untitled

Transcription

1 Hilbert (1),, (2), Fourier, Schmidt (3),, (4), Riesz, (5), Fredholm (6), Hilbert-Schmidt Banach Hahan-Banach,,.,,,.,,. Riemann Riemann Dirichlet Riemann.. Riemann,. Dirichlet R n. { u = 0 in u = f on f. Dirichlet.. S = {v C 2 (); v = f} I(v) = 1 2 v 2 dx. Dirichlet inf v S I(v) v

2 2. Dirichlet, Dirichlet...!.., inf v S I(v) v Weierstrass. Riemann, H.A.Schwarz C. Neumann Dirichlet Dirichlet,. Hilbert, Dirichlet. Fredholm. Fredholm Dirichlet,. Schmidt Hilbert l J. von Neumann Hilbert.. S. Banach Théorie des opérations linéaires Banach ?.,..,..., ),..

3 0.2. Hilbert Hilbert. Hilbert norm Cauchy Schwarz Hilbert,.. C n z,w (z,w) =,. (1) (z,z) 0, (z,z) =0 z =0 (2) (z,w) =(w, z) n z i w i (3) (αz + βz,w)=α(z,w)+β(z,w), α, β C C H u, v (u, v), (inner product). u = (u, u) u H (norm). (1) (u, αv + βv )=α(u, v)+β(u, v ) (2) (u, v) = 1 ( u + v 2 u v 2 + i u + iv 2 i u iv 2) (1) u 0, u =0 u =0. (2) αu = α u, α C. (3) (Cauchy-Schwarz ) (u, v) u v. (4) ( ) u + v u + v.. (3) t R,u,v H tu + v 2 = t 2 u 2 + 2Re (u, v)+ v 2 0 Re (u, v) u v. (u, v) =re iθ u e iθ u. (1) d(u, v) = u v, d(, ). (2) (u, v) = u v (3) u + v = u + v (complete), i.e. u m u n 0(m, n )= u Hs.t. u n u 0 H Hilbert. i=1

4 l 2 ( ). l 2 = {x =(x 1,x 2, ); x n C, x n 2 < } (x, y) = x ny n. x n y n 1 2 ( x n 2 + y n 2 ),.. x (i),i=1, 2, l 2. ) x (i) n x(j) n ( k=1 x (i) k x(j) k 2 = x (i) x (j) n x n,i=1, 2,, C. x n s.t. x (i) n x n. x =(x 1,x 2, ) lim x (i) n i x n 2 = lim i x (i) x 2 =0.... x (i),i=1, 2, H x (i),i=1, 2, R Hilbert. ) M >0 s.t. x (i) <M i. m m, m x l 2. x (i) n 2 x (i) n 2 M 2 ɛ >0 N s.t. x (i) x (j) <ɛ i, j > N m i, j > N m. j x (i) n m m x (i) x ɛ i >N. m,. x(j) n 2 x (i) x (j) 2 ɛ 2 x (i) n x n 2 ɛ 2 m, ɛ L 2 ()... R n { } L 2 () = f; f(x) 2 dx < (f,g) = f(x)g(x)dx f(x)g(x) dx 1 2 ( f(x) 2 dx + g(x) 2 dx).

5 0.2. Hilbert H Hilbert. u + v 2 + u v 2 =2( u 2 + v 2 ), 2.2 Let S be a closed subspace of H. Then for any u H, there exists a unique v S such that u v = inf u w w S.. H A A = {u H;(u, v) =0 v A} A (orthogonal complement). 2.3 A is a closed subspace. H = L 2 (0, 1), A= {f H; f(x) =00<x<1/2} A?. 2.4 ) Let M be a closed subspace of H. Then for any u H, there exist unique v M and w M such that u = v + w. H = M M ( d = inf w S u w. w n S s.t. u w n 0. w m w n 2 =2 w m u 2 +2 w n u 2 2( w m + w n 2 u) 2 (w m + w n )/2 u d. w m w n 2 2( w m u 2 + w n u 2 ) 4d 2 lim sup m,n w m w n 2(d 2 + d 2 ) 4d 2 =0. {w n } Cauchy S closed v S s.t. w n u 0.

6 6 ( v,v v v 2 = 2 v u 2 +2 v u 2 2( v + v u) 2 2 2d 2 +2d 2 4d 2 = ( u = v 1 + w 1 = u 2 + w 2, v 1,v 2 M, w 1,w 2 M v 1 v 2 = w 2 w 1. u M M = u =0 ( ) u d = u v = inf h M u h v M. w = u v. h M, λ R v + λh M d 2 u (v + λh) 2 = w λh 2 = w 2 2λRe(w, h)+λ 2 h 2 = d 2 2λRe(w, h)+λ 2 h 2 λ 2 h 2 2λRe(w, h) 0 λ R. Re(w, h) =0. h ih Im(w, h) =0. (w, h) =0 h M w M C n. e i =(0,, 0, i 1, 0,, 0), 1 i n, (e i,e j )=δ ij, C n z z = n i=1 z ie i. Fourier. H = L 2 ( π, π), ϕ n (x) = 1 e inx,n Z, 2π (ϕ m,ϕ n )=δ mn f L 2 ( π, π) f Fourier. ˆf n =(f,ϕ n )= 1 2π π π f(x)e inx dx 3.1 S N = N n= N ˆf n ϕ n ( π 1/2 f S N = f(x) S N (x) dx) 2 0 as N π. f = n= ˆf n ϕ n. Fourier 1, 2 L 2. Fourier. 1,2,

7 0.3. 7,. L Fourier. Hilbert H {ϕ n } orthonormal system (O.N.S.) (ϕ m,ϕ n )=δ mn m, n. {ϕ n } O.N.S. = {ϕ n} 3.2 (Bessel ) {ϕ n } O.N.S. f H ˆf n =(f,ϕ n ) ˆf n 2 f 2.. N N f ˆf n ϕ n 2 = f 2 ˆf n 2. O.N.S. {ϕ n } complete ( f N ˆf n ϕ n 2 0 as N f H f = ˆf n ϕ n. 3.3 O.N.S. {ϕ n }. (1) {ϕ n } complete (2) (Parseval ) f 2 = ˆf n 2. (3) (f,ϕ n )=0 n = f =0 (4) ϕ n H dense. (1) (2) 3.2. (1) = (3). (3) = (1) S N = N ˆf n ϕ n S m S n 2 = m ˆf n ϕ n 2 = i=n n ˆf n 2 {S n } Cauchy. g Hs.t. S n g 0. (f g, ϕ n )=0 n (3) f g =0. f S n 0. (1) = (4). (4) = (1) α n C N 0 i=m N N N f α n ϕ n 2 = f ˆf n ϕ n 2 + ˆf n α n 2

