Lax Solution Part 4. October 27, 2016

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1 Lax Solution Part 4 October 27, 2016 Textbook: Functional Analysis by Peter D. Lax Exercises: Ch 16: Q2 4. Ch 21: Q1, 2, 9, 10. Ch 28: 1, 5, 9, Chapter 16 Exercise 2 Let h = χ [0,1], the characteristic function of [0, 1]. We have χ [0,1] = 1, so χ [0,1] L. Then, (Hh)(x) = 1 π x t dt = 1 π [ ln x t ]1 0 = 1 π ln x x 1. As x 1, (Hh)(x). Thus, Hh is an unbounded function, so H is not bounded as a map: L L. Suppose to the contrary H is bounded as a map: L 1 L 1. In the textbook (pg 183) it is proved that the norm of H : (L p ) (L p ) is equal to the norm of H : L p L p. Since (L 1 ) = L, this implies that H is bounded as a map: L L which is a contradiction. 1

2 Thus H is not bounded as a map: L 1 L 1. Exercise 3 Consider f(t) = e at, where a > 0. Note that f L p (R + ) for all 1 p, since f p = ( 1 ap )1/p for 1 p <, f = 1. For p > 1, (Lf)(s) = 0 = 1 a + s Lf p L p = 0 = a p+1 p 1. e (a+s)t dt 1 (a + s) p ds Case 1) If 1 < p < 2, consider f n (t) = e nt, i.e. a = n. Then, Lf p L p p = sup f 0 f p p lim n Lf n p p f n p p = lim ( n p+1 n p 1 np) p = lim ( n p 1 n p+2 ) =. Case 2) If 2 < p <, consider f n (t) = e t/n, i.e. a = n 1. Then similarly, L p lim ( np 1 n p 1 n 1 p) p = lim ( n p 1 np 2 ) = 2

3 Case 3) If p = 1, f(t) = e at, Lf L1 = 0 1 a + s ds = lim r [ln a + s ] r 0 = lim r ln = a + r a Case 4) If p =, f(t) = e at, 1 Lf = inf{c 0 C for almost every s} =. a + s Thus L is not bounded as a map of L p (R + ) L p (R + ), except for p = 2 which is shown in the textbook (pg 183, Theorem 9). Exercise 4 Let s = t, then f( s) 0 g(r) = 0 r s ds = f( s) r s ds. Let h(s) := f( s) for s 0, h(s) = 0 otherwise. Then, 0 h(s) g(r) = r s ds h(s) = P V r s ds = π 1 h(s) π P V r s ds = π(hh)(r) 3

4 where H is the Hilbert transform, which is proved to be a bounded map of L p (R) L p (R) for all 1 < p <. Thus, L 2 f L p (R + ) = g L p (R + ) = πhh L p (R) π H h L p (R) = π H f L p (R + ) since h L p (R) = ( = ( = ( 0 0 = f L p (R + ). h(s) p ds) 1/p f( s) p ds) 1/p f(t) p dt) 1/p 2 Chapter 21 Exercise 1 (c) Let (x n + y n ) be a sequence in C 1 + C 2, where x n C 1, y n C 2. Since C 1 is precompact, (x n ) has a Cauchy subsequence (x nk ). Consider (y nk ), which is a sequence in C 2, thus has a Cauchy subsequence (y nkl ). Then (x nkl + y nkl ) is a Cauchy subsequence of (x n + y n ), since it is the sum of two Cauchy sequences, as shown below. Lemma 2.1. If (z n ) and (w n ) are Cauchy sequences in a normed linear space X, then (z n + w n ) is Cauchy. Proof. z n +w n (z m +w m ) z n z m + w n w m 0 as n, m. 4

5 Method 2: Let C 1 and C 2 be precompact subsets of a Banach space X. Since C 1 is precompact, C 1 k i=1 B ɛ/2(x i ), where x i X. Similarly, C 2 n j=1 B ɛ/2(y j ), where y j X. Lemma 2.2. n k C 1 + C 2 B ɛ (x i + y j ). j=1 i=1 Proof. Let c 1 + c 2 C 1 + C 2. We have c 1 B ɛ/2 (x i ) for some x i, i.e. c 1 x i < ɛ/2. Similarly, c 2 B ɛ/2 (y j ) for some y j, i.e. c 2 y j < ɛ/2. Thus, c 1 + c 2 (x i + y j ) c 1 x i + c 2 y j < ɛ. Therefore c 1 + c 2 B ɛ (x i + y j ) n j=1 k i=1 B ɛ(x i + y j ). Thus, C 1 + C 2 is precompact. (d) Let C be a precompact set in a Banach space X, and let Conv(C) denote its convex hull. Since C is precompact, C can be covered by a finite number of balls of radius ɛ, i.e. C m i=1 B ɛ(y i ), where y i X. Let Y := {y 1, y 2,..., y m }. Consider the continuous map f : R m X m (a 1, a 2,..., a m ) a j y j. Let S = {(a 1,..., a m ) a j 0, m j=1 a j = 1}. S is closed and bounded thus compact in R m, by the Heine-Borel theorem. Note that f(s) = Conv(Y ), thus Conv(Y ) is compact because continuous functions map compact sets to compact sets. Thus, in particular Conv(Y ) is precompact. 5 j=1

