Functional Analysis I Solutions to Exercises. James C. Robinson

Transcription

1 Functionl Anlysis I Solutions to Exercises Jmes C. Robinson

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3 Contents 1 Exmples I pge 1 2 Exmples II 5 3 Exmples III 9 4 Exmples IV 15 iii

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5 1 Exmples I 1. Suppose tht v α j e j nd v m β k f k. with α j, β k K nd e j, f k E. Relbel β k nd f k so tht f j e j for j 1,..., n, nd f j / {e 1,..., e n } for j > n, i.e. v β k e k + n k1 m kn +1 k1 β k f k, with the understnding tht the first sum is zero if n, nd the second zero if n m. It follows tht or n n m α j e j β k e k β k f k, k1 kn +1 (α j β j )e j m α j e j + β k f k. kn +1 kn +1 Since E is linerly independent, it follows tht α j β j for j > n, from which n n, nd tht α j β j for j 1,..., n. So the expnsion is unique. 2. If x, y l lim (C) then x + y l lim (C) nd αx l lim (C), since lim (x n + y n ) lim x n + lim y n nd lim αx n α lim x n. n n n n n 1

6 2 1 Exmples I 3. Write x p l p j x j p j x j p α x j α. Hölder s inequlity with exponents nd b yields 1/ x p l x p j (p α) x j αb j j 1/b x p α l (p α) x α l αb. We wnt to choose α, nd conjugte exponents, b ( 1 + b 1 1) so tht (p α) q nd αb r, so tht the fctors on the right re powers of x l q nd x l r. Thus b r/α; since, b re conjugte it follows tht r/(r α), nd then the condition tht (p α) q gives α (p q)r/(r q), (r q)/(r p), b (r q)/(p q). The result follows s stted on substituting for α. 4.(i) If there exist, b with < b such tht x 1 x 2 b x 1 then while (ii) if 1 b x 2 x 1 1 x 2, x 1 x 2 b x 1 nd α x 2 x 3 β x 2 then α x 1 x 3 bβ x Since (by the previous question) being equivlent is n equivlence reltion (!), we only need to show tht 2 nd re equivlent to 1.

7 Exmples I 3 We hve x j 2 x j 2 n x j 2, which gives x 2 x 1 n x 2, nd mx x j,...,n x j n mx,...,n x j, i.e. x x 1 n x. 6. First we show tht if f is continuous nd K is closed then f 1 (K) {x X : f(x) K} is closed. (In fct this is in the proof of Corollry 3.12 in the notes.) Tke x n f 1 (K), nd suppose tht x n x; since f is continuous, f(x n ) f(x). But f(x n ) K, nd K is closed, so f(x) K. It follows tht x f 1 (K), nd so f 1 (K) is closed. Now suppose tht whenever K is closed, f 1 (K) is closed. Tke x n X with x n x, nd suppose tht f(x n ) f(x). Then for some subsequence x nj, we must hve f(x nj ) f(x) > ɛ. So f(x nj ) is contined in the closed set Z Y \ B(f(x), ɛ), where B(f(x), ɛ) {y Y : y f(x) Y < ɛ} is open. It follows tht f 1 (Z) is closed. Since f(x nj ) Z, x nj f 1 (Z); since x nj x nd f 1 (Z) is closed, it follows tht x f 1 (Z), so tht f(x) f(x) > ɛ, clerly contrdiction. 7. Let Y be closed subset of the compct set K. If y n Y then y n K, so there is subsequence y nj such tht y nj y K. Since Y is closed, y Y, nd so Y is compct. 8. If {u n } is Cuchy sequence in U then, since U V nd the norm on U is simply the restriction of the mp u u V to U, it follows tht {u n } is lso Cuchy sequence in V. Since V is complete, u n u for some u V. But since U is closed, we know tht if u n u then u U. So U is complete.