8 8 L 2 ( π, π), f(x), Fourier. {ϕ n } O.N.S.. (1) {ϕ n } complete (2) (f,ϕ n )=(g, ϕ n ) n = f = g (3) (f,g) = ˆf n ĝ n, f,g H Complete Orthnormal System C.O.N.S.. Weierstrass. ( Step 1. C([0,π]) 1, cos x, cos 2x,. (u(x) C([0,π]) v(y) =u(arccos y) C([ 1, 1]) v(y) y = cos x. [ ( ) ] cos n x = 1 n 2 n 1 cos nx + cos(n 1)x + 1.) Step 2. C 0 ((0,π)) sin x, sin 2x,. (u C 0 ((0,π)) v(x) =u(x)/ sin x C([0,π]) v(x) k=1 na k sin x <ɛ., u(x) k=1 na k sin x cos kx <ɛ 2 sin x cos kx = sin(k +1)x sin(k 1)x.) Step 3. u C([ π, π]), u( π) =u(π) u(x) e inx. Step 4. e inx / 2π(n Z) L 2 (( π, π))., Lebesgue.... C 0 ((0,π)). Lebesgue,,,. I =(0, 1) Lebesgue. Lebesgue,. (,.) f(x) L 2 (I) 1 0 ϕ n (x) f(x) 2 dx 0 {ϕ n (x)}...

9 f L 2 (I) ɛ>0 f g L2 (I) <ɛ g L 2 (I) ( ɛ >0 step function ϕ s.t. f ϕ <ɛ. ϕ C 0 (I) f C(I) supp f I. I =(0, 1) K I, δ>0, K (δ, 1 δ). f C 0 ((0, 1)) f Lebesgue, Friedrichs molifier (. Friedrichs ρ(x) C0 (R), ρ(x)dx =1 I C 0 (I) =C 0 (I) C (I). R f ρ ɛ f(x) = ρ ɛ (x y)f(y)dy, ρ ɛ (x) = 1 ɛ ρ(x ɛ ).. R (1) f L p (R) ρ ɛ f L P (R) C (R) ρ ɛ f f in L p (R). (2) supp f (a, b) supp ρ ɛ f (a ɛ, b + ɛ) Lebesgue, Schmidt. Hilbert H {f n }. {ϕ n }. (1) {ϕ n } O. N. S. (2) n f 1,,f n ϕ 1,,ϕ n.,. (Schmidt ϕ 1 = f 1 / f 1. ϕ 2 ψ 2 = f 2 + aϕ 1 (ψ 2,ϕ 1 )=0 a, ϕ 2 = ψ 2 / ψ 2. ψ n = f n n 1 k=1 (f n,ϕ k )ϕ k,ϕ n = ψ n / ψ n. R Hilbert H (separable) H. i.e. f 1,f 2, H s.t. u H, ɛ >0, n s.t. u f n <ɛ. 3.5 Hilbert C.O.N.S.

10 10 H {f n } dense in H. {f n } {g n } {g n }, g n H dense.. {g n } Schmidt C.O.N.S..

11 l 2 ( ). l 2 = {x =(x 1,x 2, ); x n C, (x, y) = x n y n x n 2 < }. x n y n 1 2 ( x n 2 + y n 2 ),.. x (i),i =1, 2, l 2. x (i) n x(j) n ( k=1 x (i) k x(j) k 2 ) = x (i) x (j) n x n,i=1, 2,, C. x n s.t. x (i) n x n. x =(x 1,x 2, ) lim x (i) n i x n 2 = lim i x (i) x 2 =0.... x (i),i=1, 2, H x (i),i=1, 2, R Hilbert. ) M >0 s.t. x (i) <M i. m m, m x l 2. x (i) n 2 m i, j > N x (i) n 2 M 2 ɛ >0 N s.t. x (i) x (j) <ɛ i, j > N m. j x (i) n m m x (i) x ɛ i >N. m,. x (j) n 2 x (i) x (j) 2 ɛ 2 x (i) n x n 2 ɛ 2 m, ɛ l 2 C.O.N.S. (1, 0, ), (0, 1, 0, ), 4.2 L 2 ()... R n L 2 () = { } f; f(x) 2 dx <

12 12 f(x)g(x) dx 1 2 ( (f,g) = f(x)g(x)dx f(x) 2 dx + g(x) 2 dx) Legendre L 2 (( 1, 1)) 2n +1 P n (x), P n (x) = 1 ( d ) n(x n 1) n, n =0, 1, 2, n! dx Hermite L 2 ((, ) (2 n n!) 1/2 π 1/4 H n (x)e x2 /2, H n (x) =( 1) n e x2( d ) ne x 2, n =0, 1, 2, dx Laguerre L 2 ((0, ) (n!) 1 L n (x)e x2 /2, L n (x) =e x2( d ) n(x n e x2 ), n =0, 1, 2, dx 4.3 Hilbert space of analytic functions. C D H 2 (D) ={f(z) :analytic ind, f(x + iy) 2 dxdy < }. H 2 (D) f,g (f,g) = f(x + iy)g(x + iy)dxdy.. D D f(z) z z 0 <R analytic f(z 0 ) 2 1 πr 2 z z 0 <R f(z) 2 dxdy D = { z < 1} { n +1 π zn} C.O.N.S. n=0 5 Hilbert H f 1,f 2, f n f 0(n ) f n f (strong convergence). f n f in H. g H (f n,g) (f,g) (n ) f n f (weak convergence)., f n f weakly in H....

