2 Fundamentals of Functional Analysis


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1 Fchgruppe Angewndte Anlysis und Numerik Dr. Mrtin Gutting 22. October Fundmentls of Functionl Anlysis This short introduction to the bsics of functionl nlysis shll give n overview of the results nd nottions used in the course of the lecture. The listed results re by fr not complete nd will be supplemented in the lecture if necessry. For resons of shortness proofs re ccomplished only in prts. It is left to the students to crry out the missing ones. For more informtion nd for most of the missing proofs the reder is referred to the following stndrd textbooks of functionl nlysis. H.W. Alt: Linere Funktionlnlysis, 4. Auge, Springer Lehrbuch, Berlin, L. W. Kntorowitsch, G. P. Akilow: Funktionlnlysis in normierten Räumen, Akdemie Verlg, Berlin, Hirzebruch, Schrlu: Funktionl Anlysis I, BI 296, Heuser: Funktionlnlysis, Teubner, M. Reed, B. Simon: Functionl Anlysis I, Acdemic Press, New York, Werner: Funktionlnlysis, 4. Auge; Springer, Berlin, Normed Spces Denition 2.1. A vector spce X over eld K is clled normed spce if there exists mpping : X [0, ], which fullls the xioms of norm (i) x 0 for ll x X nd x = 0 x = 0, (ii) αx = α x, α K, x X, (iii) x + y x + y, x, y X. If we wnt to indicte the norm tht corresponds to the spce X, we use the nottion X. Using the norm xioms we cn esily dene metric on X by d(x, y) = x y. Denition 2.2. Let X be normed spce. (i) A sequence {x n } X is clled (norm) convergent in X, if there exists n element x X such tht lim n x n x X = 0,
2 (ii) A sequence {x n } X is clled Cuchy sequence in X, if, for every ε > 0 there exists n N N such tht x n x m X < ε for ll n, m > N. If X is complete with respect to its cnonicl metric, i.e. {x n } n N X is convergent if nd only if it is Cuchy sequence, then X is clled Bnch spce. Exmple 2.3. Let U R d be n open set. The spce C (k) (U), k N 0, consists of ll functions F : U K, which re continuously dierentible up to the order k. Equipped with the norm F C (k) (U) = m F (t) t m, m <k C (k) (U) is Bnch spce. Note tht m is multiindex, m N d 0, m = m m d. Exmple 2.4. Let D R n be mesurble nd 1 p <. The spce of ll mesurble functions F : D K for which ( 1/p F L p (D) = F (t) dt) p < D is denoted by L p (D). The mpping F L p (D) is just seminorm on Lp (D) becuse F L p (D) = 0 only implies F = 0 lmost everywhere, i.e. everywhere up to set of mesure zero. Let now nd dene sup t U N = {F L p (D) F L p (D) = 0}, L p (D) = L p (D)/N, 1 p <. The spce L p (D) consists of equivlence clsses nd two functions F nd G belong to the sme clss (i.e. they re identied) if F G N, i.e. if F nd G re equl lmost everywhere. The spce L p (D) equipped with the norm F L p (D) = F L p (D) = ( D F (t) p dt ) 1/p is Bnch spce. Exmple 2.5. The spce l p (N), 1 p <, is the spce of ll rel or complex sequences x = {x n } n N for which n=1 x n p is nite. Together with the norm x l p (N) = ( n=1 x n p ) 1/p the spce l p (N) is Bnch spce. The spce l (N) which is the set of bounded rel or complex sequences equipped with the norm x l (N) = sup n N x n is lso Bnch spce. An incomplete normed spce X cn be completed by dding ll limits of Cuchy sequences, i.e. in mthemticl terms Theorem 2.6. To every normed spce there exists (up to norm isometry) unique Bnch spce X, such tht X is dense subspce in X. X is clled the completion of X with respect to X. If we wnt to emphsis completion of n incomplete normed spce with respect to specic norm we write X. Exmple 2.7. Since continuous functions re integrble, it is cler tht C (0) (D) L p (D) for ll 1 p <. The completion of C (0) (D) with respect to L p (D) is Lp (D), i.e. C (0) (D) L p (D) = L p (D), 1 p <. 2