8 4 1 Exmples I 9. As it stnds the question is mbiguous, since one needs to specify norm on c (K). To be interesting the question requires the l norm; ny element of l p with p < must be n element of c (K). To show tht c (K) (with the l norm) is complete, we will first show tht l (K) is complete (this cse ws omitted from the proof of Proposition 4.5 in the notes, put the proof is simpler), nd then tht c (K) is closed, which will show tht c (K) is complete by the previous question. So first suppose tht x k (x k 1, xk 2, ) is Cuchy sequence in l (K). Then for every ɛ > there exists n N ɛ such tht x n x m p l sup x n j x m j < ɛ for ll n, m N ɛ. (1.1) j In prticulr {x k j } k1 is Cuchy sequence in K for every fixed j. Since K is complete (recll K R or C) it follows tht for ech k N x k j k for some k R. Set ( 1, 2, ). We wnt to show tht l nd lso tht x k l s k. First, since {x k } is Cuchy we hve from (1.1) tht x n x m l < ɛ for ll n, m N ɛ, nd so Letting m we obtin sup x n j x m j ɛ. j sup x n j j ɛ, j nd so x k l. But since l is vector spce nd x k l, this implies tht l nd x k l ɛ. Now tke x n c (K) such tht x n x in l. Suppose tht x / c (K). Then there exists δ > nd sequence n j such tht x nj > δ for every j. Now choose N lrge enough tht x n x l < δ/2 for ll n N. In prticulr it follows tht x n n j > δ/2 for ll n N (nd every j); but then x n / c (K), contrdiction.

9 2 Exmples II 1. If x (x 1, x 2,...) l 2 (R), then x j 2 <. So x (x 1, x 2,..., x n,,,...) 2 l x 2 j 2 jn+1 s n. So l f (R) is dense in l 2 (R). 2. If x X Y then there exist ɛ X nd ɛ Y such tht {y B : y x < ɛ X } X nd {y B : y x < ɛ Y } Y. Tking ɛ min(ɛ X, ɛ Y ) it follows tht {y B : y x < ɛ} X Y, nd so X Y is open. Now, given z B nd ɛ >, since X is dense there exists n x X such tht x z < ɛ/2. Since X is open, there is δ < ɛ/2 such tht {x B : x x < δ} X. Since Y is dense, there is y Y such tht y x < δ/2. By the bove, it follows tht we lso hve y X. So we hve found y X Y such tht y z y x + x z < δ/2 + ɛ/2 < ɛ, nd so X Y is dense. A refinement of this rgument llows one to prove the powerful Bire Ctegory Theorem: countble intersection of open nd dense sets is dense. 5

10 6 2 Exmples II 3. We hve to ssume tht (Y, Y ) is Bnch spce. Then if x n X nd x n x (with x V ) we know tht {x n } is Cuchy in V. So, since F (x n ) F (x m ) Y L x n x m it follows tht {F (x n )} is Cuchy sequence in Y. Since Y is complete, we know tht lim n F (x n ) exists nd is n element of Y. If x n v, nd y n v, then F (x n ) F (y n ) L x n y n. Tking limits s n on both sides implies tht lim F (x n) lim F (y n). n n Tht F so defined is continuous is cler: for ny v 1, v 2 V, one cn find sequences in X tht converge to v 1, v 2, nd then from the definition F (v 1 ) F (v 2 ) L v 1 v Let V R, X Q, nd expnd ech q Q to the intervl I q (q d(q), q + d(q)), where d(q) q 2 /2. Then 2 / I q for ny q, so q Q I q does not cover R. 5. Fix n N, nd cover A j with collection of intervls I (j) k such tht k I(j) k < 2 (n+j). Then j A j is covered by I (j) k, j,k nd j,k I(j) k < jn+1 2 j 2 n. If P j occurs lmost everywhere then it fils on set A j of mesure zero. Since the union of the A j still hs mesure zero, every P j occurs simultneously lmost everywhere. 6. Use φ n f to men tht φ n is n incresing sequence tht converges lmost everywhere to f. (i) is essentilly Lemm 5.5. (ii) If f, g L inc (R) then there re sequences {φ n }, {ψ n } L step (R) such tht φ n f nd ψ n g. Then {φ n + ψ n } is n incresing sequence of step functions with φ n + ψ n f + g. So f + g L inc (R), nd since the integrl is dditive on L step (R), f + g lim n φ n + ψ n lim ( n φ n + ψ n ) f + (iii) If {φ n } L step (R) with φ n f then λφ n L step (R) nd λφ n f. The result follows from the fct tht λφ n λ φ n. g.