13 f n f weakly in H = f lim inf f n.. (f n,f) f n f.. (1) f n f in H = f n f weakly. (2) C N f n f weakly = f n f in C n.. Hilbert H O. N. S. {ϕ n } ϕ n 0 weakly. ϕ n H.. Bessel. m n ϕ m ϕ n 2 =2. Hilbert H S S dim H <.,. H Hilbert, H. H S M>0 f <M, f S S f 1,f 2, g H, f n g weakly.. {ϕ n } H C.O.N.S.. {(f, ϕ 1 ); f S} C, {(f n,ϕ 1 ); n =1, 2, }. {(f n,ϕ 2 ); n =1, 2, } C. (f n2(1),ϕ 2 ), (f n2(2),ϕ 2 ), (f n2(3),ϕ 2 ), ; n 2 (1) <n 2 (2) <n 2 (3). k f nk (l) (l =1, 2, ) {(f nk (l),ϕ k )} l=1. g k = f nk (k) n (g k,ϕ n ),k =1, 2,,. f H {g k,f)} Cauchy. ( f = α nϕ n S N = N α nϕ n ɛ >0, N s.t. f S N <ɛ. (g k,f) (g l,f)=(g k,s N ) (g l,s N )+(g k,f S N ) (g l,f S N ) (g k,f) (g l,f) (g l,s N ) (g l,s N ) +2ɛM T (f) = lim n (f,g n ). T (f) f linear T (f) M f. Riesz g Hs.t. (f.g) =T (f) f H. (f,g) = lim n (f,g n ) f H.

14 14 6 C n n n. A =(a ij ) n n n Ax a ij 2 i,j=1 1/2 x, x C n H 1, H 2 Hilbert, 1, 2. H 1 H 2 map T bounded linear operator T (αf + βg) =αt (f)+βt(g), f,g H 1, α, β C,. M >0 s.t. T (f) 2 M f 1, f H 1 Linear operator T (f) Tf. T : H 1 H 2 linear operator T conti. M>0 s.t. Tf 2 M f 1, f H 1 T (bounded).. = T bounded N f N H 1 s.t. f N 1 =1, Tf N 2 >N. g N = f N / Tf N 2 g N 0, Tg N 2 =1.. B(H 1 ; H 2 )=H 1 H 2 bounded linear operator (L(H 1 ; H 2 ) ). T,S B(H 1 ; H 2 ), α,β C B(H 1 ; H 2 ) linear space (αt + βs)(f) =αt f + βsf, f H 1 T B(H 1 ; H 2 ) sup f 1=1 Tf 2 T T. 1, 2 1, 2. Tf T = sup Tf = sup f 1 f 0 f = inf{m; Tf M f, f H 1 }. (, multiplication operator)

15 H 1 = H 2 = L 2 (), a(x) L (), (Af)(x) =a(x)f(x), f L 2 () = A B(L 2 (); L 2 ()), A = ess.sup x a(x). (, integral operator) H 1 = H 2 = L 2 (), (Af)(x) = a(x, y)f(y)dy, f L 2 () sup a(x, y) dy = M 1 <, x sup a(x, y) dx = M 2 < y = A B(L 2 (); L 2 ()), A M 1 M 2 H C bounded linear functional. H = B(H; C) =bounded linear functional (Riesz) T H g H Tf =(f,g), f H. T = g. Proof. (g uniqueness) (f,g 1 )=(f,g 2 ) f H g 1 = g 2. (g T =0 g =0 T 0. N = {f H; Tf =0}. N closed. H = N N. Claim dim N =1. N {0} f 0 N s.t. f 0 0. f H f Tf Tf 0 f 0 N f = N f 0.) ( f Tf Tf 0 f 0 ) + Tf Tf 0 f 0 N + N f 1 = f 0 / f 0 f H f = f +(f,f 1 )f 1, f N. Tf =(f.f 1 )Tf 1. g = Tf 1 f 1 f H Tf =(f,g). ( T = g proof). Tf = (f,g) f g T g. Tg = (g, g) = g 2 Tg g = g. T g. H 1, H 2 : Hilbert, T n,t B(H 1 ; H 2 ), n =1, 2, T n T in norm T n T 0, T n T strongly T n f Tf 0 f H 1, T n T weakly (T n f,g) (Tf,g) f,g H 1. = = T n B(l 2 ; l 2 ) T n (x 1,x 2, )=(x n,x n+1, ) define. T n 0 strongly. T n T in norm T =0. T n =1 T n.

16 16 H = L 2 (R 1 ), (T n f)(x) =f(x n) T n 0 weakly (f,g (T n f,g) 0 ). T n f = f, T n. (adjoint operator) S, T B(H; H). S = T (Tf,g)=(f,Sg) f,g H S T. f H f = sup (f,g) g =1 (1) (T ) = T (2) T = T (2) T = sup T f = f =1 sup (T f,g) = f = g =1 sup (f,tg) = T f = g =1 H = L 2 (), (Af)(x) = sup x a(x, y)f(y)dy a(x, y) dy <, = (A f)(x) = sup a(x, y) dx < y a(y, x)f(y)dy 7 ( f n H, sup n f n < = {f ni } subsequence s.t. f ni f weakly. f n H, f n f weakly = sup n f n < f =0. sup n f n =. f n. Claim {f nj } subsequence s.t. f nj 1, (f ni,f nj ) 4 i if j>i, f ni f nj 4 j if j>i (n 1 >> 1 f ni 1( i). (f n1,f n ) 0, f n n 2 >n 1 s.t. (f n1,f n2 ) < 4 1 f, n1 f n ) g = j=1 2j f nj / f nj j f n j f nj 2 2 j f n i f nj 2 1 f ni 1 2 j 4 j < f n1 j=1 j=1 j=1

17 j i 1 j i +1 (g, f ni )=2 i + j i (f ni,f nj ) f nj 2 (f ni,f nj ) f nj 2 2 j (f n j,f ni ) f nj 2 (f ni,f nj ) 4 j f n i f nj 4 j (g, f ni ) 2 i j i 2j 4 j 2 i j=1 2 j (i ). K B(H; H). K compact operator ( f n M n 1 = f ni subsequence s.t. {Kf ni } is convergent in H ) K. K B(C n ; C n ) K ( ) f n 0 weakly K compact operator = Kf n 0 strongly. = f n 0 weakly. Lemma sup n f n < f ni subsequence s.t. Kf ni g in H. g =0. ((Kf ni,h)=(f ni,k h) n i (g, h) = 0). {f n } {f n } ( {f n } Kf n 0 Kf n 0 = sup n f n <. Lemma f ni subsuquence s.t. f ni f weakly. K(f ni f) 0 Kf ni. (Hilbert-Schmidt ) H = L 2 () (Kf)(x) = K(x, y)f(y)dy, K(x, y) 2 dxdy < = K compact operator. Φ(x) = K(x, y) 2 dy Φ(x) L 1 (). f n 0 weakly f n M ( n 1). (Kf n )(x) 2 K(x, y) 2 dy f(y) 2 dy MΦ(x) x g x (y) = K(x, y) g x L 2 () Kf n = (f n, g x ) Kf n (x) 0 a.e. Lebesgue Kf n (x) 2 dx 0 (1) K n (n =1, 2, ) compact, K n K 0 = K compact (2) K 1,K 2 compact = αk 1 + βk 2, α,β C, compact (3) K compact, T = TK, KT compact (4) K compcat = K compact. (1) f n 0 weakly. Kf n Kf n K p f n + K p f n K K p f n + K p f n M K K p + K p f n

