Functional Analysis: Assignment Set # 11 Spring 2009 Professor: Fengbo Hang April 29, 2009

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1 Eduardo Corona Functional Analysis: Assignment Set # Spring 2009 Professor: Fengbo Hang April 29, (i) Every convolution operator commutes with translation: ( c u)(x) = u(x + c) (ii) Any two convolution operators commute. Solution (i) So, using the change of variables z = y + c; we have that: c (f u)(x) = c f(x y)u(y)dy = f(x + c (f c (u))(x) = f(x y)u(y + c)dy = f(x + c y)u(y) z)u(z)dz (ii) We have that: f (g u)(x) = f g(y)u(x y)dy = f(z) g(y)u(x z y)dy dz = f(z)g(y)u(x z y)dydz Since this expression is symmetric with respect to f and g (does not change if we exchange f and g) we conclude any two convolution operators commute Show that for all f in L 2 ; Af is continuous and is zero at 0 and : (Hint: use the estimate ku x k L 2 2 kfk L 2) Solution 2 The estimate 2 kuk L +ku xk 2 L 2 2 kfk L shows that f 2 2 L2 =) Af 2 H (0; 2). However, by the Sobolev Embedding (or the Rellich) Theorem, we know that H is continuously and compactly embedded in C(0; 2). Hence, Af 2 C(0; 2) (in fact it is Holder continuous). An easier way to show this involves using the Fourier Transform: Extending u by zero, we have: jbu()j d = jbu()j jj d 2 jbuj 2 d =2 d 2 < Since = 2 is integrable, and bu = cu x 2 L 2. Since bu 2 L ; it follows that u is uniformly continuous. Finally, to conclude the boundary conditions hold, we can make an approximation argument using test functions (to argue that, since u is a weak solution to the ODE, u 2 H 0 and the boundary conditions are ful lled). 30. A bounded linear operator K mapping a Hilbert space H into itself belongs to the HS class if for some orthonormal basis fe j g of H, kke j k 2 < j= (a) Show that if K satis es this for one orthonormal basis, then it satis es it for every orthonormal basis, and the sum is independent of the basis. The square root of this sum is called the HS-norm, denoted as kk HS : (b) Show that kkk kkk HS (c) Show that if K 2 HS; so is its adjoint K and kkk HS = kk k HS (d) Show that the HS operators form a complete normed space in the HS-norm. (e) Show that if K 2 HS and B any bounded operator, then BK and KB are HS, and kkbk HS ; kbkk HS are both kbk kkk HS : (g) Show that every HS operator is compact.

2 Solution 3 (a) and (c) Let fe j g be the orthonormal basis for which we know this to be true, and ff k g some other orthonormal basis. First, using Parseval, we show a result that will lead to (c): Hence, we have that Now, we also have: And so: Kf k = hkf k ; f m i f m jhkf k ; f m ij 2 = k= jhf k ; K f m ij 2 kk f m k 2 jhkf k ; e m ij 2 = k= jhf k ; K e m ij 2 kk e m k 2 = kke j k 2 < Hence, we see that the kkk HS does not depend on the basis, and that if K 2 HS; so does its adjoint, and their norms are equal. (b) For any unit vector kxk = ; we can use the Gram-Schmidt process to make it the rst of an orthonormal basis: fx; f 2 ; f 3 ; :::g: But then:! =2 kkxk kkk HS k= Taking the sup over all such vectors, we obtain our inequality. (d) It is immediate, by the properties of the norm in H; that kkk HS = jj kkk HS 8 2 C. - Also, kkk HS = 0 () kke j k = 0 8j () K 0 (by completeness of the orthonormal basis). - P j= k(b + C)e jk 2 P j= kbe jk P j= kbe jk kce j k + P j= kce jk 2 kbk 2 HS + 2 kbk HS kck HS + kck 2 HS = (kbk HS + kck HS )2 Hence, the HS class of operators is a normed vector space. Completeness follows from the inequality in (b) and completeness of the space of bounded operators with respect to the norm topology: Let fk n g be a Cauchy sequence with respect to the HS norm. Then: j= kk n K m k kk n K m k HS And hence, K n! K in the operator norm. Now, 8" > 0; 9M > 0 s.t. n; m M =) kk n K m k HS < " NX k(k n K m )e j k 2 kk n K m k 2 HS < "2 8N 2 N j= Hence, taking n! ; we can conclude that P N j= k(k K m)e j k 2 < " 2 =) kk K m k HS! 0. (e) 8j; we have that kbke j k kbk kke j k and adding these up, we nd: kbkk 2 HS = X j kbke j k 2 kbk 2 X j kke j k 2 = kbk 2 kkk HS and by (c) and the previous result, B K 2 HS; and kkbk 2 HS = kb K k 2 HS kb k 2 kk k 2 HS = kbk2 kkk 2 HS 2