11 Exmples II 7 (iv) Follows from the sme properties for step functions, long similr lines to the bove. 7. For x [n, n + 1) we hve x n 1 f(r) dr + x n f(r) dr n 1 ( 1) j j + (x n) ( 1)n n, nd so x lim x f(x) dx ( 1) j j! <. But the sme clcultion with f(x) gives x n 1 f(r), dr n 1 which diverges s n. [Functions cn be integrble in some sense even when they re not Lebesgue integrble.] 8. Consider f n n f. Then f n is monotonic sequence, nd f n n f for every n. So there exists g L 1 such tht f n g lmost everywhere. The limit s n of f n is zero where f, nd + where f n. Since g is defined lmost everywhere, it follows tht f lmost everywhere. 9. If f L inc (R) then there exists sequence {φ n } L step (R) such tht φ n f (lmost everywhere), nd we define f lim φ n. n 1 j, Since f φ n, f φ n f φ n, which tends to zero s n by the definition of f. Now tke step function φ(x) c j χ Ij (x) with the endpoints of I j being j nd b j. Consider the continuous version

12 8 2 Exmples II of χ I (x), for n intervl I with endpoints, b, x < δ then (x + δ)/δ δ x X I (x; δ) 1 < x < b Now if δ < ɛ/(n mx j c j ) nd (b + δ x)/δ b x b + δ x > b + δ. φ ɛ (x) c j X Ij (x; δ), φ ɛ (x) φ(x) dx < ɛ. So if g L 1 (R), we hve g f 1 f 2 L inc (R). Approximte f 1 nd f 2 to within ɛ/4 (wrt the L 1 norm) by step functions h 1 nd h 2, nd then pproximte h 1 nd h 2 by continuous functions c 1 nd c 2 to within ɛ/4 (gin wrt the L 1 norm). So c 1 c 2 C (R) pproximtes g to within ɛ, i.e. C (R) is dense in L 1 (R). 1. With g n ( n k1 f k ) 2 we hve g n ( f k ) 2 2 ( 2 ( ) 2 f k f k L 2) f k L 2 k1 k1 L 2 using the definition of the L 2 norm nd the tringle inequlity. It follows tht g n g lmost everywhere for some g L 1. It follows tht k1 f k is bsolutely convergent lmost everywhere to some f L Note tht we hve 2 f 2 ( 2 f k f k u k ) f 2. k1 kn+1 k1 kn+1 So we cn use the DCR to deduce tht lim f f k 2 lim f u k 2. n n k1 k1 k1

13 3 Exmples III 1. Expnding the right-hnd side gives (x + y, x + y) (x y, x y) + i(x + iy, x + iy) i(x iy, x iy) x 2 + (x, y) + (y, x) + y 2 x 2 + (x, y) + (y, x) y 2 + i( x 2 + i(y, x) i(x, y) + y 2 x 2 + i(y, x) i(x, y) y 2 ) 4 rel(x, y) + 4 img(x, y) 4(x, y) 2. Write ( t 2 ( t f(r) dt) s s ( t s ) 2 r 1/2 r 1/2 f(r) dr ) ( t ) r 1 dr r f(r) 2 dr s K(log t log s) 1/2 using the Cuchy-Schwrz inequlity. [K 2 in exmples sheet is misprint.] 3. Suppose tht E is orthonorml nd tht α j e j for some α j K nd e j E. Then tking the inner product of both sides with e k (with k 1,..., n) gives ( α j e j, e k ) α k, so α k for ech k, i.e. E is linerly independent. 9

