURE NOTES, FUNCTIONAL ANALYSIS Theorefore, int (T (B X (0, 1))) due to Theorem 4.7 (Baire theorem). Thus, there exists y o Y and ε > 0 such that B Y (y o, ε) T (B X (0, 1)). Since T is surjective, there exists x o such that T x o = y o and B Y (0, ε) = B Y (y o, ε) y o T (B X (0, 1)) T x o T (B X (0, 1)) T (B X (0, x o X )) = T (B X (0, 1)) + T (B X (0, x o X )) = T (B X (0, 1 + x o X )). Thus, form linearity of T it follows that B Y (0, Proof of Theorem 5.3. ε (1+ x o X ) ) T (B X(0, 1)). Due to Lemma 5.4 there exists ε o > 0 such that B Y (0, ε o ) T (B X (0, 1)). Let y B Y (0, ε o ). Then, there exists ε > 0 such that y Y < ε < ε o. Define ȳ = (ε o /ε)y. Then, ȳ Y < ε o and ȳ T (B X (0, 1)). There also exists y o such that y o = T x o T (B X (0, 1)), ȳ y o < αε o, where 0 < α < 1 and α fulfills ε 1 ε o 1 α < 1. Note that ȳ y o α B Y (0, ε o ). By the same argument we can find y 1 = T x 1 T (B X (0, 1)) such that Therefore, Iterating the procedure we obtain which is equivalent to ȳ y o α y 1 Y < αε o. y (y o + αy 1 ) Y < α 2 ε o. ȳ n i=0 n α i y i Y α n+1 ε o ȳ T ( αi) i Y < α n+1 ε o. i=0 Note that {x i } i N B X (0, 1) and 0 < α < 1, hence n i=0 α i x i converges. Since X is a Banach space we set x = + i=0 α i x i X

LECTURE 5. OPEN MAPPING AND CLOSED GRAPH THEOREMS 31 and by the continuity of T we obtain ȳ = T x. Denote x = (ε/ε o ) x. Then, y = T x and x X = ε x ε X ε + α i x i o ε X < ε + α i = ε 1 o i=0 ε o i=0 ε o 1 α < 1. Remark 15. Let X, Y be Banach spaces and T α(x, Y ) be bijective. continuous. Then, T 1 is Remark 16. If (X, 1 ) and (X, 2 ) are Banach spaces and norms 1, 2 are such that x 1 M x 2, x X, then 1 and 2 are equivalent. Definition 5.5. Let X, Y be normed spaces, D be a subspace of X and T D Y be a linear functional. Then, T is closed if for every sequence {x n } n N D such that x n x and {T x n } n N Y, T x n y, it holds that x D and T x = y. (If T is continuous, then it is also closed?) Definition 5.6. For a linear functional T D Y we define its graph as following graph(t ) = {(x, T x) x D} X Y. Lemma 5.7. Let X, Y, D, T, graph(t) be defined as above. Then, (a) graph(t ) is a linear subspace of X Y, (b) T is closed if and only if graph(t ) is closed in X Y. Lemma 5.8. Let X, Y be Banach spaces and D X be a linear subspace. If T D Y is closed, then (a) D with a norm x = x X + T x Y is a Banach space, (b) T is continuous as a mapping from (D, ) to Y. Proof of Lemma 5.8. a) Since T is closed, we know that for every sequences {x n } n N D such that x n x X and {T x n } n N Y such that T x n y it holds that x D and T x = y. {x n } n N D is a Cauchy sequence with respect to the norm, since {x n } n N is a Cauchy sequence in X and {T x n } n N is a Cauchy sequence in Y. Moreover, its limit in X is an element from the subspace D. Thus, x n x = x n x X + T (x n x) Y 0, which implies that (D, ) is complete. b) T is bounded, which holds due to the following inequality Thus, T is also continuous. T x Y x X + T x Y = x.

32 LECTURE NOTES, FUNCTIONAL ANALYSIS Theorem 5.9. Let X, Y be Banach spaces and T X Y be a linear and closed. Then, T is continuous. Proof of Theorem 5.9. T is continuous with respect to the norm defined in the previous lemma. Note that if we set D = X in the previous lemma, then X is a Banach space with norms X and. Moreover, x X x for all x X. Then, according to Remark 16 the norm X is equivalent to. Therefore, T is continuous with respect to the norm X as well. 5.1. Application of Banach -Steinhaus Theorem Lemma 5.10. For a subset M of a normed space (X, X ), the following conditions are equivalent: (a) M is bounded, (b) for every x X, x (M) K is bounded. Proof of Lemma 5.10. The implication i) ii) is trivial. To show ii) i) consider i X X, such that i(x) is given by (4.1) for all x X. Note that sup x M By Banach-Steinhaus theorem we have x (x) = sup i(x)(x ) < +, x X. x M sup x M x = sup i(x) < +. x M Remark 17. Any weakly convergent sequence is bounded. Lemma 5.11. Let X be a Banach space, Y be a normed space and T n α(x, Y ) for all n N. If for all x X there exists T x = lim n + T n x, then T α(x, Y ).