3 2.2 Opertors A mpping between two normed spces X nd Y is clled n opertor. With N (A) = {x X Ax = 0} X nd R(A) = {y = Ax x X} Y we denote the kernel of A or the null spce of A, resp. the imge of A. Denition 2.8. Let X nd Y be normed spces. The opertor A : X Y is clled liner, if (i) A(x + y) = Ax + Ay for ll x, y X, (ii) A(αx) = αax for ll x X nd α K. The opertor A is clled bounded, if there exists constnt C 0 such tht Ax Y C x X for ll x X. Theorem 2.9. Let X nd Y be normed spces nd A : X Y be liner opertor. Then the following sttements re equivlent. (i) A is continuous on X. (ii) A is continuous in 0 X. (iii) A is bounded on X. Beweis. The equivlence of (i) nd (ii) esily follows from the linerity of the opertor A. Now, let A be bounded with bound C > 0 und let {x n } n N X be sequence which is convergent to x X. Then by the boundedness of A we hve Ax n Ax Y = A(x n x) Y C x n x X 0 for n. But this shows the continuity of A in x. Let on the other hnd A be continuous und ssume A is not bounded. Then, for every n N there exists n x n X \ {0} such tht Ax n Y n x n X. Now we set y n = x n / x n X which implies y n X = 1 for ll n N nd lim n Ay n Y =. If we now dene z n = y n / Ay n Y for ll n N, then z n is sequence convergent to 0 X which fullls Az n Y = 1 for ll b N. But this is contrdiction to the continuity of A in 0. The spce of ll continuous liner opertors between X nd Y is denoted by L(X, Y ). If Y = X we set L(X) = L(X, X). A norm on L(X, Y ) is given by Ax A = sup Y = sup Ax x 0 x Y = sup Ax Y. (1) X x X 1 x X =1 We lso use the nottion A X Y for the norm of n opertor A : X Y. Theorem Together with the norm (1) the spce L(X, Y ) is normed spce. If X is normed spce nd Y is Bnch spce, then L(X, Y ) is Bnch spce. For combintions of two liner opertors we hve the following result. 3
4 Theorem Let A L(X, Y ) nd B L(Y, Z), then BA L(X, Z) nd we hve BA X Z B Y Z A X Y. For A L(X) we get itertively A n A n for ll n N. For sequences of opertors two dierent terms of convergence re used. Denition Let {A n } n N L(X, Y ) be sequence of opertors between X nd Y nd let A L(X, Y ). (i) A sequence of opertors {A n } n N is clled pointwise convergent to n opertor A if lim A nx Ax n Y = 0 for ll x X. (ii) A sequence of opertors {A n } n N is clled uniformly convergent to n opertor A if lim A n A n X Y = 0. Uniform convergence implies pointwise convergence but the converse is in generl not true. Since K = R or K = C, the spce L(X, K) is clerly Bnch spce. It is symbolized by X nd clled dul spce of X. The elements of X re clled liner functionls. 2.3 Min Theorems Theorem 2.13 (Uniform Boundedness Principle). Let X be Bnch spce nd Y be normed spce. Let the fmily {A i } i I L(X, Y ) be pointwise bounded, i.e. for every x X there exists constnt C(x) > 0 such tht A i x Y C(x) for ll i I. The fmily of norms { A i } i I is bounded. Theorem 2.14 (Theorem of BnchSteinhus). Let X nd Y be Bnch spces. A sequence {A n } n N L(X, Y ) is pointwise convergent to A L(X, Y ) if the following conditions re fullled. 1) The sequence of norms { A n } n N is bounded. 2) The sequence {A n x} n N Y is pointwise convergent in Y on dense subset of X. Theorem 2.15 (Inverse Mpping Theorem). Let X nd Y be Bnch spces. If A L(X, Y ) is bijective, then A 1 is continuous, i.e. A 1 L(Y, X). Theorem 2.16 (Closed Grph Theorem). Let X nd Y be Bnch spces, let A : X Y be liner opertor. Then grph(a) = {(x, Ax) X Y ; x X} is closed in X Y if nd only if A L(X, Y ). 4