14 1 3 Exmples III 4.(i) Suppose tht d. Then there exists sequence y n Y such tht lim x y n, n i.e. y n x. Since Y is closed liner subspce, it follows tht x Y but this contrdicts the fct tht x / Y. (ii) By definition there exists sequence y n Y such tht d lim n x y n. So for n sufficiently lrge, clerly x y n < 2d. Let z be one such y n. (iii) We hve ˆx y x z x z y 1 x z x (z + x z y) > 1 2d d 1 2, since z + x z y Y nd d inf{ x y : y Y }. 5. Tke x 1 B with x 1 1. Let Y 1 be the liner subspce {αx 1 : α K}. Using the result of question 7 there exists n x 2 B with x 2 1 such tht x 2 x 1 > 1 2. Now let Y 2 Spn(x 1, x 2 ), nd use question 7 to find x 3 B with x 3 1 with x 3 y > 1 2 for every y Spn(x 1, x 2 ) in prticulr x 3 x 1 > 1 2 nd x 3 x 2 > 1 2. Continuing in this wy we obtin sequence {x n } B such tht x n 1 nd x n x m > 1 2 for ll n m. It follows tht the unit bll is not compct, since no subsequence of the {x n } cn be Cuchy sequence. 6. The unit bll is closed nd bounded. If B is finite-dimensionl then this implies tht the unit bll is compct. We hve just shown, conversely, tht if B is infinite-dimensionl then the unit bll is not compct. 7. Tke x, y A nd suppose tht λx + (1 λ)y A for ll λ [, 1] such tht 2 k λ N. Then given µ with 2 k+1 µ N, 2 k µ / N we hve with l odd, so tht µ (l 1) 2 µ l2 (k+1) 2 (k+1) + (l + 1) 2 2 (k+1). With λ 1 (l 1)2 (k+1) nd λ 2 (l + 1)2 (k+1) we hve 2 k λ 1, 2 k λ 2 N, nd so µx + (1 µ)y 1 2 [(λ 1x + (1 λ 1 )y) + (λ 2 x + (1 λ 2 )y)],

15 with λ j x + (1 λ j )y A, i.e. µx + (1 µ)y A. Exmples III 11 It follows by induction tht λx + (1 λ)y A for ny λ [, 1] such tht 2 k λ N for some k N. Since ny λ [, 1] cn be pproximted by sequence λ j such tht 2 j λ j N, it follows since A is closed tht λx + (1 λy) A for ll λ [, 1], i.e. A is convex. 8. We hve x (X ) if for every y X, (x, y). So clerly X (X ). If we do not hve equlity here then there exists z / X with (z, y) for every y X. But we know tht z x + ξ with x X nd ξ X ; but then (x + ξ, ξ), which implies tht xi 2, i.e. x X. So (X ) X s climed. 9. We hve e 4 x 3 ( x 3, ( x 3, ) ( ) (3x2 1) 8 (3x2 1) x 3, 2 x 2 x ) x (3x2 1) 1(3t 5 t 3 ) dt 3x 2 x 3 3x 5. 1 t 4 dt t 3 dt Then nd so e ( t 3 3t ) 2 [ ] t 7 1 dt 5 7 6t t e (5x3 3x). 1. Approximtion of sin x by third degree polynomil f 3 (x): note tht (sin x, e 1 ) nd (sin x, e 3 ) will be zero since sin x is odd nd e 1 nd e 3 re even, so f 3 (x) 3x 2 1 t sin t dt + 7(5x3 3x) 8 1 (5t 3 3t) sin t dt. Now, 1 t sin t dt [ t cos t] cos t dt [sin t] sin 1 1

16 12 3 Exmples III nd 1 t 3 sin t dt [ t 3 cos t ] [ 3t 2 sin t ] sin 1, 1 3t 2 cos t dt t sin t dt giving it follows tht f 3 (x) sin 1 1 (5t 3 3t) sin t dt 36 sin 1; [ 3x + 7(27x ] 45x3 ) sin 1 2 [ 195x 315x First wy: given orthonorml polynomils ẽ j (x) on [ 1, 1] put since then e j (x)e k (x) dx 2 Doing this gives e j (x) 2 ẽ j (2x 1), ẽ j (2x 1)ẽ k (2x 1) dx e 1 (x) 1, e 2 (x) 3 (2x 1), 1 2 ]. ẽ j (y)ẽ k (y) dy. nd e 3 (x) 5 2 [3(4x2 4x + 1) 1] 5 (6x 2 6x + 1), 7 [ e 4 (x) 5(8x 3 12x 2 + 6x 1) 3(2x 1) ] 2 7 [2x 3 3x x 1]. Or the pinful wy: e 1 (x) 1, then nd then e 2 2 e 2(x) x (x 1 2 )2 dx t dt x 1 2, [ ] (x )

