MA5206 Homework 4. Group 4. April 26, ϕ 1 = 1, ϕ n (x) = 1 n 2 ϕ 1(n 2 x). = 1 and h n C 0. For any ξ ( 1 n, 2 n 2 ), n 3, h n (t) ξ t dt

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1 MA526 Homework 4 Group 4 April 26, 26 Qn 6.2 Show that H is not bounded as a map: L L. Deduce from this that H is not bounded as a map L L. Let {ϕ n } be an approximation of the identity s.t. ϕ C, sptϕ (, ), ϕ, ϕ, ϕ n (x) n 2 ϕ (n 2 x). Let h n (t) χ [ n 2,+ n 2 ] ϕ n (t), then h n and h n C. For any ξ ( n, 2 n 2 ), n 3, Hh n (ξ) π P V π π 2 n 2 h n (t) ξ t dt h n (t) ξ t dt + ln(n) π. ξ t dt /n + t dt Therefore, Hh n ln(n) for n 3, lim π Hh n n. H is an unbounded map : L L. h n (t) + 2 ξ t dt + n 2 h n (t) ξ t dt Note that if H: L L is bounded, then H : L L is bounded for L (L ). From the kernel of H we have: H l(t) π P V l(ξ) ξ t dt, l L. H(L L ) H (L L ). H : L L is also bounded, which is a contradiction. Lecturer comment The key idea in the above is the fact that for any ε >, there exists h ε C (indeed C ) such that χ [,] h ε χ ( ε,+ε ). It is then clear that for any ξ ε, we have Hh ε (ξ) P V h ε (t) π P V +ε h ε(t) π ε π P V +ε ξ t dt dt ξ t h ε (t) dt ε t ξ dt π t ξ ln ξ. π

2 In particular, for all ε >, we have Hh ε (ξ) π ln ε. But h ε. End of comment Qn 6.3 Prove that the Laplace transform L is not bounded as a map of L p (R + ) L p (R + ), except for p 2. (Hint: Try f(t) e at.) Let f(t) e at with a >. For p <, /p f p (e at ) dt) p (ap) /p ; Lf(s) Obviously, Lf(s) / L (R + ). For < p <, Lf p a 2 p p Since lim a + p except p 2. ( a + s L Lf p f p + for < p < 2 and lim For p, let f(t), then f < but Lf(s) therefore L is not an operator from L L. For p, let f(t) e t, then f < but Lf(s) therefore L is not an operator from L L. e (a+s)t dt a + s ; ) p ) /p ( a p ds p ( a 2 p ) /p p. p a 2 p p a + p e st dt s / L, e (s+)t dt s + / L, ) /p ; + for p > 2, we know L Qn 6.4 Prove that the Hilbert-Hankel operator is a bounded map of L p (R + ) L p (R + ) for < p <. Hilbert-Hankel operator: f(t) Mf(r) t + r dt. By Minkowski s inequality, we have p ) f(t) t + r d t /p d r ) p /p f(t) t + r d t d r) ) p ) /p f(rz) + z d r, f L p ) p /p f(rz) + z d r) ) /p f(rz) p ( + z) p d r f(rz) p d(rz) z z /p ( + z) <. 2 ) /p + z

3 The last step is for z /p ( + z) z + /p z /p+ p p + p <. Hence the Hilbert-Hankel operator is a bounded map of L p L p, for < p <. 2 For any f L p (R + ), < p <, let { f(t), t, g(t) f( t), t <, { f(t), t >, h(t), otherwise. g, h L p (R), and g L p (R) 2 f L p (R + ), h L p (R) f L p (R + ), Mf(r) π(hg( r) + Hh(r)) L p. From the fact that Hilbert transform H : L p L p is bounded, we have M is bounded. M 3π H, Qn 2. Prove statements (c),(d) and (e). ( c ) If C and C 2 are precompact subsets of a Banach space X, then C + C 2 is precompact. (d) If C is a precompact in a Banach spaces, so is its convex hull. ( e ) If C is a precompact subset of a Banach space X, M a liner bounded map of X into another Banach space U, then MC is a precompact subset of U. ( c ) Let {x n + y n } be a sequence of C + C 2 with x n C, y n C 2. Since C is precompact set, we know {x n } has a convergent subsequence {x nk } converging to x. Since {y nk } is a sequence of C 2 we know it has a convergent subsequence {y nk } converging to y. {x nk + y nk } is a a convergent subsequence of {x n + y n } converging to x + y. Since {x n + y n } is arbitrary, we know C + C 2 is precompact. (d) Since C is a precompact, for any ϵ >, there exists a finite ϵ net A{x, x 2,, x n } of C: C n B(x i, ϵ). i Since Conv(A) is of finite dimension and bounded, it is precompact. It has a finite ϵ net B. For any y Conv(C), it is a convex combination of elements in C: y m λ k c k, k where λ k >, λ k and c k C. Since A is an ϵ net of C, there exists {x ik } s.t. i.e. and then y x ik c k ϵ. m λ k x ik ϵ, k d(y, Conv(A)) ϵ, d(y, B) 2ϵ, Since y can be arbitrary, B is a finite 2ϵ net of Conv(C). 3