3 Hence, the HS class is a two sided ideal in B(H). (g) Now, for some orthonormal basis, let P N be the projection into the linear span of the rst N functions, and K N = P N K. Then, K N is degenerate, and we have that: k(k K N )xk 2 = j=n+ jhe j ; Kxij 2 kxk 2 X j=n+ kke j k 2! 0 Hence, K is a compact operator (since kk K N k! 0; and K N is a sequence of degenerate maps). 3.2 Show that a positive symmetric operator has a unique positive square root. How many square roots does it have that are not positive? Solution 4 Let A be the positive symmetric operator, C a positive square root. Since C 2 = A; it follows that C commutes with A; and hence it also commutes with p A. Now, for all x 2 H; Dp p p E D A( A C)x; ( A C)x + C( p A C)x; ( p E D A C)x = ( p A + C)( p A C)x; ( p E A C)x D = (A C 2 )x; ( p E A C)x = 0 Now, since both p D pa( p p E D A and C are positive, it follows that A C)x; ( A C)x = C( p A C)x; ( p E A C)x = 0: This in turn implies p A( p A C)x = C( p D pby; p E A C)x = 0 ( for B 0, hby; yi = 0 =) By = 0 =) p By = 0 =) By = 0). Hence, by symmetry of these two operators: ( p A C)x 2 D = ( p E A C) 2 x; x = 0 8x 2 H And so, C = p A. Non-positive roots: If H has nite dimension r (or our operators have nite rank), we know that A has 2 r distinct non-positive roots if all its eigenvalues are distinct (since we can choose p i for each i 2 (A)), and an in nite amount of non-positive square roots if it has a repeated eigenvalue (using re ections inside the eigenspace of dimension ). If A has an in nite dimensional range, then in any case it has an in nite amount of non-positive square roots by any of these two arguments. It can also be proven that A; B symmetric, and such that A 2 = B 2 =) 9 U unitary such that A = (2U I)B. 3.3 (a) Show that E(S) is countably additive in the strong topology (Hint: use the orthogonality of the ranges of E(S) and E(T ) when S and T are disjoint. (b) Show that E(S) is not countably additive in the norm topology. Solution 5 (a) Let fs i g i= B set of disjoint subsets, S = [ S i. Then, we know by using the properties of E that E(S i )E(S j ) = E(S i ) ij ; and that he(s i )x; E(S j )xi = 0 8x 2 H; i 6= j. Then, i= he(s j )x; yi = he(s)x; yi j= Hence, we have that P j= E(S j)x = E(S)x 8x 2 H. (b) Let N; M 2 N, N > M: Then, we see that: NX MX E(S j ) E(S j ) = j= j= NX j=m+ 8y 2 H E(S j ) Which is 0 only if E(S j ) 0 8j 2 fm + ; ::; Ng: In any other case, since they are orthogonal projections, and the ranges are orthogonal to each other, the norm is always. Hence, in general, we cannot even nd a Cauchy subsequence (hence convergent) of partial sums in the norm topology. 3