17 Exmples III 13 which gives e 2 (x) 3 (2x 1). Now set ( ) e 3(x) x 2 3(2x 1) t 2 (2t 1) dt nd then e 3 2 [ ] t x (2x 1) 2 t x 2 2x x2 x + 1 6, (t 2 t )2 dt [ t 5 5 t t3 9 t2 6 + t , nd so e 3 (x) 5(6x 2 6x + 1). Finlly, we hve with t 2 dt t 4 2t 3 + 4t2 3 t dt ] 1 ( e 4(x) x 3 5(6x 2 6x + 1) e ) t 3 (6t 2 6t + 1) dt ( ) 3(2x 1) t 3 (2t 1) dt x 3 6x2 6x (2x 1) 1 4 2x3 3x x 1 2 (2t 3 3t t 1) 2 dt t 3 dt 4t 6 + 9t t t t 4 4t 3 72t 3 +6t 2 24t dt 1 [ ] , nd so e 4 (x) 7 (2x 3 3x x 1) s bove.

18 14 3 Exmples III 12. Given x, y l 2 w, set ˆx j w 1/2 j x j nd ŷ j w 1/2 j y j. Then (x, y) l 2 w (ˆx, ŷ) l 2, so the fct tht (, ) l 2 w is n inner product on l 2 w follows from the fct tht (, ) l 2 is n inner product on l 2. Similrly the completeness of l 2 w follows from the completeness of l 2 if {x n } is Cuchy sequence in l 2 w then {ˆx n } is Cuchy sequence in l 2, so there exists y l 2 such tht ˆx n y. Setting z j y j /w 1/2 j gives z l 2 w such tht x n z in l 2 w.

19 4 Exmples IV 1. If x X 1 then Ax Y x X A x x X A x x X, nd so since (x/ x X ) X 1 we hve sup Ax Y x X 1 sup Az Y z X 1 Rerrnging the bove equlity we hve Ax Y x X A x x X, Y nd so, gin since (x/ x X ) X 1, sup Az Y. z X 1 2. Given x l 2, we hve Ax Y sup x x X T x 2 l 2 j sup Ax Y. x 1 x j 2 j 2 j x j 2, so T 1. Now, for ny fixed n one cn tke x n (1,, 1,, ), n element of l 2 whose first n entries re 1 nd whose others re zero. Then T x n (1, 1/2, 1/3,..., 1/n,, ) which is in l 2. {T x n } forms sequence in l 2 tht converges to (1, 1/2, 1/3, 1/4,...), which is n element of l 2 since n (1/n2 ) is finite. However, the preimge 15

20 16 4 Exmples IV of this would be the sequence consisting ll of 1s, which is not n element of l 2. So the rnge of T is not closed. 3. Clerly f φ is liner. Given u C ([, b]) we hve f φ (u) φ(t)u(t) dt ( mx u(t) t [,b] ) φ(t) dt u φ(t) dt, (4.1) nd so f φ φ(t) dt. Now since φ is continuous we cn find sequence of continuous functions u n tht pproximte the sign of φ (+1 if φ >, 1 if φ < ) incresingly closely in the L 2 norm: if φ(s) then there is n intervl round s on which φ hs the sme sign. Since this intervl hs non-zero length, nd the union of these intervls lies within [, b], there re t most countble number, {I j }. Let u n be the continuous function defined on I j (l j, r j ) s (t l j )/δ j,n l j t l j + δ j,n u n (t) sign(φ(t)) 1 l j + δ j,n < t < r j δ j,n (r j t)/δ j,n r j δ j,n t r j where δ j,n min(2 (n+j), (r j l j )/2), nd zero when φ(t). Then u n (t) sign(φ(t)) 2 dt 2 (n+j) 2 (n 1), i.e. u n sign(φ) in L 2 (, b). Note tht u n 1. It follows tht f φ (u n ) φ(t)u n (t) dt φ(t) dt φ(t)(u n (t) sign(φ(t))) dt φ(t) dt φ(t)(u n (t) sign(φ(t))) dt φ(t) dt φ L 2 u n sign(φ) L 2. Since φ C ([, b]) we know tht φ L 2 <, nd we hve just shown