4 ( e ) For any sequence of {u k } of MC, there exists x k C s.t. Mx k u k, k N. Since C is precompact, {x k } has a convergent subsequence {x kn } converging to x X. u kn u km M x kn x km, as m, n. {u k } has a convergent subsequence {u kn } that converges to Mx. Since {u k } is arbitrary we know MC is precompact. Qn 2.2 Prove that a degenerate bounded linear map D (dimr D < ) is compact. Since D is bounded, it maps bounded sets to bounded sets. Since dimr D <, R D is isometric to R n for some n N. From Weierstrass theorem we know any bounded set in R n is precompact, thus D is precompact. Qn 2.9 Show that if C is compact and M n tends strongly to M, then M n C tend uniformly to MC. And give an counter example for the case CM n. (a) Suppose M : X U, C : U Y. Since lim M n x Mx, x X, we know M n x is bounded for each x X. From the principle of uniform boundedness, we know { M n M } is bounded. Suppose { M n M } ρ, n. Let B be the unit ball in X. Since C is compact, for any ϵ > there exists a finite ϵ net A of CB. (M n M)C sup (M n M)u sup (M n M)u + ρϵ, u CB u A Since A is finite, there exists N s.t. Therefore Since ϵ is arbitrary positive number, we know (b) Let M n be the left shift operator: sup (M n M)u ϵ, n > N. u A (M n M)C ( + ρ)ϵ, n > N. lim (M n M)C. M n : l 2 l 2 C : (x, x 2, ) (x n+, x n+2, ); l 2 R (x, x 2, ) x. M n converges to M strongly, but CM and CM n. CM n does not tend to uniformly. Qn 2. ( i ) Prove theorem 9. A compact map C : X U maps every weakly convergent sequence into one that converges strongly. (ii) What about its converse? ( i ) From the principle of uniform boundedness, we know for any weakly convergent sequence {x k } of X, { x k } is bounded. Suppose x k x. Since C is compact, any subsequence of {Cx k } is a precompact subset of U and this subsequence has a convergent {Cx kn } subsequence. 4

5 For any l U, (Cx kn Cx, l) (x kn x, C l) as n. From the uniqueness of weak limit and the fact that strong limit is weak limit, we know lim Cx k n n Cx. Since any subsequence of {Cx k } has a subsequence converges to Cx. we know lim Cx k Cx. n (ii) Since the closed unit ball in a Banach space X is weakly sequentially compact iff X is reflective, we know that the converse is true if and only if X is reflective. Qn 28. Show that a symmetric operator A is closed. Show that A is bounded. (a) Suppose {x k } is a sequence of H s.t. and for any y H, lim x k x H, k lim Ax k u H. k (A(x k x), y) (x k x, Ay) x k x Ay as k. This means Ax k Ax. Since weak limit is unique and strong limit is weak limit, we know Ax u. From the arbitrariness of {x k } we know A is closed. (b) Since A is closed, from the closed graph theorem we know A is bounded. Qn 28.5 Show that (in Hilbert space) if w lim x n x and lim x n x, then x n converges strongly to x. Since x n x, we have So, lim (x n x, x n x) n Qn 28.9 Show that a unitary map U satisfies U U I. For any x H, Subtract RHS from LHS, For any a, b H lim n ( x n 2 + x 2 (x n, x) (x, x n )) 2 x 2 lim n ((x n, x) + (x, x n )) lim(x n, x) (x, x) x 2. lim (x n x, x n x). n Ux x, (U Ux, x) (Ux, Ux) (x, x). ((U U I)x, x), x H. ((U U I)(a + b), (a + b)), ((U U I)a, b) + ((U U I)b, a) ((U U I)a, b) + (b, (U U I)a) 2Re(((U U I)a, b)). Take b ib we have i.e. ((U U I)a, b) a, b H, U U I. 5