18 18 lim sup n Kf n M K K p 0(p ) (4) f n 0 weakly K f n 2 =(f n,kk f n ) K f n 2 M KK f n 0 (KK compact). H = Hilbert space, K = compact. Null(A) ={u H; Au =0}, Ran(A) ={Au; u H} (1) dim Null(1 K) < (2) Ran(1 K) closed. (1) f Null(1 K) f = Kf. S = {f; f =1,f Null(1 A)}, S = K(S). K compact K(S) compact set. S compact set dim Null(1 K) <. (2) V = Null(1 K). V closed. H = V V. Claim C >0 such that f C (1 K)f f V. ( f n V such that f n =1, (1 K)f n 0. f ni (subsequence) s.t. Kf ni g. (1 K)f ni 0 f ni g Kf ni Kg. Kg = g g V. f ni V, f ni g g V. g V V g =0. f ni 0, f ni =1.) Ran (1 K) closed. f n Ran (1 K) f n =(1 K)g n. g n V. Claim g n g m C (1 K)(g n g m ) = C f n f m 0 g n g. f n =(1 K)g n (1 K)g. 1 K onto = 1 K 1:1. T = 1 K. T 1 to 1. M n = {f H; T n f = 0}. M n closed subspace. M 0 = {0} M 1 M 2 M n. T 1to1 f 1 0 s.t. Tf 1 = 0 i.e. f 1 M 0,f 1 M 1. T onto f 2 s.t. f 1 = Tf 2 i.e. f 2 M 1, T 2 f 2 = Tf 1 =0. f 2 M 2. f n s.t. f n M n 1,f n M n. n M n M n+1. M n,m n+1 closed M n+1 = M n Mn, M n+1. ϕ n+1 M n+1 s.t. ϕ n+1 M n, ϕ n+1 =1. Kϕ n Kϕ m = (1 T )ϕ n (1 T )ϕ m = ϕ n (Tϕ n + ϕ m Tϕ m ). n>m Tϕ n + ϕ m Tϕ m M n 1. ( T n 1 (Tϕ n + ϕ m Tϕ m )=T n ϕ n + T n m 1 T m ϕ m T n m T m ϕ m = 0). Kϕ m Kϕ n 2 = ϕ n 2 + Tϕ n + ϕ m Tϕ m 2 1 {Kϕ n }. 1 K 1to1= 1 K onto. T Ran T dense T 1to1(. T =1 K 1to1= Ran T dense. T =1 K K compact. Ran T closed Ran T = H. T 1to1 Ran T dense closed. Ran T = H. (1) T (Ran T ) = Null T, Ran T = (Null T )

19 (2) K compact Ran (1 K) = (Null (1 K) ). (1) (Tf,g)=(f,T g) g Ran T (f,t g)=0 f g Null T. (Ran T ) = Null T. ((Ran T ) ) = (Null T ). H subspace (closed ) (1) S = ( S ) (2) (S ) = S. (1) exercise. (2) S closed. H = S S x S x S. = x = x 1 + x 2, x 1 S, x 2 S 0=(x, x 2 )=(x 2,x 2 ) x 2 =0. = x S (x, y) =0 y S x S. Fredholm K compact operator (1 K)u = f. Case 1 : 1 K 1to1 1 K onto. Case 2 : 1 K 1to1 Ran (1 K) = Null (1 K )) f Null (1 K ). 8 Hilbert space H closed subspce V H = V V. f H g V h V, f = g + h. f g V orthogonal projection, g = Pf. (1) P 1 (2) P 2 = P (3) P = P (4) (Pf,f) 0 f H. (1) Pf 2 + h 2 = f 2. (2) Pf = g = g +0 V + V P 2 f = Pg = g = Pf. (3) f = f 1 + f 2 V + V, g = g 1 + g 2 V + V (Pf,g)=(Pf,g 1 )=(Pf,Pg), (f,pg) =(f 1,Pg)=(Pf,Pg). (4) (Pf,f)=(P 2 f,f) =(Pf,Pf) 0. T B(H; H) (self-adjoint) T = T. H = L 2 () A a(x, y) A a(x, y) =a(y, x) T B(H; H) Tf = λf, 0 f H, λ C λ T, f. K (1) K.. (2) 0. (3) 0.

