Analysis IV : Assignment 3 Solutions John Toth, Winter ,...). In particular for every fixed m N the sequence (u (n)

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1 Analysis IV : Assignment 3 Solutions John Toth, Winter 203 Exercise (l 2 (Z), 2 ) is a complete and separable Hilbert space. Proof Let {u (n) } n N be a Cauchy sequence. Say u (n) = (..., u 2, (n) u (n), u (n) 0, u (n), u (n) 2,...). In particular for every fixed m N the sequence (u (n) m ) n= is Cauchy in (C, ), so!u m C such that u (n) m u m as n. We claim that u = (..., u 2, u, u 0, u, u 2,...) l 2 (Z) and that u (n) u in (l 2 (Z), 2 ). Since the sequence {u (n) } n N is Cauchy, it is bounded by say N. Also for all n and M we have ( ( M m= M M m= M u m 2 ) /2 u m u (n) m 2 ) /2 +N. Taking first the limit as n and then the limit as M shows that u 2 N. Now let ɛ > 0 be given. Choose N such that u (n) u (n ) 2 < ɛ whenever n, n N. In particular for every M M N, and n, n N we have u (n) m u (n ) 2 < ɛ 2. Taking first the limit as n and then the limit m= M as M shows that u (n) u 2 < ɛ whenever n N. m Now for n N, let D n = {u (l 2 (Z) : u i Q for i n, u i = 0 for i > n}. We claim that countable dense set of l 2 (Z). Indeed given u l 2 (Z) and ɛ > 0, first choose N such that choose 2N + rationals q N,...q N such that we certainly have u q 2 2 < 2ɛ. n N n >N n 0 D n is a u n 2 < ɛ and then u n q n 2 < ɛ. Then letting q = (..., 0, q N,..., q N, 0,...) D, Exercise 2. In general, L (R n ) L 2 (R n ) 2. In general, L 2 (R n ) L (R n ). 3. If f L 2 (R n ) and f is supported on a set E of finite measure, then f L (R n ). 4. If f L (R n ) and f is bounded by M (almost everywhere), then f L 2 (R n ). Proof We are working in n dimensions.. Consider x α dx = 0 r α r n dr dω n = 0 rn α dr dω n where dω n = n/2. Γ(n/2) B(0,) Choose α so that n α = 2. Then B(0,) x α dx = 4πn/2 Γ(n/2), but x n+/2 B(0,) belongs to L (R n ) but doesn t belong to L 2 (R n ). i B(0,) x 2α dx =. i.e. the function

2 2. Consider but B(0,) c x α dx = r n α dr dω n. Choose α so that n α =. Then B(0,) c x α dx =, B(0,) c x 2α dx = n/2 Γ(n/2) i.e. the function x n B(0,) c belongs to L 2 (R n ) but doesn t belong to L (R n ). 3. f = f E f 2 E 2 = f 2 m(e) /2 4. f 2 f M /2 = f /2 M /2. Exercise 3 L 2 (R n R n ). Proof Let {φ k } k= be an orthonormal basis for L2 (R n ). Then {φ k (x)φ j (y)} k,j= is an orthonormal basis for By Fubini s theorem, φ R n R n k (x)φ j (y)φ k (x)φ j (y)dxdy = ( ) φ R n R n k (x)φ k (x)dx φ j (y)φ j (y)dy = Kronecker(k,k ) Kronecker(j,j ). So the φ k (x)φ j (y) are mutually orthogonal. To show that they span L 2 (R n R n ), it is equivalent to show that if f(x, y) L 2 (R n R n ) is such that f(x, y)φ R n R n k (x)φ j (y)dxdy = 0, then f(x, y) = 0. (In other words, the only vector that is simultaneously orthogonal to all the basis elements is the 0 vector.) Indeed, by Fubini, 0 = f(x, y)φ R n R n k (x)φ j (y)dxdy = ( ) f(x, y)φ R n R n k (x)dx φ j (y)dy. Since {φ j } j= is an orthonormal basis for L 2 (R n ), this implies that for every k f(x, y)φ R n k (x)dx = 0 for Leb R n - a.e. y R n. That is, for every k, there is a set M k R n with m(m k ) = such that for y M k, f(x, y)φ R n k (x)dx = 0. Then obviously m( M k ) = k and for y M k, f(x, y) = 0 for Leb R n - a.e. x R n. Hence f(x, y) = 0 for Leb R n R n - a.e. (x, y) Rn R n. k Exercise 4 Let L 2 ([, π]) be the Hilbert space of functions F (e iθ ) on the unit circle with inner product F, G = π F (e iθ )G(e iθ )dθ. Then the mapping U : L 2 ([, π]) F f L 2 (R), where f is given by f(x) = π(i+x) F ( i x) i+x is unitary. Proof Let C + = {z C : Im(z) > 0} (open upper half plane) and D = {z C : z < } open unit disc. The well-known conformal map C + z i z D extends to a homeomorphism C i+z + z i z D \ { }. i+z Clearly if z = x R, then i z i x =, and the map x takes 0, i, i, etc... The inverse of i+z i+x the map R x i x i( ω) D \ { } is D \ { } ω R. i+x +ω f, g = f(x)g(x)dx = F ( i x i x )G( )dx = F ( i x i x )G( )dx π (i+x)(i+x) i+x i+x π +x 2 i+x i+x Performing the change of variable x = tan( θ 2dx ), we have dθ =, sin(θ) = 2x, cos(θ) = x2, and i x = 2 +x 2 +x 2 +x 2 i+x eiθ Hence f, g = F, G. We also need to check that U is surjective: ii