21 Exmples IV 17 tht u n sign(φ) L 2 s n. It follows since u n 1 tht we cnnot improve the bound in (4.1) nd therefore 4. Tke x, y L 2 (, 1), then (T x, y) f φ ( t t t ( t (x, T y) φ(t) dt. ) K(t, s)x(s) ds y(t) dt K(t, s)x(s)y(t) ds dt K(t, s)x(s)y(t) dt ds ) K(t, s)y(t) dt x(s) ds (drw picture to see how the limits of integrtion chnge) where 5. If x (rnge(t )) then T (y)(s) t K(t, s)y(t) dt. (x, y) for ll y rnge(t ), i.e. (x, T z) for ll z H. Since (T ) T, this is the sme s (T x, z) for ll z H. This implies tht T x, i.e. tht x Ker(T ). This rgument cn be reversed, which gives the required equlity. Now with pologies for the mistke in the question: we will show tht λ is n eigenvlue of T suppose tht rnge(t λi) H. Then there exists some non-zero u (rnge(t λi)). By the equlity we hve just proved, this gives non-zero u in Ker((T λi) ), i.e. non-zero u such tht (T λi) u. Since (T λi) T λi, this gives non-zero u with T u λu. So λ is n eigenvlue of T.

22 18 4 Exmples IV 6.(i) If T B(H, H) then T B(H, H) nd T op T op. So if {x n } is bounded sequence in H, with x n M, sy, it follows tht T x n T op x n M T op. So {T x n } is lso bounded sequence in H. Since T is compct, it follows tht {T (T x n )} {(T T )x n } hs convergent subsequence, which shows tht T T is compct. (ii) Following the hint, we hve T x 2 (T x, T x) (T T x, x) T T x x. Now, if {T T x n } is Cuchy then given ny ɛ >, there exists N such tht for ll n, m N, T T x n T T x m T T (x n x m ) ɛ. It follows tht for ll n, m N, T x n T x m 2 T (x n x m ) 2 T T (x n x m ) x n x m. Since {x n } is bounded sequence, with x n M, sy, we hve T x n T x m 2 Mɛ for ll n, m N. It follows tht {T x n } is Cuchy. (iii) So suppose tht {x n } is bounded sequence in H. Prt (i) shows tht T T is compct, so {T T x n } hs subsequence {T T x nj } tht is Cuchy. But prt (ii) shows tht this implies tht {T x nj } is Cuchy too. So T is compct. 7.(i) If for every z H we hve (x n, z) (x, z) nd (x n, z) (y, z) then clerly (x, z) (y, z) for every z H. But then (x y, z) for every z H; in prticulr we cn tke z x y, which shows tht x y 2, i.e. tht x y. (ii) Tke y H, nd suppose tht (e n, y) does not tend to zero s n. Then for some ɛ >, thee exists sequence n j such tht (e nj, y) > ɛ. But then y 2 n (y, e j ) 2 j (y, e nj ) 2, contrdicting the fct tht y H with {e j } n orthonorml bsis.