20 20. (2) V λ = {f; Kf = λf}. dim V λ =. V λ O.N.S. {ϕ n }. ϕ n 0 weakly K Kϕ n 0. Kϕ n = λ 0. (4) λ 1,λ 2, λ n λ 0. ϕ n λ n {ϕ n } O.N.S. (3). A (Af, f) =0 f H A =0. Hint (A(f + g),f + g). A A 0 sup f =1 (Af, f) > 0 inf f =1 (Af, f) < 0. K = sup (Kf,f) > 0, f 0 H s.t. f 0 =1, (Kf 0,f 0 ) = sup (Kf,f) f =1 f =1 = ( ) f 0 g = Kf 0 g. g ɛ =(f 0 + ɛg)/ 1+ɛ 2 g 2. g ɛ =1 (Kg ɛ,g ɛ )=(Kf 0,f 0 )+2ɛRe(Kf 0,g)+ O(ɛ 2 ). Re(Kf 0,g) 0 ɛ K(g ɛ,g ɛ ) > (Kf 0,f 0 ). (Kf 0,g)=0. g e iθ g,(kf 0,g)=e iθ (Kf 0,g) (Kf 0,g)=0., f 0 s.t. f 0 =1, (Kf 0,f 0 ) = sup f =1 (Kf,f) f 0 K. M = {tf 0 ; t C}. M. f 0 M = Kf 0 M = Kf 0 M = Kf 0 = αf, α C., K. S 1 = sup f =1 (Kf,f) (S 1 0 ) f n s.t. f n = 1, (Kf n,f n ) S 1. {f n } ( ) s.t. f n ϕ 1 weakly. Kf n Kϕ 1 strongly S 1 = (Kϕ 1,ϕ 1 ). ϕ 1 =1. ( ϕ 1 lim inf f n =1. ϕ 1 < 1 g = ϕ 1 / ϕ 1 g =1 (Kg,g) = (Kϕ 1,ϕ 1 ) / ϕ 2 = S 1 / ϕ 1 2 >S 1 ). (Kϕ 1,ϕ 1 ) = sup f =1 (Kf,f) ϕ 1 K. Aϕ 1 = λ 1 ϕ 1 λ 1 = (Aϕ 1,ϕ 1 ) = S 1. λ 1 λ 2 λ n ϕ 1,ϕ 2,,ϕ n. M n ϕ 1,,ϕ n. M n K- (i.e. f M n = Kf M n ). H = M n Mn M n K-. M n K λ n+1 ϕ n+1. (Case 1) m (Kf,f)=0 f M m. Kf =0 f M m. (Case 2) Case 1 λ n, ϕ n. λ 1 λ 2, λ n 0. M ϕ 1,ϕ 2, closure. M = Mn. f M f Mn (Kf,f) λ n f 2. λ n 0 (ϕ n 0 weakly λ n = (Kϕ n,ϕ n ) 0). M (Kf,f)=0 Kf =0, f M.

21 (Hilbert-Schmidt ) K λ n, ϕ n, ϕ n =1, f = (f,ϕ n )ϕ n, f (Null K), n Kf = n λ n (f,ϕ n )ϕ n f H. 9 L = d2 + q(x), dx2 q(x) C(I), I =[0, 1] q(x) > 0onI. (L λ)u = f on I, u(0) = u(1) = 0. Green ϕ 1, ϕ 2 : Lϕ 1 = Lϕ 2 =0 on I, ϕ 1 (0) = 0, ϕ 1(0) = 1, ϕ 2 (1) = 0, ϕ 2(1) = 1 ϕ 1, ϕ 2.. ϕ 1 = cϕ 2 ϕ 1 (0) = ϕ 1 (1) = 0 0= 1 0 ( ϕ 1 + qϕ 1 ) ϕ 1 dx = 1 0 ( ϕ q ϕ 1 2) dx w := ϕ 1 (x)ϕ 2(x) ϕ 1(x)ϕ 2 (x) =Const. 0. { 1 w G(x, y) = ϕ 2(x)ϕ 1 (y) 1 w ϕ 1(x)ϕ 2 (y) x>y x<y L Green. (1) G(x, y) C(I I), G(0,y)=G(1,y)=0 0< y <1. (2) G(x, y) =G(y, x) real-valued ( ) (3) x y 2 x 2 + q(x) G(x, y) =0 (4) G x (x, y) = G(x, y) x G x (x +0,x) G x (x 0,x)= 1 0 <x<1 (Gf)(x) = 1 u C 2 (I), u(0) = u(1) = 0 GLu = u 0 G(x, y)f(y)dy

22 22 (L λ)u = f, u(0) = u(1) = 0 G (1 λg)u = Gf G Hilbert-Schmidt Hilbert-Schmidt.. f C(I) (1 λg)u = Gf (L λ)u = f, u(0) = u(1) = 0 1 H n Hilbert (, ) n n. H = {f =(f 1,f 2, ); f n H, f n 2 < } H f.g (f,g) = (f n,g n ) n H Hilbert. 2 A (Af, f) =0 f H A =0 3 Hilbert closed 4 f n f in H f n f weakly in H, f n f 5 (Af)(x) = 1 0 a(x, y)f(y)dy, f L 2 (I), I =(0, 1) a(x, y) C x y α (1) 0 <α<1 A L 2 (I) (2) <α<1/2 A Hilbert-Schmidt (3) 0 <α<1 A 6 H Hilbert V. P V P dim V< 7 H Hilbert K closed convex set f H f g = inf h K f h g K. 8 L 2 () (Af)(x) = a(x, y)f(y)dy. ρ(x, y) > 0 a(x, y) ρ(x, y)dy M <, sup x sup a(x, y) ρ(x, y) 1 dx M < y A (MM ) 1/2.

23 L 2 (0,a) (Af)(x) = 1 x x 0 f(y)dy. (Hint. 8 a(x, y) =(x/y) 1/2.) 10 Banach X R C. : X R X (norm). (1) x 0 x X. (2) x =0 x =0. (3) αx = α x α C or R, x X. (4) x + y x + y.. R n, C() =, f = sup f(x) x R n 1 p< p = L p () = {f(x) : ( 1/p f(x) p dx < }, f Lp () = f(x) dx) p. L () = {f(x) : }, f L () = ess sup x f(x). 1 p< l p = {x =(x 1,x 2, ); x n C, ( ) 1/p x n p < }, x = x n p. p = l = {x =(x 1,x 2, ); x n C, sup n x n < }, x = sup x n. n X complete X Banach. i.e. x n x m 0(m, n )= x X s.t. x n x 0 Banach. X, Y Banach A X Y. A (bounded operator) M >0 s.t. Ax M x x X

24 24 A, A. Ax sup x 0 x B(X; Y ) X Y. B(X; Y ). i.e.. (1) A 0 A B(X; Y ) (2) A =0 A =0 (3) αa = α A α C, A B(X; Y ) (4) A + B A + B X, Y Banach, B(X; Y ) Banach.. Completeness. A n B(X; Y ), A n A m 0(m, n ). x X {A n x} Y Cauchy. Y y Y s.t. A n x y 0. y x. x y map linear. A. i.e. Ax = s-lim A n x. A n A m 0 { A n Cauchy. A B(X; Y ). ɛ >0 N s.t. A m A n <ɛ, m, n > N. A m x A n x ɛ x, m, n > N. m Ax A n x ɛ x, n >N. A A n ɛ, n >N. (dual space) B(X; C) X dual space (, X. (X X = B(X; R)). X = X C, (continuous linear functional). X Banach. Hilbert H H H (Riesz ). y H f y H f y (x) =(x, y), x H H y f y H, onto,. ( f y = y ) Banach. 1 p< 1 p + 1 q =1 (lp ) l q. l q y =(y 1,y 2, ) f y (x) =, x =(x 1,x 2, ) l p y f y l q (l p ), onto,. ( f y = y l q). Hölder ( ) 1/p ( x n y n x n p n n n y n q ) 1/q f y l p. f y = y l q. f y y l q. f y y l q. p = 1,q = y l = sup n y n ɛ >0, n s.t. y n > y l ɛ. z n =(0,, 0, 1, 0, ) f y (z n ) = y n > y n l ɛ, z n l 1 = 1. f y > y l ɛ. ɛ