3 Given a function f L 2 (R), we have for all x R, f(x) = f Also i + x = 2i = i x i+x + ( ) i( ω), where ω = i x. +ω i+x ( 2i. Consider the function D \ { } ω F (ω) = 2i π f i( ω) ω+ ω+ +ω ) C. By construction UF = f and so U is surjective. Note: technically, as a function, F should be defined on D. But since { } is a set of measure zero, we have defined the equivalence class in which F belongs, and that is what is relevant. Since {e inx : n Z} form an orthonormal basis for L 2 ([, π]), automatically {Ue inx : n Z} = { π(i+x) ( i x i+x) n : n Z} form an orthonormal basis for L 2 (R). Exercise 5 Let H be a Hilbert space; A : H H be a linear operator. Then AA = A A = A 2 = A 2 Remark If A = sup x H, x Ax <, we say that A is a bounded operator with norm A. The following are equal to A as defined above:. sup Ax Ax 2. sup x x H,x 0 3. inf{m 0 : Ax M x for all x H} 4. sup{ x, Ay : x, y H, x, y } 5. sup{ x, Ay : x, y H, x =, y = } The proof of these is obvious for the most part. By Cauchy-Schwarz, we have sup{ x, Ay : x, y H, x, y } A. Conversely, sup{ x, Ay : x, y H, x, y } sup{ Ay, Ay : y H, y } Ay = sup{ Ay : y H, y } = A. Proof Clearly x, Ay = y, A x for all x, y so this gives A = A. Next, A A = sup ( A (Ax) sup A Ax = A A. Conversely A 2 = = sup Ax, Ax = 2 sup Ax ) = sup Ax 2 (since x x 2 is increasing and continuous on [0, )) x, A Ax A A by Cauchy-Schwarz. sup iii

4 Exercise 6 Let H be an infinite dimensional (separable) Hilbert space, (f k ) k= a sequence in H, with f k = for all k. Then there exists a subsequence (f kj ) j= of (f k ) k= and a unique f H such that f k j w f, that is f kj, g f, g for all g H. Proof This is a special case of the famous theorem by Alaoglu. Let {g n } be an orthonormal basis for H. We are assuming that H is separable so that the basis is countable. Consider the (infinite) array with k, n th entry equal to f k, g n. We describe the diagonalization procedure which produces a subsequence (f kj ) j= of (f k ) k= such that f kj, g n converges in (C, ) as j for every n N. Consider the first column of the array, namely the sequence ( f k, g ) k=. This is a bounded sequence (C-S) in C so it has a convergent subsequence denoted by ( f kj, g ) j=. Denote this limit c. Now we move on to the second column of the array, and consider the sequence ( f kj, g 2 ) j=. Notice that we are restricting our attention only to the indices of the subsequence obtained in the previous column. This is a bounded sequence and we may therefore extract a converging subsequence ( f k2j, g 2 ) j=. Denote the limit c 2. In general, suppose that in the n th column we have found a subsequence ( f knj, g n ) j= of ( f k, g n ) k= converging to c n. Then we pass to the (n + ) th column and extract a subsequence ( f k(n+)j, g n+ ) j= of ( f knj, g n+ ) j= converging to c n+. Now set k j = k jj for j =, 2, 3,... Then for every n, ( f kj, g n ) j= is a subsequence of ( f knj, g n ) j= starting from j = n. Hence for every n, ( f kj, g n ) j= c n. Finally letting f = c n g n, we have that ( f kj, g n ) j= f, g n for every n. If g = N a i g i, Then by linearity of n= the inner product ( f kj, g ) j= f, g. i= Exercise 7 Consider a Hilbert space H and the collection of bounded operators on H equipped with the operator norm.. Strong convergence doesn t imply convergence in norm. 2. Weak convergence doesn t imply strong convergence 3. For any bounded operator T on H there exists a sequence of finite rank operators (T n ) n= such that T n T strongly. Proof. Consider the Hilbert space l 2 (N), 2 ) and the shift operator to the left (Lx) n = x n+ for n 0, x l 2 (N). We claim that the sequence (L n ) n= (L composed with itself n times) converges strongly to the 0 operator but does not converge in norm to the 0 operator. Given ɛ > 0 and x l 2 (N) choose N such that x i 2 < ɛ. Then L n x 2 2 < ɛ for every n N. Since ɛ was arbitrary we have L n x 2 0, i.e. L n x 0. On the other hand we clearly have L n and L n δ n+ 2 =, where δ n+ is the element (0, 0,...,, 0, 0,...) where the occurs in the n + position. Hence L n = for all n, and L n cannot converge in norm to the 0 operator. i=n iv