LECTURE 6 Adjoint Operator Definition 6.1. Let (X, X ), (Y, Y ) be normed spaces and T α(x, Y ). T Y X is defined by (T y )(x) = y (T x). One can check that T α(y, X ). Then, Theorem 6.2. Let X, Y be Banach spaces and T α(x, Y ). Then, the following conditions are equivalent where (a) Im(T ) is closed, (b) Im(T ) = ker(t ), (c) Im(T ) is closed, (d) Im(T ) = ker(t ), U = {x X x (x) = 0, x U}, V = {x X x (x) = 0, x V }. Lemma 6.3. Let X, Y be Banach spaces and T α(x, Y ) have a closed image. Then, there exists K 0 such that y Im(Y ) x X s.t. T x = y and x X K y Y. Proof of Lemma 6.3. factorization?) Define a canonical factorization ˆT (what is precisely a canonical ˆT X/ker(T ) Im(T ). Note that X/ker(T ) and Im(T ) are Banach spaces and ˆT is bijective. Then, according to Remark 15 there exists ˆT 1 α(im(t ), X/ker(T )) and K 0 such that ˆT 1 K. 33

34 LECTURE NOTES, FUNCTIONAL ANALYSIS Lemma 6.4. Let (X, X ),(Y, Y ) be Banach spaces and T α(x, Y ). If there exists a constant C such that for all y Y then T is surjective and open. C y α(y,k) T y α(x,k), Proof of Lemma 6.3. In order to prove that T is open we need to show that B Y (0, C) T (B X (0, 1)) (according to Lemma 5.2, claim (a)). In fact it is enough to prove B Y (0, C) T (B X (0, 1)) = D. Let y o B Y (0, C) and assume that y o D (note that D is a convex set). According to Hahn-Banach theorem, there exists y Y, α R such that for each y D Re y (y) α < Re y (y o ) y (y o ). Note that 0 D, which implies Re y (0) = 0 α. We can assume that α > 0. Define ỹ = y /α. Then Thus, Re ỹ (y) = Re (y (y)/α) 1 < Re (y (y o )/α) = Re ỹ (y o ), y D. ỹ (y) 1 < ỹ (y o ), y D (note that y D λy D, λ such that λ > 1). Moreover, for all x such that x X < 1, T x = y D and from the definition of T which implies that T ỹ α(x,k) 1, but T ỹ (x) = ỹ (T x) 1, ỹ (y o ) ỹ α(y,k) y o Y < C ỹ α(y,k) and 1 < C ỹ α(y,k), which leads to the contradiction. Lemma 6.5. Im(T ) = (ker(t )). Proof of Lemma 6.5.. Let y Im(T ), which means that there exists x X such that y = T x. If y ker(t ), then T y is a zero functional on X and thus 0 = (T y )(x) = y (T x) = y (y). Therefore, for each y Im(T ) it holds that y (y) = 0, if y ker(t ). Since (ker(t )) = {y Y y (y) = 0, y ker(t )}, we conclude that Im(T ) (ker(t )).. Let U = Im(T ). Thus, U is a closed subspace of Y. We will show that y U y (ker(t )).