25 f y y l. 1 <p<, 1 <q<, y k = y k e iθ k,z k = y k q 1 e iθ k, z =(z 1,z 2, ) n z n p = n y n (q 1)p < z l p z p l = p y q lq. f y (z) = n y n y n q 1 e iθn = n y n q = y q l = y q l q z l p. f y y l q. f (l p ) y l q f = f y. e n = (0,, 0, 1, 0, ). y n = f(e n ),y =(y 1,y 2, ). y l q. p =1,q = y n = f(e n ) f e n l 1 = f y l. 1 <p<, 1 <q< y n = y n e iθn,z n = y n q 1 e iθn, z (n) =(z 1,z 2,,z n, 0, ). f(z (n) )= k n z kf(e k )= k n y k q 1 e iθ k y k = k n y k q k n y k q f z (n) l p. z (n) l p = k n z k p = k n y k p(q 1) z (n) l p = ( k n y k q ) 1/p. ( k n y k q ) 1 1/p f. n k=1 y k q <. f = f y. l p x = (x 1,x 2, ) x (n) = k n x ke k. f(x (n) ) = k n x kf(e k )= k n x ky k = f y (x (n) ) n x (n) x in l p. R n, 1 p< (L p ()) L q (),. 1 p + 1 q =1 Hahn-Banach H Hilbert, S. S f H O.K. f S unique. H = S (S) H = x = y + z,y S,z (S) F (x) =f(y). Banach space 11 Hahn-Banach 11.1 (Hahn-Banach) X = real vector space, p : X R functional s.t. p(x + y) p(x)+p(y), p(αx) =αp(x), x, y X, α 0. Y = X subspace, f : Y R linear functional f(x) p(x) x Y. = F : X R linear functional s.t. F (x) =f(x), x Y, F(x) p(x), x X.. ( ) Y X. z X \ Y. Y 1 = {x = y + αz; y Y,α R}. Y Y 1 X. f 1 (x) =f(y)+αc x = y + αz Y 1 c. f 1 : Y 1 R linear functional f 1 (x) =f(x) ifx Y. f 1 (x) p(x) x Y 1 c. x = y + αz, f 1 (x) =f(y)+αc p(y + αz) αc p(y + αz) f(y) y Y α R (1) = c p(y/α + z) f(y/α) y Y α >0 c inf {p(y + z) f(y)} (2) y Y

26 26 αc p(y + αz) f(y) y Y α R = c p( y/α z) f( y/α) y Y α <0 c sup {f(y) p(y z)} (3) y Y (1) (2) (3). (4). c satisfying (2) and (3) sup {f(y) p(y z)} inf {p(y + z) f(y)} y Y y Y f(y) p(y z) p(y + z) f(y ) y, y Y f(y)+f(y ) p(y + z)+p(y z) y, y Y. (4) f(y)+f(y )=f(y + y ) p(y + y )=p(y z + y + z) p(y z)+p(y + z). (X ). Zorn Lemma. A (g, Z) : Z X subpspace, Z Y, g : Z R linear functional, g(y) =f(y) ify Y, g(x) p(x) x Z. A (g, Z), (g,z ). (g, Z) (g,z ) Z Z, g(z) =g (z) if z Z. A. A (f,y ) A. A 1 A. A 1. (Z 0 = (g,z) A1 Z. Z 0 X subspace. Z 0 x 1,x 2,x 1 Z 1,x 2 Z 2. A 1 Z 1 Z 2. x 1,x 2 Z 2 x 1 + x 2 Z 2. ) Z 0 z (g, Z) A 1 s.t. z Z. h(z) =g(z) define. well-defined. ( (g 1,Z 1 ), (g 2,Z 2 ) A 1 s.t. z Z 1,z Z 2. (g 1,Z 1 ) (g 2,Z 2 ). g 1 (z) =g 2 (z)). h(z) linear. (z 1,z 2 Z 0. (g 1,Z 1 ), (g 2,Z 2 ) A 1 s.t. z 1 Z 1,z 2 Z 2. (g 1,Z 1 ) (g 2,Z 2 ). g 1 (z 1 + z 2 )=g 1 (z 1 )+g 1 (z 2 ) h(z 1 + z 2 )=h(z 1 )+h(z 2 ). ). (h, Z 0 ) A 1. Zorn Lemma A (F, X ). X = X. (X X F X. F ) (Hahn-Banach complex vector space version) X = vector space over C. p : X R functional s.t. p(x) 0, p(x + y) p(x)+p(y) x, y X, p(αx) = α p(x), α C x X. Y = X subspace, f : Y C linear functional s.t. f(y) p(y) y Y

27 = F : X C linear functional s.t. F (y) =f(y) y Y, F (x) p(x) x X.. X real vector space. X R. Y real vector space X R real vector subspace. Y R. g(y) =Ref(y), y Y R. g : Y R R linear functional g(y) f(y) p(y) y Y R. Hahn-Banach G : X R R linear functional G(x) p(x) x X R, G(y) =g(y) y Y R. F (x) =G(x) ig(ix). y Y F (y) =G(y) ig(iy) =g(y) ig(iy) =Ref(y) ire f(iy) =Ref(y)+iIm f(iy) =f(y) Im f(y) = Re (if(y)) = Re f(iy). F real linear F (x 1 + x 2 )=F(x 1 )+F(x 2 ), F(αx) =αf (x) α R. F (ix) =G(ix) ig( x) =G(ix) +ig(x) =if (x) a, b R F ((a + ib)x) =F (ax + ibx) =F (ax)+f(ibx) =af (x)+ibf (x) =(a + ib)f (x) F complex linear. F (x) = F (x) e iθ. F (x) = e iθ F (x) =F (e iθ x). F (e iθ x)=g(e iθ x) ig(ie iθ x). G real linear real F (e iθ x)=g(e iθ x) p(e iθ x)=p(x) F (x) p(x), x X X = normed space, Y =X subspace, f 0 : Y C bounded linear functional = f X s.t. f 0 (y) f(y) =f 0 (y) y Y, f = f 0 = sup 0 y Y y Complex Hahn-Banach p(x) = f 0 x X = normed space = 0 x X f X s.t. f(x) = x, f = Y = {αx; α C}, f 0 (αx) =α x x, y X, x y = f X s.t. f(x) f(y) x X. x = sup f(x) = f X, f 1 sup f(x) = f X, f =1 f(x) sup 0 f X f