4 3.0 Show that the spectrum of a unitary operator lies on the unit circle. Solution 6 Let U be a unitary operator. Since U is normal, by the spectral theorem, U is unitarily equivalent to a multiplication operator T m. Also, since U U = I = UU ; by the properties of this equivalence, we have that m(x)m(x) = jm(x)j 2 = 8x. Hence, (U) S. 3. Show that if M is a symmetric operator, and k any real number, U = (M + iki) (M iki) is unitary. How about the converse? Solution 7 Assuming (M + iki) for k 2 R (which we know for k 6= 0; since for M symmetric, M invertible), we have the following: i is U = (M iki) (M + iki) U = (M + iki)(m iki) We notice that U and U have the same factors, so it will su ce to show that they commute. First, we notice that: (M + iki)(m iki) = M 2 + k 2 I = (M iki)(m + iki) And so (M + iki) and (M iki) commute. Also, we know that in general, if AB = BA; it follows that AB = B BAB = B ABB = B A. Hence, the two factors commute, and we have U = U (U is unitary). About the converse: let k 6= 0; and assume U = (M + iki) (M iki) is unitary. this means U = U ; so: U = (M + iki)(m iki) = (M iki) (M + iki) = U Multiplying on the right by (M iki) and on the left by (M iki) (to cancel the inverses), we obtain: (M iki)(m + iki) = (M + iki)(m iki) MM ikm + ikm + k 2 I = MM + ikm ikm + k 2 I 2ikM = 2ikM M = M Hence, M is symmetric. (This converse is false for k = 0; since then this only tells us I is unitary). 3.2 Combine exercises 0 and to show that the spectrum of a bounded symmetric operator is real. Solution 8 We observe that, for a certain k 6= 0 (say k > 0), U = (M); where (z) = z ik z + ik Now, is a linear fractional transformation (Mobius transformation), which we know well from complex analysis. It is analytic everywhere (except at z = ik), and maps bijectively and conformaly (depending on the sign of k) either the lower or upper half plane to the unit disc. We also know that it maps the real line to the unit circle. Since we know ik =2 (M); we have that ((M)) = ((M)). From 0, we know that ((M)) S ; and so (M) (S ) = R. 4

5 4 Let H be a Hilbert space over C, A 2 L(H) symmetric, then prove that: kak = sup jhax; xij Solution 9 We rst note that: And so, Now, to prove the other inequality, we observe that: By the polatization identity, jhax; xij kaxk kxk kak kxk 2 = kak 8 kxk = kak sup jhax; xij kak = sup kaxk = sup sup kyk= jhax; yij hax; yi = [ha(x + y); (x + y)i ha(x y); (x y)i + i ha(x iy); (x iy)i i ha(x + iy); (x + iy)i] 4 Since A is symmetric, hax; xi 2 R 8x; and so: Re(hAx; yi) = [ha(x + y); (x + y)i ha(x y); (x y)i] 4 Now, we can always y such that hax; yi is real. Then, we have: jhax; yij 2 = (x + y) (x + y) [kx + yk2 A ; kx 6 kx + yk kx + yk 6 sup jhax; xij 2 [kx + yk 2 + kx yk 2 ] 2 yk 2 (x y) A kx yk ; (x y) ] 2 kx yk = 4 sup jhax; xij 2 [kxk 2 + kyk 2 ] 2 = sup jhax; xij 2 for kxk = kyk =. 5 Let H be a Hilbert space, A n ; A 2 L(H); A n 0 : (a) If A n! A.uniformly, then p A n! p A uniformly. (b) If A n! A.strongly, then p A n! p A strongly. Solution 0 (a) Since ka n Ak! 0; we know that 9C > 0 such that ka n k and kak are bounded above by C. This in turn, implies that (A n ); (A) [0; ] for some 0. Now, we can approximate uniformly the function p (or any continuous function f(), to that e ect) with polynomials. Let p n a sequence of polynomials such that kf p m k ;[0;] < m. Now, from previous homework assignments, we know that for any polynomial p; it is true that: kp(a n ) p(a)k! 0 Since ka n Ak! 0 and kb n Bk! 0 =) ka n B n ABk! 0; and we can iterate this result as many times as it is needed. Now, from Theorem 6, we have that: p p p m (A n ) An pm;[0;]! 0 as m! p p p m (A) A pm;[0;]! 0 as m! 5

6 Hence, combining these three results, we conclude that p p A n A! 0. (In fact, this shows kf(an ) f(a)k! 0 for all f continuous in [0; ]). (b) The approach for strong convergence is quite similar. We also know that p(a n )x! p(a)x (again, by iterating the corresponding result for a product of two bounded operators). Since a strongly (and weakly) convergent sequence of operators is uniformly bounded, it still follows that (A n ); (A) [0; ] for some 0; and so: ( p p ( p A n A)x An p m (A n ))x + k(p m (A n ) p m (A))xk + ( p A p m (A))x! 0 As m and n!. 6