5 2. Now consider the shift operator to the right (Rx) n+ = x n for n 0, (Rx) 0 = 0, x l 2 (N). We claim that the sequence (R n ) n= converges weakly to the 0 operator but does not converge strongly to the 0 operator. Clearly R n x 2 = x 2 for all x and so (R n ) n= cannot converge strongly to 0 operator. However, since (l 2 (N)) = l 2 (N), every linear functional on l 2 (N) is of the form a, = a n n for some a l 2 (N). Hence n=0 given ɛ > 0, a, x l 2 (N), choose N such a i 2 < ɛ. Then for n N, a, R n x a i x i N i=n i=n a i 2 x 2 ɛ x 2. i=n 3. Let (g k ) k= be a orthonormal basis of H. Let T ng k = T g k for k n and T n g k = 0 for k > n. Then each T n has finite rank (i.e. dimension of range is finite) and T n g k T g k = 0 for n sufficiently large. For an arbitrary vector g, applying the triangle inequality gives T n g T g = 0 for n sufficiently large. Exercise 9 Suppose w is a measurable function on R d with 0 < w(x) < for a.e. x, K is a measurable function on R 2d that satisfies:. R d K(x, y) w(y)dy Aw(x) for almost every x R d 2. R d K(x, y) w(x)dx Aw(y) for almost every y R d Then the integral operator L 2 (R d ) f(x) T f(x) = R d K(x, y)f(y)dy L 2 (R d ) is bounded with operator norm less than A. Proof First, K(x, y) f(y) dy = K(x, y) /2 w(y) /2 K(x, y) /2 f(y) w(y) /2 dy R d R ( d ) /2 ( K(x, y) w(y)dy K(x, y) f(y) 2 w(y) dy R d R d A /2 w(x) /2 (R d K(x, y) f(y) 2 w(y) dy ) /2 ) /2 v

6 Next, T f 2 2 = K(x, y)f(y)dy 2 dx R d R d ( ) Aw(x) K(x, y) f(y) 2 w(y) dy dx R d R d ( ) = (Fubini) A f(y) 2 w(y) K(x, y) w(x)dx dy R d R d A R 2 f(y) 2 w(y) w(y)dy d = A 2 f 2 2 Somehow it all works out nicely and T A. Exercise 0 The operator L 2 (0, ) f(x) T f(x) = 0 f(y)dy π(x+y) L2 (0, ) is bounded and T. Proof Let w(x) = x. By the previous exercise it is enough to check that By the change of variable y = u one verifies that 0 A =. π(x+y) 0 π(x+y) w(y)dy Aw(x). y dy = x. So the desired inequality holds with Exercise Consider the sawtooth function defined on [, π) by K(x) = i(πsgn(x) x) and extended to R with period. Suppose that f L ([, π]) is extended to R with period, and consider the operator T f(x) = K(x y)f(y)dy. Then π. F (x) = T f(x) is absolutely continuous and F (x) = if(x) a.e. x whenever π f(y)dy = 0 2. The map f T f is compact and symmetric on L 2 ([, π]). 3. ϕ(x) L 2 ([, π]) is an eigenfunction for T iff ϕ(x) e inx for some integer n 0 with eigenvalue /n or ϕ(x) = with eigenvalue {e inx } n Z is an orthonormal basis of L 2 ([, π]) Remark First we remind two important results. If f : [a, b] R is absolutely continuous then f exists for a.e. x [a, b]. vi

