Math 70 Problem Set 0 Solutions Unless otherwise stated, X is a complex Hilbert space with the inner product h; i the induced norm k k. The null-space range of a linear map T W X! X are denoted by N.T / R.T /. Problem. This exercise proves a family of polarization identities in inner product spaces. (i) Let D e i=n, where n is an integer. Show that ( nx kj if j D 0 D n 0 if j n : kd If j D 0 we have kj D for every k the result is immediate. If j n, then j the geometric sum formula gives nx kj D.nC/j j D 0 j kd since.nc/j D nj j D j. (ii) Use (i) to show that for every x; y in an inner product space, hx; yi D nx kx C k yk k n provided that n 3. The expansion shows that kd kx C k yk D hx C k y; x C k yi D kxk C kyk C k hy; xi C k hx; yi kx C k yk k D.kxk C kyk / k C k hy; xi C hx; yi: Averaging over k n, the first second terms vanish by part (i) (here we use the condition n 3), while the third term gives hx; yi (iii) Prove the following continuous version of the above identity: hx; yi D Z 0 kx C e it yk e it dt: This can be obtained from (ii) by letting n! since the average in (ii) is a Riemann sum approximation of the above integral average. It can also be proved directly as
follows: The expansion shows that kx C e it yk D hx C e it y; x C e it yi D kxk C kyk C e it hy; xi C e it hx; yi kx C e it yk e it D.kxk C kyk / e it C e it hy; xi C hx; yi: Averaging over 0 t, the first second term clearly vanish, while the third term gives hx; yi. Problem. Assuming Y is a closed subspace of X, prove the following statements: (i).y? /? D Y. If y Y, then hy; zi D 0 for every z Y?, so y.y? /?. This proves Y.Y? /?. The orthogonal decompositions X D Y? Y D Y?.Y? /? show that Y cannot be a proper subspace of.y? /?. (ii) The natural map Y?! X=Y given by z 7! z C Y is an isometric isomorphism (recall that the norm on X=Y is defined by kx C Y k D inf yy kx C yk). Recall that every x X can be written uniquely as P.x/ C Q.x/ where P.x/ Y Q.x/ Y?. Moreover, P.x/; Q.x/ are the nearest points of Y; Y? to x. Let us denote the natural map Y?! X=Y by. If.z/ D Y for some z Y?, then z Y \ Y?, so z D 0, showing is injective. Moreover, for every x X, x C Y D Q.x/ C Y D.Q.x//; which shows is surjective. Finally, since P.z/ D 0 Q.z/ D z whenever z Y?, k.z/k D kz C Y k D distance from z to Y D kq.z/k D kzk; so is an isometry. (iii) If x X, () min kx yk W y Y D max jhx; zij W z Y? ; kzk D : Let us assume x Y, the result being trivial otherwise. Consider the orthogonal projections P; Q as above. Then the left side of () is kq.x/k. For every z Y? with kzk D, the Cauchy-Schwarz inequality shows that jhx; zij D jhp.x/; zi C hq.x/; zij D jhq.x/; zij kq.x/k; with equality iff z D Q.x/=kQ.x/k (notice that Q.x/ 0 since x Y ). It follows that the right side of () is also kq.x/k.
Comment. The quantity in () can also be interpreted as the operator norm of the functional z 7! hz; xi on Y?. Problem 3. Suppose f is a non-zero bounded linear functional on X. Prove the following statements: (i).ker f /? has dimension. According to Riesz, f.x/ D hx; ai for some non-zero a X. Hence ker f D Y?, where Y is the subspace of X spanned by a. It follows from problem (i) that.ker f /? D.Y? /? D Y is -dimensional. (ii) ker f D ker g for a bounded linear functional g if only if f D g for some scalar. If f D g for some (necessarily non-zero), then clearly ker f D ker g. Conversely, suppose ker f D ker g. Then g is not identically zero, so g.x/ D hx; bi for some non-zero b X. If Z is the subspace of X generated by b, it follows from (i) that Z D.ker g/? D.ker f /? D Y; so a D cb for some scalar c. Thus, f D g, where D c. Comment. The results of this problem hold more generally in any vector space if (i) is formulated appropriately. If f is a non-zero linear functional on a vector space X, then X=.ker f / is -dimensional. To see this, choose x 0 X such that f.x 0 / 0. For every x X there is a scalar c such that f.x/ D cf.x 0 /. Then f.x cx 0 / D 0 H) x cx 0 ker f H) x C ker f D c.x 0 C ker f /: This shows X=.ker f / is spanned by x 0 C ker f. If ker f D ker g, choose x 0 as above set D f.x 0 /=g.x 0 /. For every x X there is a c such that x cx 0 ker f D ker g, so This proves f D g. f.x/ D cf.x 0 / D cg.x 0 / D g.x/: Problem 4. Take advantage of the fact that L Œ ; is a Hilbert space to compute 3 min a;b;cc max g jx 3 a bx cx j dx ˇ x 3 ˇ g.x/ dxˇ;