6 Qn 28. Let U be a unitary operator of form I +C, C compact. Show that U has a complete se of orthogonal eigenvectors, and that all eigenvalues have absolute value. Since U is one to one and onto, for any y H, there exists x s.t. then y x. so, This means that U is also a unitary matrix. From we have : Ux y, (U y, U y) (U Ux, U Ux) (x, x), (U y, U y) (y, y), y H. U U I UU (I + C )(I + C) (I + C)(I + C ), C C CC. Since C is a compact normal operator, it has a complete set of orthonormal eigenvectors. Obviously these eigenvectors are eigenvectors of I, therefore they are eigenvectors of U. Suppose ξ s.t. Uξ λξ. (ξ, ξ) (Uξ, Uξ) λ 2 ξ 2, λ 2. Therefore all eigenvalues of U have absolute value. Qn Extra Generalized Arzela-Ascoli theorem: Let (X, d ) be a separable metric space and (Y, d 2 ) be a complete metric space. Suppose F C((X, d ), (Y, d 2 )) such that F is equicontinuous and the range R F is precompact in (Y, d 2 ), then F is precompact in C((X, d ), (Y, d 2 )). First note that since (X, d ) is separable, so there is a countable dense subset {x k } X. By hypothesis, R F is precompact and (Y, d 2 ) is complete, hence every sequence in R F has a convergent subsequence. Let {f n } F. We will show that {f n } has a subsequence that converges in C((X, d ), (Y, d 2 )) under the compact convergence topology, that is, there exists a subsequence {f nk } such that f nk converges uniformly to a function f C((X, d ), (Y, d 2 )) when restricted to any compact subset of X. Consider the sequence {f n (x )} in R F. This sequence has a convergent subsequence, which we may denote as {f,n (x )}. Next we consider the sequence {f,n (x 2 )} which is also in R F. This sequence also has a convergent subsequence {f 2,n (x 2 )}. Note that the sequence {f 2,n } converges at x and x 2. By induction, we can repeat the process and obtain sequences {f m,n } for every m which converges at x, x 2,, x m, and {f m,n } is a subsequence of {f m,n }. we look at the sequence {f n,n }, which is a subsequence of {f n } which converges at x k for all k. To complete the proof, we will prove that this sequence converges on every compact subset of C X. Let ϵ >, then by the equicontinuity of F, there exists δ > such that d 2 (f n,n (x), f m,m (y)) < ϵ 3 () for all x, y X, n N whenever d (x, y) < δ. Furthermore, by the compactness of C, there exist open balls B,, B K which cover C and have radius δ/2. Since {x k } is dense in X, we can find x, x 2,, x K {x k} such that x i B i for every i K. We know that {f n,n } converges at x i for every i K, hence there is N N such that d 2 (f n,n (x i), f m,m (x i)) < ϵ 3 (2) 6

7 for all m, n > N, i K. Let x C, then x B i for some i K, thus d (x, x i ) < δ, which implies that d 2 (f n,n (x), f m,m (x)) d 2 (f n,n (x), f n,n (x i)) + d 2 (f n,n (x i), f m,m (x i)) +d 2 (f m,m (x i), f m,m (x)) < ϵ 3 + ϵ 3 + ϵ 3 ϵ, for all m, n > N, hence {f n,n (x)} is Cauchy for every x C and therefore is convergent by the completeness of (Y, d 2 ). Let y x : lim n f n,n (x), then we can construct a function f : (X, d ) (Y, d 2 ), f(x) y x. By taking limit m in the inequality above, we obtain d 2 (f n,n (x), f(x)) < ϵ for all n > N, with N being independent of the value of x, which implies that f n,n f uniformly on C. Moreover, uniform convergence preserves continuity, so f is continuous and hence lies in C((X, d ), (Y, d 2 )). Since every sequence in F has a convergent subsequence, F is precompact in C((X, d ), (Y, d 2 )), and the proof is complete. 7