5 2.4 Inner Product Spces Denition Let X be vector spce over K. A mpping, : X X K is clled n inner product, if (i) x, x 0 for ll x X nd x, x = 0 if nd only if x = 0, (positive denite) (ii) x, y = y, x for ll x, y X, (symmetric) (iii) αx + βy, z = α x, z + β y, z for ll x, y, z X, α, β K, (biliner). A vector spce X together with n inner product, X is clled n inner product spce or prehilbert spce. In every inner product spce X cnonicl norm is induced by x = x, x X for x X. For this norm in X we hve the following results. Theorem Let X be n inner product spce, then, for ll x, y X, we hve (i) x, y X x X y X (CuchySchwrz inequlity), (ii) x + y 2 X + x y 2 X = 2 x 2 X + 2 y 2 X (Prllelogrm equlity). If X is complete with respect to the cnonicl norm, then X is clled Hilbert spce. Every pre Hilbert spce cn be completed to Hilbert spce. Theorem To every incomplete inner product spce X there exists unique (up to norm isometry) Hilbert spce X, such tht X is dense in X. Exmple The spce C (0) (D) where D R d together with the inner product F, G L 2 (D) = F (t)g(t) dt is n incomplete inner product spce. Its completion with respect to the L 2 (D) norm is the spce L 2 (D), i.e. D C (0) (D) L 2 (D) = L 2 (D). Getting the dul spce of Hilbert spce is very esy becuse it is the Hilbert spce itself. Theorem 2.21 (Representtion Theorem of Riesz). To every liner functionl l : X K on Hilbert spce X there exists unique element x l X such tht l( ) =, x l nd l = x l X. Every Hilbert spce X is norm isomorphic to its dul spce X. With the id of n inner product nother term of convergence cn be dened. Denition Let X be Hilbert spce. (i) A sequence {x n } n N X is clled wekly convergent to n element x X if lim x n, z = x, z, for ll z X. n (ii) A sequence {x n } n N X is clled strongly convergent to n element x X if lim x x n n X = 0. Strong convergence which is lso clled norm convergence implies wek convergence nd in nitedimensionl spces, wek nd strong convergence re equivlent. 5
6 2.5 Orthogonlity Using the inner product in n inner product spce X we re ble to dene when two elements of X re orthogonl. Denition Two elements x, y X of n inner product spce X re clled orthogonl (symbolized by x y) if x, y X = 0. Two subsets U, V X of X re clled orthogonl (symbolized by U V ), if x, y X = 0 for ll x U nd for ll y V. The set U = {x X {x} U} is clled orthogonl complement of U in X. Lemm Let X be n inner product spce. (i) We hve {0} = X nd X = {0}. (ii) For every subset U X the orthogonl complement U is closed subspce of X. (iii) If U V X then X V U. Theorem Let U be closed subspce of the inner product spce X. Then we hve ( U ) = U nd the orthogonl decomposition X = U U, holds, i.e. every element x X cn be uniquely decomposed in the form x = u + v with u U nd v U. Denition Let U be closed subspce of the inner product spce X. Then the continuous liner opertor P U : X U is dened by P U : x = u + v u with u U nd v U. The opertor P U is clled orthogonl projection of X to U. Theorem Let U be closed subspce of the inner product spce X. The orthogonl projection P U hs the following properties. (i) P U x, y X = x, P U y X for ll x, y X, (ii) P 2 U = P U, (iii) P U = 1, (iv) x P U x X = min u U x u U, i.e. P U x is the best pproximting element to x in U, (v) I P U is the orthogonl projection to U, (vi) R(P U ) = N(P U ). Denition Let X be n inner product spce nd I be n index set. A fmily {u i } i I X is clled n orthogonl system in X, if u i, u j X = u j 2 X δ i,j, for ll i, j I. The orthogonl system {u i } i I is clled n orthonorml system (ONS), if u i X = 1 for ll i I. 6