6 Lemma 2.3. Conv(C) Conv(Y ) + B ɛ (0). Proof. Let z = k j=1 b jz j Conv(C), where b j 0, z j C, k i=1 b j = 1. We have that z j B ɛ (y i(j) ) for some y i(j) (where 1 i(j) m depends on j), i.e. z j y i(j) < ɛ. Write z j = y i(j) + (z j y i(j) ). Then, z = k j=1 b jy i(j) + k j=1 b j(z j y i(j) ). Note that k j=1 b jy i(j) Conv(Y ). k b j (z j y i(j) ) j=1 Thus k j=1 b j(z j y i(j) ) B ɛ (0). < = ɛ k b j z j y i(j) j=1 k b j ɛ Since Conv(Y ) and B ɛ (0) are both precompact, Conv(Y ) + B ɛ (0) is precompact. Conv(C) is a subset of Conv(Y ) + B ɛ (0) and thus precompact, j=1 since a subset of a precompact set is clearly precompact. (e) Let (Mx n ) be a sequence of points of MC. Since (x n ) is a sequence in C, it has a Cauchy subsequence (x nk ). Mx nk Mx nl = M(x nk x nl ) M x nk x nl 0 as k, l. (Mx nk ) is a Cauchy subsequence of (Mx n ), thus MC is precompact. 6

7 Exercise 2 Let D : X U be a bounded linear map such that dim R D <. X and U are Banach spaces. Let B be the closed unit ball in X. Since R D is a finite-dimensional subspace of U, it is complete. Hence, we may consider D : X R D, where X and R D are both Banach spaces. Let Dx DB. Then, Dx D x D. Thus DB {u R D u D } := G. Thus, the closed ball G is compact since R D is finite-dimensional. This means G is precompact, thus DB is precompact since DB G. Hence D is a compact map. Exercise 9 Counterexample to CM n tend uniformly to CM: Proof. Let X = l 2. Consider C : X X, Cx = (x, e 1 )e 1, i.e. C(x 1, x 2,... ) = (x 1, 0, 0,... ). Since dim R C = 1 <, thus C is compact. Consider M n : X X, M n x = (x, e n )e 1, i.e. M n (x 1, x 2,... ) = (x n, 0, 0,... ). Since x l 2, x n 0. Thus, M n x 0x 0 as n, i.e. {M n } tends strongly to 0. However, CM n C0 = CM n = sup CM n x x =1 CM n (e n ) = C(1, 0, 0,... ) = (1, 0, 0,... ) = 1. Thus CM n does not tend uniformly to CM. 7

8 Proof of M n C tends uniformly to MC: Assume C : X U is compact and {M n } tends strongly to M : U V, i.e. M n u Mu 0 for all u U. Rewriting, (M n M)u 0u 0 for all u U. This means {M n M} converges strongly (thus weakly) to the zero operator. By the Principle of Uniform Boundedness, there exists c > 0 such that M n M c for all n N. Fix n N. By definition, we have M n C MC = sup (M n M)Cx. x 1 Let {x k } be a sequence such that x k 1 and (M n M)Cx k M n C MC 1 k. Since C is compact, it maps the closed unit ball B to a precompact set in U. Thus CB N i=1 B ɛ(u i ). Each Cx k Bɛ(u i(k) ) for some i(k) depending on k. Let ɛ > 0. There is a single N 1 N such that for all n N 1, M n u i Mu i < ɛ for all 1 i N. For n N 1, choose k > 1/ɛ. Then, M n C MC (M n M)Cx k + 1 k (M n M)(Cx k u i(k) ) + (M n M)u i(k) + 1 k < cɛ + ɛ + ɛ. Since ɛ > 0 is arbitrary, M n C MC 0 as n. Thus M n C tends uniformly to MC. 8

9 Exercise 10 (i) Let {x n } be a weakly convergent sequence, i.e. lim n l(x n ) = l(x) for every l in X. By the Principle of Uniform Boundedness, there exists c > 0 such that x n c for all n N. Let {Cx nk } be any subsequence of {Cx n }. Since C maps bounded set to precompact set, {Cx nk } has a Cauchy subsequence {Cx nkl }. Since U is complete, {Cx nkl } converges strongly to some u U, i.e. lim Cx n kl u = 0. l Since strong convergence implies weak convergence, Cx nkl u. On the other hand, for any φ U, lim l φ(cx nkl ) = lim l (C φ)(x nkl ) = (C φ)(x) = φ(cx). By uniqueness of weak limit, u = Cx. Thus every subsequence of {Cx n } has a further subsequence that converges to Cx. Lemma 2.4. Let {y n } be a sequence in a normed linear space. Suppose every subsequence of {y n } has a further subsequence that converges (strongly) to y. Then {y n } converges (strongly) to y. Proof. Suppose not. Then there exists ɛ > 0 such that y n y ɛ for infinitely many n. Then there exists a subsequence {y nk } with y nk y ɛ for all k N so {y nk } has no further subsequence that converges to y. Contradiction. Cx. Apply Lemma 2.4 to {Cx n }, we see that {Cx n } converges strongly to (ii) The converse of theorem 9 is: Let M : X U map every weakly convergent sequence into one that converges strongly. Then M is compact. 9