LECTURE 6. ADJOINT OPERATOR 35 By Hahn-Banach theorem there exists y Y such that y U = 0 and y (y) 0. In particular 0 = y (T x) = (T y )(x) for all x X, that is, T y is a zero functional on X and thus y ker(t ). This proves that y (ker(t )), since the opposire claim would imply that y (y) = 0. Proof of Theorem 6.2. (b) (a) is straightforward. (b) (a) follows from Lemma 6.5. Now, we shall prove (a) (d). Clearly Im(T ) (ker(t )) (for this we do not need (a)) and T y (x) = y (T x) = 0 for x ker(t ). Let x (ker(t )) and define a linear functional z Im(T ) K, z (y) = x (x), for y = T x. Note that z is continuous. Indeed, using one of the previous lemmas and the open mapping theorem we have that y Im(T ) x X s. t. y = T x, and z (y) = x (x) x α(x,k) x X x α(x,k) K y Y. By Hahn-Banach theorem we can define y Y, which is an extension of z. Then, x (x) = z (T x) = y (T x) = (T y )(x), x X, that is, x = T y. This finishes the proof, since we assumed that x (ker(t )). (d) (c) is straightforward. (d) (a). Let Z = Im(T ) Y and define S α(x, Z) by setting Sx = T x. For y Y and x X we have (T y )(x) = y (T x) = y Z (Sx) = (S (y Z ))(x), x X. Then, T y = S (y Z ). Therefore Im(T ) Im(S ). To prove the opposite inclusion assume that S z Im(S ) for some z Z and consider any extension y of z (by Hahn-Banach theorem such extension exists). Then, S z = T y Im(T ) = Im(S ). Then, by the assumption we have Im(S ) = Im(S ). Moreover, Im(S) is dense in Z and thus S is injective (?). S is then a continuous bijection between Z and Im(S ). In fact it is an in isomorphism, in particular C z α(z,k) S z α(x,k), z Z. Then, by one of the Lemma above, S is surjective (and open). Thus, Im(S) = Z = Im(T ) (?) and Im(T ) = Im(T ).

36 LECTURE NOTES, FUNCTIONAL ANALYSIS Lemma 6.6. Let P be a continuous projection on a normed space X (P is a projection if P 2 = P ). Then, (a) P = 0 or P 1, (b) ker(p ) and Im(P ) are closed, (c) X = ker(p ) Im(P ). Proof of Lemma 7.11. (a) P = P 2 P 2, which implies P = 0 or P 1. (b) ker(p ) = P 1 ({0}) is closed (since P is continuous), Id P is a continuous projection and Im(P ) = ker(id P ) is closed. (c) x X it holds that x = (Id P )x + P x.

LECTURE 7 Hilbert Spaces Definition 7.1. Let X be a K- linear space. A mapping is called a scalar product if <, > X X K (a) < x 1 + x 2, y > = < x 1, y > + < x 2, y >, x 1, x 2, y X, (b) < λx, y > = λ < x, y >, x, y X, λ K, (c) < x, y > = < y, x >, x, y X, (d) < x, x > 0, x X, (e) < x, x > = 0 x = 0. Lemma 7.2. (Cauchy - Schwarz inequality). product. Then, < x, y > 2 < x, x > < y, y >, x, y X. The equality holds if and only if x = λy for λ K. Lemma 7.3. Define a mapping X X R by x = < x, x >. Then, is a norm. In particular, it holds that < x, y > x y, x, y, X. Let X be a K-linear space with a scalar Definition 7.4. A normed space (X, X ) is called a prehilbert space, if there exists a scalar product <, > such that x X = < x, x >, x X. Complete prehilbert space is called a Hilbert space. Lemma 7.5. Let (X, ) be a prehilbert space and U be its dense linear subspace. If for all u U it holds that < x, u >= 0, then x = 0. 37

38 LECTURE NOTES, FUNCTIONAL ANALYSIS Proof of Lemma 7.5. Define a set Y = {y X < x, y >= 0}. Y is closed since the map y < x, y > is continuous. It also contains a dense subspace U. This implies that Y = X. In particular, x Y and thus x 2 = < x, x > = 0. If (X, X ) is a normed space, we can introduce a scalar product by the norm X. More precisely, we set < x, y > = 1 4 ( x + y 2 X x y 2 X ), for K = R and < x, y > = 1 4 ( x + y 2 X x y 2 X + i x + iy 2 X i x iy 2 X ), for K = C. Lemma 7.6. Scalar product <, > X X K is a continuous mapping. Proof of Lemma 7.6. Let x 1, x 2, y 1, y 2 X. Then, < x 1, y 1 > < x 2, y 2 > = < x 1 x 2, y 1 > + < x 2, y 1 y 2 > x 1 x 2 y 1 + y 1 y 2 x 2. Theorem 7.7. (Parallelogram equality) A normed space (X, X ) is a prehilbert space if and only if, for all x, y X the following inequality holds x + y 2 X + x y 2 X = 2 x 2 X + 2 y 2 X. Proof of Theorem 7.7. (for K = R). We introduce a scalar product as following < x, y > = 1 4 ( x + y 2 X x y 2 X ). Clearly, it holds that x X = < x, x > 2. We need to prove the properties of a scalar product form the Definition 7.1. (a) Let x 1, x 2 X and define Therefore, α = x 1 + x 2 + y 2 X = 2 x 1 + y 2 X + 2 x 2 2 X x 1 x 2 + y 2 X, β = x 1 + x 2 + y 2 X = 2 x 2 + y 2 X + 2 x 1 2 X x 1 + x 2 + y 2 X. x 1 + x 2 + y 2 = α + β 2 Similarly one gets = x 1 + y 2 + x 1 2 + x 2 + y 2 + x 2 2 1 2 ( x 1 x 2 + y 2 + x 1 + x 2 + y 2 ). x 1 + x 2 y 2 = x 1 y 2 + x 1 2 + x 2 y 2 + x 2 2 1 2 ( x 1 x 2 y 2 + x 1 + x 2 y 2 ).