28 28.. f f(x) f x x x sup f X, f 1 f(x)., 11.4 f X ast s.t. f =1,f(x) = x. x sup f X, f 1 f(x). 12 Baire Category (Baire) X = complete metric space, X n (n =1, 2, ) X closed set, X = X n = X n. i.e. n 0, x X n0, r >0 s.t. {y X; y x <r} X n0.. B n (n =1, 2, ) open ball of radius ρ n > 0 s.t. B 1 B 2, B n X n =, ρ n 1/n. (X 1 open ball X 1 X. x 1 X1 C. XC 1 open r 1 > 0 s.t. {y; y x 1 <r 1 } X1 C. B 1 = {y; y x 1 <ρ 1 }, ρ 1 = min{1,ρ 1 /2} B 1 X 1 =. B 1 B 2 B n. X n+1 open ball X n+1 B n. i.e. B n Xn+1 C. x n+1 B n Xn+1 C. B n Xn+1 C open r n+1 > 0 s.t. {y; y x n+1 < r n+1 } B n Xn+1. C ρ n+1 = min{1/(n +1),r n+1 /2}, B n+1 = {y; y x n+1 <ρ n+1 }, B n+1 B n, B n+1 X n+1 =, ρ n+1 1/(n + 1).) B n x n Cauchy. (m>n B m B n = {y; y x n <ρ n }, x m x n ρ n 1/n 0asn.) X complete lim n x n = x in X. m>n x m B m B n. m x B n. B n X n = x X n, n. X = X n. ( ) X Banach space, Y normed space, T n B(X; Y ),n =1, 2,, sup n T n x <, x X = sup T n < n. X n = {x X; sup k T k x n}. X = n X n, X n closed. (x i X n,x i x. T k x i n, k, i i T k x n. sup k T k x n.) Baire Category n 0, x 0 X n0, r > 0 s.t. {y; y x 0 <r} X n0. i.e. y x 0 <r= sup k T k y n 0. z <r T k z = T k (x 0 + z) T k x 0 T k z T k (x 0 + z) + T k x 0 2n 0. z < 1 T k z 2n 0 /r. sup k T k 2n 0 /r. X, Y Banach spaces, T n B(X; Y ),, 2,. x X T n x Y. = Tx= lim n T n x T B(X; Y ). X, Y, Z Banach spaces, T n B(X; Y ), S n B(Y ; Z), n =1, 2,, T n T strongly, S n S strongly = S n T n ST strongly.. sup n T n <, sup n S n <. x X y = Tx S n T n x STx S n (T n Tx) + (S n S)Tx C T n x Tx + S n y S y. n. 1 p. f R n. fg L p (), g L p () = f L ().

29 k = {x ; f(x) k}, χ k k. f k = χ k f. X L p () T k g = f k g T k g L p fg L p sup k T k g L p <. sup k T k <. T k = ess.sup x k f(x) ess.sup x f(x) <. 13 Banach ( ) X, Y Banach spaces, T B(X; Y ), T onto = T open map i.e. O open set in X = TO open set in Y.. X n = {x X; x <n}. Step 1. ρ >0 s.t. TX 1 {y Y ; y <ρ}. (X = n X n, T onto Y = n TX n = n TX n. Baire Category n 0, y 0 Y, r 0 > 0 s.t. {y; y y 0 <r 0 } TX n0. y; y <r 0 } TX 2n0. y <r 0 y =(y + y 0 ) y 0 TX n0 TX n0 y k,z k X n0 s.t. Ty k y + y 0,Tz k y 0. y k + z k < 2n 0 y k z k X 2n0. T (y k z k ) TX 2n0. k y + y 0 y 0 TX 2n0.) ρ = r 0 /(2n 0 ) {y Y ; y <ρ} TX 1. y <ρ= 2n 0 y <r 0 = 2n 0 y TX n0 = y 1 2n 0 TX n0 TX 1. Step 2. TX 1 {y Y ; y <ρ/2}. (y Y, y <ρ/2 x X s.t. x < 1,Tx= y. ɛ k > 0 k=1 ɛ k =1/2. η k = ɛ k ρ. y <ρ/2 y TX 1/2. x 0 X 1/2 s.t. y Tx 0 <η 1. Step 1 y Tx 0 TX η1/ρ = TX ɛ1. x 1 X ɛ1 s.t. y Tx 0 Tx 1 <η 2. x k X ɛk s.t. y Tx 0 Tx 1 Tx k <η k. k x k x 0 + k ɛ k < 1 x = k=0 x k X. ɛ k 0 y = Tx.) Step 3. proof. X O = open, x 0 O. δ >0 s.t. {x X; x x 0 <δ} O. T {x; x x 0 <δ} = Tx 0 + TX δ TO Step 2 TX δ {y Y ; y <ρδ/2}. Tx 0 O. X, Y Banach spaces, T B(X; Y ), T 1to1,onto= T 1 B(Y ; X).. T open map T 1 conti. (Banach closed ( graph theorem) X, Y Banach ) spaces, T X Y linear operator x n x, T x n y = Tx= y = T B(X; Y ).. X Z x Z = x X + Tx Y. X Z Z. Z Banach space. ( x n x m Z 0. x n x m X + Tx n Tx m Y 0, x n x in X, Tx n y in Y. Tx = y. x n x Z = x n x X + Tx n Tx Y 0.) Z X S Sx = x. S B(Z; X). ( Sx X = x X x Z.) S 1 to 1, onto. S 1 B(X; Z). S 1 x Z = x Z = x X + Tx Y C x X. Tx Y (C 1) x X. (Cauchy Problem)