7 2. A function f : R R satisfies the Lipschitz condition with Lipschitz constant M iff f is absolutely continuous and f (x) M for a.e. x. Proof. To show that F (x) = T f(x) is absolutely continuous we could hope to show that F (x) is Lipshitz. We will do a first calculation and realize that F (x) may not necessarily be Lipschitz, but the calculation will shed some light on how to get around the problem. T f(x) T f(y) = WLOG we assume that x < y then π π πsgn(x s) πsgn(y s) + y x = (K(x s) K(y s))f(s)ds π(sgn(x s) πsgn(y s) + y x f(s) ds { y x if s < x or s > y + y x if x < s < y We don t need to specify the function at s = x and s = y since this doesn t change the integration.so T f(x) T f(y) x 2 y x y x f(s) ds + π y x f(s) ds + + y x f(s) ds + y x π + y x f(s) ds y y x f(s) ds Now f L ([, π]), so from here we see here that we almost have Lipschitzity, but the second term on the RHS with the is problematic. Indeed as the distance between x and y gets smaller, the first term on the RHS gets better while the while the second term gets worse. This is precisely what happens to the difference of two identical sawtooth functions as the phase between them gets smaller. In general the difference gets smaller, but it gets larger at the point where there is a jump discontinuity. This problem is inherent to the sawtooth function so to bypass the problem we have to use the fact that f L ([, π]). Indeed we choose a bounded (simple) function g such that π f(s) g(s) ds < ɛ. Then y + y x f(s) ds y f(s) g(s) ds + M y x where M is an upper bound for g. x x Let ɛ > 0 be given and suppose that (x i, y i ) are disjoint intervals of [, π] such that n x i y i < ɛ. Choose g such that g(x) M and π f(s) g(s) ds < ɛ. Then n T f(x i ) T f(y i ) n C x i y i + i= i= i (x i,y i ) i= f(s) g(s) ds Cɛ + ɛ, where C = π So F (x) = T f(x) is absolutely continuous and its derivative exists almost everywhere. Suppose that π T f(x+h) T f(x) f(s)ds = 0. We want to calculate lim h vii = lim π π f(s) ds + M. K(x+h s) K(x s) f(s)ds. h

8 K(x + h s) K(x s) = i (πsgn(x + h s) πsgn(x s) h) = { ih i( h) if s < x or s > x + h if x < s < x + h T f(x+h) T f(x) π So lim = lim i (x,x+h)(s) f(s)ds i π i x+h f(s)ds = lim f(s)ds = if(x) by the Fundamental h h h x T f(x+h) T f(x) theorem of calculus. Similarly lim = if(x). So (T f) = if. Another way of deriving this way would h 0 h have been to use DCT and the fact that the derivative of the sign function is two times the delta function. 2. The integral operator T is easily seen to be Hilbert-Schmidt. Indeed, since the sawtooth function is bounded and we are on a finite measure space, π π K(x, y) 2 dxdy <. From the general theory of bounded operators, Hilbert-Schmidt operators are compact (see p.90 of Stein & Shakarchi). We also check that T is symmetric: T f, g = = = = π π π π π π π T f(x)g(x)dx = f, T g i(πsgn(x s) (x s))f(s)ds g(x)dx i(πsgn(s x) (s x))f(s)ds g(x)dx i(πsgn(s x) (s x))g(x)dx f(s)ds 3. One may compute the integrals to verify that {e inx } n Z\{0} are indeed eigenfunctions of T with eigenvalue n and that is an eigenfunction with eigenvalue 0. Conversely, if T f = λf, then π π K(x y)f(y)dy dx = λ π f(x)dx = π f(y) π K(x y)dx dy = 0. Hence by., (T f) = λf = if, so f(x) = e i λ x, when λ 0. Going back to eigenvalue problem will give us a condition on λ: π K(x y)e i λ y dy = λe λ x λ2 e λ x sin π. Hence we must have sin π = 0, that is λ =. For the case λ = 0, π λ λ n we have (T f) = 0 = if, so f is a constant function. 4. T has been checked to be compact and symmetric. By the spectral theorem, a subset of the eigenvectors of T forms an orthonormal basis for L 2 ([, π]). Since {e inx } n Z are the only eigenvectors of T and are mutually orthogonal, they must be a basis for L 2 ([, π]). viii