10 It is false. Consider the identity operator I : l 1 l 1. We will prove below that in l 1, every weakly convergent sequence is strongly convergent. Thus I maps every weakly convergent sequence into one that converges strongly. However, in l 1 the closed unit ball B is not compact since l 1 is infinitedimensional. Thus IB = B = B is not compact. Thus I is not a compact operator. Theorem 2.5. In l 1, every weakly convergent sequence is strongly convergent. Proof. Let (x k ) be a sequence in l 1. Suppose x k x. WLOG, by subtracting x, we may assume that x k 0 but x k 1 ɛ for some ɛ > 0. From this assumption we are going to derive a contradiction by constructing a f l with f = 1 and a subsequence (x k l ) such that f, x k l does not converge to zero. (Notation: We write f, x := m=1 f mx m ). We initialize j 0 = 0, set k 1 = 1, choose j 1 N such that j>j 1 x 1 j ɛ/6 and define the first j 1 entries of f as f j = sgn(x 1 j) for 1 j j 1. Now proceed inductively and assume that for some l 1 the numbers j 1,..., j l the subsequence {x 1,..., x k l } and the entries f1,..., f jl have already been constructed and fulfill for all m l: j j m 1 f j x km j ɛ/6 (1) j m j=j m 1 +1 f j x km j 2ɛ/3 (2) j>j m x km j ɛ/6 (3) Note that for l = 1 these conditions are fulfilled: (1) is fulfilled since the sum is empty, (2) is fulfilled since j 1 j=1 f jx 1 j = x 1 j>j 1 x 1 j > 5ɛ/6 and (3) is fulfilled by definition. To go from a given l to the next one, we first 10

11 observe that x k 0 implies that for all j it holds that x k j 0. Hence, we may take k l+1 such that j j l x k l+1 j ɛ/6 and take k l+1 > k l. Since x k l+1 is a summable sequence, we find jl+1 such that j>j l+1 x k l+1 j < ɛ/6 and again we may take j l+1 > j l. We set f j = sgn(x k l+1 j ) for j l j j l+1 and observe j l+1 j=j l +1 f jx k l+1 j = j l+1 j=j l +1 xk l+1 j > x k l+1 1 ɛ/3 > 2ɛ/3. By construction, the properties (1), (2) and (3) are fulfilled for l + 1, and we continue the procedure ad infinitum. For the resulting f l, we have f = 1, and f, x k l = j f j x k l j = j j l 1 f j x kl j j 1 j + f j x k l j + j l j=j l 1 +1 ɛ/6 + 2ɛ/3 ɛ/6 ɛ/3. jl f j x k l j j=j l j>j l f j x k l j f j x k l j j>j l x k l j 3 Chapter 28 Exercise 1 Let A : H H be a symmetric operator. Let {x n } be a sequence in H such that x n x and Ax n u, for some x, u H. Since strong convergence implies weak convergence, we have (x n x, y) 0 and (Ax n u, y) 0 for 11

12 all y H. (Ax u, y) = (A(x x n ) + Ax n u, y) (A(x x n ), y) + (Ax n u, y) = (x x n, Ay) + (Ax n u, y) 0 as n. This means (Ax u, y) = 0 for all y H, thus Ax = u. Hence A is closed. It is clear that A is linear, using the symmetric property of A and linearity of the inner product. By the closed graph theorem, A is continuous and hence bounded. Exercise 5 Let H be a Hilbert space. Assume {x n } converges weakly to x H and lim n x n = x. We have that x n, y x, y for all y H. In particular, x n, x x 2. x n x 2 = x n x, x n x = x n, x n x n, x x, x n + x, x = x n 2 2Re x n, x + x 2 x 2 2 x 2 + x 2 = 0 Thus, x n converges strongly to x. 12

13 Exercise 9 Let U be a unitary map. Note that Ux = x implies U = sup x =1 Ux = sup x =1 x = 1. We also have U = U = 1. U Ux x, U Ux x = U Ux 2 U Ux, x x, U Ux + x 2 = U Ux 2 Ux, Ux Ux, Ux + x 2 = U Ux 2 2 Ux 2 + x 2 = U Ux 2 Ux 2 U 2 Ux 2 Ux 2 = 0 By positive definiteness of inner product, U Ux = x for all x. Thus U U = I. Exercise 10 Lemma 3.1. C is a compact normal operator. Proof. We have C = U I, thus C = U I. By direct computation we have C C = U U U U + I = 2I U U = CC. We use Corollary 2 which states that every compact normal operator has a complete set of orthonormal eigenvectors. Let {z n } be a complete set of orthonormal eigenvectors of C, with Cz n = α n z n. Then, Uz n = z n + α n z n = (1 + α n )z n. So {z n } is a complete set of orthonormal eigenvectors of U. 13

14 We have Uz n = 1 + α n z n = z n, thus all eigenvalues {1 + α n } have absolute value 1, since z n 0. 14