23 Exmples IV 19 (iii) For some fixed z H, consider the mp f : H K given by This mp is clerly liner, nd it is bounded, u (T u, z). u + λv (T (u + λv), z) (T u, z) + λ(t v, z), (T u, z) T u z ( ) T op z u. So f H. It follows tht there exists y H such tht (T u, z) f(u) (u, y) for every u H. So if u n u, in prticulrly for this choice of y we hve (u n, y) (u, y) (T u n, z) (T u, z). Since this holds whtever our choice of z, it follows tht T u n T u. (iv) **This prt requires you to know tht ny wekly convergence sequence is bounded, which is not t ll obvious (but true). Suppose tht T x n T x. Then there exists δ > nd subsequence T x nj such tht T x nj T x > δ for ll j 1, 2, 3,.... (4.2) Since T is compct nd {x nj } is bounded there exists further subsequence such tht T x nj w for some w H. But since convergence implies wek convergence, we must hve T x nj w. But we lredy know tht T x nj T x, so it follows from the uniqueness of wek limits tht w T x, nd so T x nj T x, contrdicting (4.2). 8. (i) For n x with K(x, ) L 2 (, b), use the fct tht {φ j } is bsis for L 2 to write K(x, y) k i (x)φ i (y). Since k i (x) i1 K(x, y)φ i (y) dy,

24 2 4 Exmples IV we hve k i (x) 2 dx K(x, y)φ i (y) dy ( 2 dx ) ( K(x, y) 2 dy φ i (y) 2 dy ) φ i (y) 2 dy dx K(x, y) 2 dy dx, nd so k i L 2 (, b). (ii) Since k i L 2 (, b), we cn write k i (x) j κ ijφ j, where κ ij k i(x)φ j (x) dx, i.e. K(x, y) i,j κ ij φ i (y)φ j (x), nd so {φ i (y)φ j (x)} is bsis for L 2 ((, b) (, b)). 9. Clerly T n x λ n j (x, e j )e j. If λ j < 1 for ll j 1,..., n then T < 1, nd so (I T ) 1 I + T + T 2 +, i.e. (I T ) 1 I + λ j (x, e j )e j + λ 2 j(x, e j )e j + I + I + (λ j + λ 2 j + )(x, e j )e j λ j 1 λ j (x, e j )e j. 1.(i) Since x f + αt x, the solution is given by x (I αt ) 1 f. So if αt op < 1 we cn write x (I + αt + α 2 T 2 + α 3 T 3 + )f (αt ) j f. So this expnsion is vlid for α < 1/ T op. (ii) Suppose tht (T n 1 x)(t) j K n 1 (t, s)x(s) ds.

25 Then i.e. with s climed. 11. We hve (T n x)(t) (T u, v) Exmples IV 21 K(t, s)(t n 1 x)(s) ds ( K(t, s) ( (T n x)(t) K n (t, r) ) K n 1 (s, r)x(r) dr ds K(t, s)k n 1 (s, r)x(r) dr ds ) K(t, s)k n 1 (s, r) ds x(r) dr, λ j (u, e j )e j, v K n (t, r)x(r) dr. K(t, s)k n 1 (r, s) ds λ j (u, e j )(e j, v) u, λ j (e j, v)e j u, λ j (v, e j )e j (u, T v) nd so T is self-djoint. Now consider T n u defined by T n u λ j (u, e j )e j. it is finite- Since the rnge of T is contined in the spn of {e j } n dimensionl, nd so T n is compct. Since T n u T u 2 2 λ j (u, e j )e j jn+1 λ j 2 (u, e j ) 2, jn+1

26 22 4 Exmples IV we hve from the fct tht λ n tht given ny ɛ > there exists n N such tht for ll n N we hve λ j < ɛ, nd hence i.e. T n u T u 2 ɛ 2 (u, e j ) 2 ɛ 2 (u, e j ) 2 ɛ 2 u 2, jn+1 T n T op ɛ, nd so T n T. It follows from Theorem 13.8 tht T is compct. 12. We hve T u K(x, y)u(y) dy λ j e j (x)e j (y)u(y) dy λ j e j (e j, u), nd so T e k λ k e k. To show tht these re the only eigenvlues nd eigenvectors, if u L 2 (, b) with u w + k1 (u, e k)e k nd T u λu then, since w e j for ll j, nd T u T w + j,k1 (u, e k )T e k k1 λ j e j (x)e j (y)(u, e k )e k (y) dy λ j (u, e j )e j (x) λu λ(u, e j )e j (x). Tking the inner product with ech e k yields so either (u, e k ) or λ λ k. λ(u, e k ) λ k (u, e k ),