LECTURE 7. HILBERT SPACES 39 Finally, < x 1 + x 2, y > = 1 4 ( x 1 + x 2 + y 2 X x 1 + x 2 y 2 X ) = 1 4 ( x 1 + y 2 X + x 2 + y 2 X x 1 y 2 X x 2 y 2 X ) = < x 1, y > + < x 2, y >. Claim (b) follows from (a) for λ N. Form the construction of <, > we also have the property (b) fulfilled for λ = 0, λ = 1 and thus for all λ Z. Let λ = m n Q. Then, n < λx, y > = n < m x, y > = m < x, y > = nλ < x, y >. n By the continuiuty of X K we can extend this result for λ R. Claims (c), (d) and (e) are straightforward. Remark 18. The following claims hold: (a) A normed space is prehilbert if and only if all its 2-dimensional subspaces are prehilbert. (b) Any subspace of a prehilbert space is prehilbert. (c) Closure of a prehilbert space is a Hilbert space. Example 11. Examples of Hilbert spaces. (a) R n and C n with < s i, t i > = n i=1 s i t i, (b) l 2 with < s i, t i > = + i=1 s i t i, (c) L 2 (Ω, µ) with < f, g > = Ω f ḡ dµ. Definition 7.8. Let X be a prehilbert space. We say that x and y are orthogonal and write x y, if < x, y >= 0. Sets A and B are orthogonal, if < x, y > = 0 for all x A and y B. The set A = {y X x y, x A} is called the orthogonal complement of A. Remark 19. For a Hilbert space X the following claims hold: (a) x y x 2 + y 2 = x + y 2, (b) A is a closed subspace of X, (c) A (A ), (d) A = (lin{a}). Theorem 7.9. Let (H, ) be a Hilbert space and K be its convex and closed subset. Let x o H. Then, there exists a unique x K such that x x o = inf y K y x o. This statement still holds, if H is a uniformly convex space.

40 LECTURE NOTES, FUNCTIONAL ANALYSIS Proof of Theorem 7.9. It is trivial if x o K (we set x = x o ). Let x o K. Without loss of generality we can assume that x o = 0. In this case we define d = inf y K y. There exists {y n } n N K and lim n + y n = d. We will prove that {y n } n N is a Cauchy sequence. This is sufficient to show, since (K, ) is a Banach space. By Parallelogram equality we have (y n + y m )/2 + (y n y m )/2 = 1 2 ( y n 2 + y m 2 ), m, n N. By convexity of K we have that 1 2 (y n + y m ) K and thus 1 2 (y n + y m ) 2 d 2. Therefore, d 2 + (y n y m )/2 1 2 ( y n 2 + y m 2 ), m, n N. Since the right hand side of the inequality converges to d 2, we have that y n y m converges to zero as n, m tend to infinity. By the comleteness of K, there exists x K such that x = lim n + y n and x = d. Now, we shall show that the element is unique. Let x, x K be such that Then, by Parallelogram equality x = x = d. (x x)/2 2 = d 2 (x + x)/2 2. Note that (x + x)/2 K, (x + x)/2 2 d 2, which implies x x = 0 and thus x = x. Lemma 7.10. Let K be a closed convex subset of a Hilbert space (H, ) and x o H. The, the following conditions are equivalent (a) x o x = inf y K x o y, (b) Re x o x, y x 0, y K. Proof of Lemma 7.10. (b) (a) x o y 2 = (x o x) + (x y) 2 = x o x 2 + 2Re x o x, x y + x y 2 x o x 2. (a) (b). Let t [0, 1], y K and define y t = (1 t)x + ty K. Then, Finally, x o x 2 x o y t 2 = x o x + t(x y), x o x + t(x y) = x o x 2 + 2Re x o x, t(x y) + t 2 x y 2. Re x o x, t(y x) t 2 x y 2, t [0, 1].