30 30 I =[0,T] T>0 (CP) ( ) m m 1 ( ) m k ( ) k u + a k (x, t) u =0, t I, x R t x t k=0 ( ) k u(x, 0) = f k (x), 0 k m 1 t. B k = {f(x) C k (R); sup f (n) (x) < } x R n k Banach space. f B k = sup f (n) (x) x R n k. Y k = C k (I; B m k ) B m k I C k -. u(t) Y k u Y k = sup t I l k ( ) l d u(t) dt Bm k. u Y := m k=0 Y k u Y = sup t I u Y k k. Y Banach space. f =(f 0,f 1,,f m 1 ) B m B m 1 B 1 =: X f X = k f k B m k. X Banach space. f X f CP u Y,. i.e. C >0 s.t. u Y C f X.. f X u S. S X linear. f (n) f in X, Sf (n) u in Y = Sf = u. (u (n) = Sf (n) u (n) f (n) CP. n u f CP. Sf = u.) Banach closed graph theorem S. II 1 (1) ( d dx ) 2u = λu, 0 <x<2π, u(0) = u(2π) =0

31 u(0) = u(2π) =0 A = d 2 /dx 2,. (2) z C, ( ( d ) ) 2 z u = f, 0 <x<2π, dx u(0) = u(2π) =0.. 2 C(T) R f(x) 2π, f(x +2π) =f(x), x R. (1) C(T) Banach. (2) a(x) C(T). A Af = f = sup f(x) x R A =. (3) x 0 [0, 2π], 2π 0 2π 0 a(x)f(x)dx a(x) dx S n f = 1 2π sin(2n +1) x x 0 2 2π 0 sin x x f(x)dx 0 2. S n,n. (4) S n f f(x) Fourier n. (5), f(x) C(T) Fourier f(x) x 0. 3 X Banach T B(X; X) T < 1.. (1) 1 T 1to1,onto. (2) (1 T ) 1 = n=0 T n,. (3) (1 T ) 1 (1 T ) 1. 4 X = C([a, b]), a(x, y) C([a, b] [a, b]) Af(x) = x a a(x, y)f(y)dy, f X. (1) M = sup (x,y) [a,b] 2 a(x, y) A n (M n /n!)(b a) n. (2) u X u(x) x a a(x, y)u(y)dy = f(x)

32 32 f X, X = normed space over C, X Y complete subspace (i.e. Cauchy ), x 0 X = Z = {λx 0 + µy ; λ, µ C,y Y } complete subspace.. x 0 Y. Z Cauchy {z n }.. z n = α n x 0 + y n, y n Y, α n C Claim : {α n }. (. {α n } {α in } {α n } α n. x 0 = z n /α n + y n /α n { z n } y n /α n x 0. y n Y Y closed x 0 Y.) Bolzano-Weierstrass {α n }. α n α. y n = z n α n x 0 Cauchy y n y Y. z n = α n x 0 + y n αx 0 + y Z normed space X Y complete ( closed).. dim Y =0 O.K. dim Y = n O.K.. dim Y = n +1 Y x 1,,x n+1, Y x 1,,x n, Y n complete. Y x n+1 Y n. X M, u X u M. inf u v = dis (u, M) v M 14.3 (1) dim M< u M attain v M. (2) u M attain.. Exercise. u M attain. X = C(I),I =[0, 1], M = {cx(t); c R}, x(t) =t. u =1 v = αt u v = max αt 1 = 0 t 1 α 1 (α 2) 1 (0 α 2) 1 α (α 0) dis (u, M) =1, u(t) αt, 0 α 2.

33 X normed space, M closed subspace, M X. (1) ɛ >0, u 0 X s.t. u 0 =1, dis (u 0,M) 1 ɛ. (2) dim M< u 0 X s.t. u 0 =1, dis (u 0,M)=1.. u X s.t. dis (u, M) > 0. u u X inf v M u v =0. u X, v n M s.t. u v n 0. M closed u X X = M ). (1) proof. d = dis (u, M). ɛ >0, v ɛ M s.t. d u v ɛ d + ɛ. u 0 = (u v ɛ )/ u v ɛ. u 0 =1, v M u 0 v = dis (u 0,M)= 1 u v ɛ (u v ɛ u v ɛ v) d u v ɛ d d + ɛ =1+O(ɛ). (2) proof. u attain v M v 0. u 0 =(u v 0 )/d dim X =. u 1 =1 u 1 X u 1 X 1. u 2 =1, dis (u 2,X 1 )=1 u 2 u 1,u 2 X 2. {u n }, u 1,,u n X n dis (x n+1,x n )=1. m<n x m X m X n 1 x n x m 1, {x n } Cauchy. 0.4 H Hilbert, H 2 = H H. A G(A) ={(u, Au); u D(A)}. (1) A = B G(A) =G(B). (2) A B G(A) G(B). A closed operator G(A) closed subspace in H 2. T, J : H 2 H 2 T :(u, v) (v, u), J :(u, v) (v, u). (1) T, J. (2) TJ = JT. (3) T 2 = J 2 = I.

34 34 (JG(A)) = J(G(A)). Proof. JG(A) (f,g) G(A) J(f,g) G(A) (g, f) (g, u) (f,au) =0, u D(A) (f,g) JG(A). (JG(A)) D(A) =H. Proof. ((0,h), (Au, u)) = (h, u), u D(A). =. D(A) h (0,h) (JG(A)).(JG(A)) (0,g) (JG(A)) = g =0. h =0.. (0,h) (JG(A)) 0= (h, u), u D(A). D(A) dense h =0. G(A )=(JG(A)) = JG(A). Proof. g D(A ), A g = g (Af, g) =(f,g ), f D(A) ( J(f,Af), (g, g ) ) =0, f D(A) G(A ) JG(A). A is densely defined and closed = D(A ) is dense. There exists (A ) = A and A = A. Proof. h D(A ) = (0,h) (A g, g), g D(A ) = (0,h) JG(A ) = (0,h) (JG(A )) =(G(A) ) = G(A) =G(A) = h =0. D(A ) dense. (A ). G(A ) = (JG(A ) = (G(A) ) = G(A) A = A. A 1,A,(A 1 ) = (A ) 1 (A ) 1 =(A 1 ) Proof. G(A 1 )=TG(A) G((A 1 ) ) = (JG(A 1 )) =(JTG(A)) =(TJG(A)) = T(JG(A)) = TG(A ) (A ) 1 (A 1 ).