7 Theorem Let {u i } i I be n orthonorml system in the inner product spce X. Then, for every x X, the inner product x, u i X is zero except for n t most countble subset of I. Furthermore, Bessel's inequlity holds, i.e. x 2 X x, u i X 2, x X. (2) i I Note tht the series in (2) hs t most countble mny summnds. Furthermore, the reder should note tht the subset of the index set I, for which the inner product x, u i X is not zero, vries with respect to x X. Denition An orthonorml system {u i } i I in n inner product spce X is clled orthonorml bsis (ONB) of X, if, for every x X, the orthogonl expnsion x = i I x, u i X u i holds in the sense of the X norm. A Hilbert spce is clled seprble if it possesses countble orthonorml bsis. Theorem Let U be closed subspce of the inner product spce X. If {u i } i I is n orthonorml bsis in U, then, for every x X, P U x = x, u i X u i. i I Exmple Let D be region nd let ω : D R + be continuous, then the spce L 2 ω(d) consists of ll functions F : D K such tht (F (t)) 2 ω(t) dt <. Equipped with the inner product F, G L 2 ω (D) = D D F (t)g(t)ω(t) dt the spce L 2 ω(d) is Hilbert spce which is clled weighted L 2 spce on D. For D = [ 1, 1] nd ω(t) = (1 t) 1/2 (1 + t) 1/2 = 1 1 t 2 the system of Tschebysche polynomils of the rst kind T n (t) = cos(n rccos(t)), t [ 1, 1], n N, forms n orthogonl bsis in L 2 ω( 1, 1). Further exmples of orthogonl bsis systems in weighted L 2 spces re discussed in course of the lecture nd in the lecture Specil Functions. 2.6 Adjoint Opertors Denition Let A : X Y be bounded liner opertor between two inner product spces X nd Y. If there exists n opertor B : Y X such tht Ax, y Y = x, By X, for ll x X nd y Y, then we cll B the djoint opertor to A which is symbolized by A. If A = A then the opertor A is clled selfdjoint. 7
8 If the djoint opertor exists, then it is unique, which cn be seen s follows: Let B 1 nd B 2 be djoint opertors to A. Then x, (B 1 B 2 )y X = (A A)x, y Y = 0. The specil choice x = (B 1 B 2 )y then yields (B 1 B 2 )y X = 0 for ll y Y which implies B 1 = B 2. For the cse of X to be Hilbert spce nd A to bounded liner opertor the existence of the djoint opertor cn be gurnteed: Theorem Let X be Hilbert spce, Y be n inner product spce nd A L(X, Y ). Then the djoint opertor A L(Y, X) of A exists nd A Y X = A X Y. Furthermore we hve (A ) = A nd A A = A 2. Beweis. Let y Y be rbitrry but xed. Hving look t the liner functionl l y : X K given by l y ( ) = A, y Y we esily see tht l y is liner nd continuous. Then by the Riesz representtion theorem, there exists unique x y X such tht l y ( ) =, x y X, which gives Ax, y Y = x, x y X for ll x X. Now we set A : Y X, A y = x y nd we just hve to check the linerity nd the boundedness of A. Theorem Let X, Y nd Z be Hilbert spces. (i) For A, B L(X, Y ) nd α K we hve (ii) For A L(X, Y ) nd B L(Y, Z) we hve Furthermore, the following reltions re true. (A + B) = A + B nd (αa) = αa. R(A) = N (A ), R(A ) = N (A), (BA) = A B. R(A) = N (A ) R(A ) = N (A) The opertor A is bijective if nd only its djoint A is bijective. Exmple Let the integrl opertor A : L 2 (, b) L 2 (, b) be dened by AF (t) = K(t, s)f (s) ds, t [, b], where the bivrite function K L 2 ([, b] [, b]) is clled the kernel of the integrl opertor. For F, G L 2 (, b) we hve AF, G L 2 (,b) = with A : L 2 (, b) L 2 (, b) given by A G(s) = = = AF (t)g(t) dt ( ) K(t, s)f (s) ds G(t) dt ( F (s) = F, A G L 2 (,b), ) K(t, s)g(t) dt ds K(t, s)g(t) dt, s [, b]. 8
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