4 where g L Π; is subject to the restrictions g.x/ dx D xg.x/ dx D x g.x/ dx D 0 jg.x/j dx D : Let Y be the subspace of L Π; spanned by f; x; x g. As a finite dimensional subspace, Y is closed. By the orthogonal decomposition theorem, L Π; D Y Y? the above minimum is achieved when a C bx C cx is the orthogonal projection P.f / of f.x/ D x 3 onto Y. This projection can be computed once we have an orthonormal basis fu ; u ; u 3 g for Y. Applying the Gram-Schmidt process to f; x; x g, we obtain such a basis: v D u D v kv k D p v D x hx; u i u D x u D v p 3 D p x kv k Z x dx D x v 3 D x hx ; u i u hx ; u i u D x u 3 D v 3 kv 3 k D x 3 dx x p 5 p.3x /: 8 x 3 dx D x 3 We compute the Fourier coefficients of f with respect to fu ; u ; u 3 g: hf; u i D p x 3 dx D 0 p Z 3 p 6 hf; u i D p x 4 dx D 5 p Z 5 hf; u 3 i D p x 3.3x / dx D 0: 8 Thus, P.f / D hf; u i u C hf; u i u C hf; u 3 i u 3 D 3 5 x min jx 3 a bx cx j dx D x 3 3 a;b;cc 5 x 8 dx D 75 :
The maximization problem amounts to finding max jhx 3 ; gij W g Y? ; kgk D : By problem (ii), this is the same as kf P.f /k D kq.f /k D p 8= p 75, is achieved when g.x/ D Q.f / p 75 D p x 3 3 kq.f /k 8 5 x : Comment. Alternatively, we could have found the projection P.f / rather easier by noting that Q.f / D x 3 a bx cx is orthogonal to ; x; x. This would give 0 D hq.f /; i D 0 D hq.f /; xi D 0 D hq.f /; x i D.x 3 a bx cx / dx D a.x 4 ax bx cx 3 / dx D 5 3 c 3 b.x 5 ax bx 3 cx 4 / dx D a c; 3 5 or a D c D 0, b D 3=5, showing as before P.f / D 3x=5. Problem 5. Let T W X! X be a bounded linear map. (i) Show that there is a unique bounded linear map T W X! X, called the adjoint of T, which satisfies () ht.x/; yi D hx; T.y/i for every x; y X. For each y Y, the assignment x 7! ht.x/; yi is clearly a bounded linear functional on X. According to Riesz, there is a unique vector T.y/ X for which this functional is expressed as x 7! hx; T.y/i. This proves the existence uniqueness of a map T W X! X satisfying (). Since hx; T. y C z/i D ht.x/; y C zi for all x; y; z X C, we have D ht.x/; yi C ht.x/; zi D hx; T.y/i C hx; T.z/i D hx; T.y/ C T.z/i T. y C z/ D T.y/ C T.z/; so T is linear. By the Cauchy-Schwarz inequality, jhx; T.y/ij D jht.x/; yij kt.x/k kyk kt k kxk kyk 5
6 for every x; y. Setting x D T.y/ then gives kt.y/k kt k kt.y/k kyk or kt.y/k kt k kyk for every y. This shows T is bounded kt k kt k. (ii) Verify the relations kt k D kt k kt T k D kt T k D kt k : By what we have seen above, kt k kt k. Since.T / D T, we also have kt k D k.t / k kt k. Hence, kt k D kt k. Next, for every x X, kt.x/k D ht.x/; T.x/i D hx;.t T /.x/i kxk k.t T /.x/k kt T k kxk ; which shows The reverse inequality holds since Finally, by what we just proved, kt k kt T k: kt T k kt k kt k D kt k : kt T k D k.t / T k D kt k D kt k : (iii) Show that N.T / D R.T /? N.T / D R.T /?. These are straightforward from the definition: y N.T / T.y/ D 0 hx; T.y/i D 0 for every x X ht.x/; yi D 0 for every x X y R.T /? x N.T / T.x/ D 0 ht.x/; yi D 0 for every y X hx; T.y/i D 0 for every y X x R.T /? : Problem 6. (i) Let Y be a closed subspace of X P W X! X be the orthogonal projection onto Y (so R.P / D Y N.P / D Y? ). Show that P D P D P kp k D. Clearly, P.x/ D x if x Y, so P.P.x// D P.x/ for all x X. This shows P D P.
On the other h, if x; y X, write x D P.x/ C Q.x/ y D P.y/ C Q.y/ notice that This proves P D P. hp.x/; yi D hp.x/; P.y/ C Q.y/i D hp.x/; P.y/i D hp.x/ C Q.x/; P.y/i D hx; P.y/i: Finally, by the Pythagorean theorem, kxk D kp.x/k C kq.x/k. So kp.x/k kxk with equality iff x Y. This shows kp k D. (We need to assume Y f0g for the last conclusion.) (ii) Let P W X! X be a bounded linear map such that P D P D P. Show that R.P / is a closed subspace of X P is the orthogonal projection onto R.P /. First notice that if y D P.x/ R.P /, then y D P.x/ D P.x/ D P.y/. Conversely, if y D P.y/, then clearly y R.P /. Thus, R.P / is precisely the kernel of the continuous linear map P I. As such, it must be a closed subspace of X. Next, let W X! X be the orthogonal projection onto R.P /. By what we have just seen, P D D I on R.P /. Also, by problem 5(iii), R.P /? D N.P / D N.P /, so P D D 0 on R.P /?. Since X D R.P / R.P /?, it follows that P D on X. Problem 7. (Bonus) Suppose T W X! X is linear ht.x/; yi D hx; T.y/i for every x; y X. Show that T is bounded (hence T D T ). By the closed graph theorem, it suffices to check that if fx n g is a sequence in X, if x n! a, if T.x n /! b, then b D T.a/. But this is easy: Letting n! in the relation ht.x n /; xi D hx n ; T.x/i using continuity of the inner product, we obtain for every x X. This shows b b T.a/ D 0, as required. hb; xi D ha; T.x/i D ht.a/; xi T.a/ is orthogonal to every vector in X. Hence, Comment. This statement, known as the Hellinger-Toeplitz theorem, is one of the oldest results in functional analysis (proved in 90). It shows that one cannot define an unbounded symmetric operator on a Hilbert space, an obstruction that had troubling implications for the mathematical formulation of quantum mechanics in its early days. 7