27 Exmples IV 23 (i) Let (, b) ( π, π) nd consider K(t, s) cos(t s). Then K is clerly symmetric, so the corresponding T is self-djoint. We hve cos(t s) cos t cos s sin t sin s; recll tht cos t nd sin t re orthogonl in L 2 ( π, π). We hve T (sin t) π π π π 2π sin t K(t, s) sin s ds cos t cos s sin s sin t sin 2 s ds nd T (cos t) π π π π 2π cos t. K(t, s) cos s ds cos t cos 2 s sin t sin s cos s ds (ii) Let (, b) ( 1, 1) nd let Then in fct K(t, s) 4 K(t, s) 1 3(t s) 2 + 9t 2 s 2. ( ) ( ) ( ) ( ) t 2 s (3t2 1) 8 (3s2 1), 3 nd so, since { 2 t, 5 8 (3t2 1)} re orthonorml (they re some of the Legendre polynomils from Chpter 6) the integrl opertor ssocited with K(t, s) hs T (t) 4t nd T (3t 2 1) 8 5 (3t2 1). 13. This eqution is the eqution in Theorem 14.2 with p 1 nd q. So it suffices to show tht (i) u 1 (x) x stisfies the eqution d 2 u/dx 2 with u() (which is cler), (ii) tht u 2 (x) (1 x) stisfies the sme eqution with u(1) (which is lso cler), nd (iii) tht W p (u 1, u 2 ) 1: W p (u 1, u 2 ) [u 1u 2 u 2u 1 ] 1(1 x) ( 1)x 1. Or one could follow the proof of Theorem 4.12 with this prticulr choice of G to show tht u(x) s defined does indeed stisfy d 2 u/dx 2 f.

28 24 4 Exmples IV 14.(i) If x then x >, from which it follows tht T x >, nd so x / Ker(T ). So Ker(T ) {} nd T 1 exists for ll y rnge(t ). Then we hve T (T 1 x) 2 α T 1 x 2, i.e. T 1 x 2 α 1 x 2, so T 1 is bounded. (ii) Let λ α + iβ. Then (T λi)x 2 (T x αx iβx, T x αx iβx) (T x αx) 2 (T x αx, iβx) (iβx, T x αx) + β 2 x 2 T x αx 2 + i(t x αx, βx) i(βx, T x αx) + β 2 x 2. Now, since T is self-djoint nd α is rel, it follows tht nd so (T x αx, βx) (βx, T x αx), (T λi)x 2 T x αx 2 + β 2 x 2 β 2 x 2, i.e. T λi is bounded below s climed. (iii) If λ C nd T λi hs densely defined inverse then (T λi) 1 B(H, H), i.e. λ R(T ). So σ c (T ) R. (iv) If the rnge of T λi is not dense in H then by Question 3 we know tht λ is n eigenvlue of T. If T is self-djoint, we must hve λ n eigenvlue of T. But ll eigenvlues of T re rel, so in fct λ is n eigenvlue of T. So λ σ p (T ) rther thn σ r (T ), which implies tht σ r (T ) is empty. 15.(i) If y l q then l(x) x j y j x p y q using Hölder s inequlity. So l p y q s climed. (ii) Any x (x 1, x 2, x 3,...) l p cn be written s x x j e j, for the elements e j defined in the question. Since l is bounded it is

29 Exmples IV 25 continuous, so l(x) l lim lim n n l lim n x j e j x j e j x j l(e j ) x j l(e j ) x j y j, if we define y j l(e j ). [Should you come cross something similr in n exm you should t lest mention continuity of l s the justifiction for tking it inside the infinite sum.] (iii) We hve x (n) p p y j q y j p y j p q 1 y j q, nd so, s climed, Also x (n) p y j q l(x (n) ) y j q y j y j 1/p. y j q. Since l(x (n) l p x (n) p, we hve y j q l p y j q 1/p,

30 26 4 Exmples IV from which it follows, since 1 1/p 1/q, tht 1/q y j q l p. Since this holds for every n, we must hve y l q with y q l p.