Functional Analysis: Assignment Set # 3 Spring 2009 Professor: Fengbo Hang February 25, 2009

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1 duardo Corona Functional Analysis: Assignment Set # 3 Spring 9 Professor: Fengbo Hang February 5, 9 C6. Show that a norm that satis es the parallelogram identity: comes from a scalar product. kx + yk + kx yk = (kxk + kyk ) Solution We know that, if the norm came from a scalar product, it would also satisfy one of the polarization identities: hx; yi = kx + yk kx yk hx; yi = X i k x + i k y k= if F = R if F = C Hence, we de ne hx; yi using these identities, and show that the parallelogram identity implies these are indeed scalar products: (i) Symmetry: hx; yi = kx + yk kx yk = ky + xk ky xk = hx; yi hx; yi = X i k x + i k y = X i k ki k (y + i k x)k = hy; xi k= (ii) Positive: This follows from the properties of the norm, since hx; xi = kxk : k= hx; xi = kxk kk = kxk hx; xi = X i k ( + i k )x = ( + i i + ) kxk k= (iii) Bilinear / Sesquilinear: We rst check additivity: hx; yi + hz; yi = kx + yk kx yk Using parallelogram identity four times, we obtain: + kz + yk kz yk Adding these, we nd: kx + yk = (kz x yk + kx + z + yk ) kzk kx + yk = (kz x + yk + kx + z yk ) + kzk kz + yk = (kz x + yk + kx + z + yk ) kxk kz yk = (kz x yk + kx + z yk ) + kxk hx; yi + hz; yi = kx + x + yk kx + z yk = hx + z; yi

2 The complex case would proceed in the same way, applying the parallelogram identity 8 times. Now, all that is left is to check is that scalars come out of the scalar product. By additivity, we have this for all rational numbers, since: hnx; yi = n hx; yi 8n N [ fg D x hx; yi = m m ; y 8m N h x; yi = k x + yk k x yk = hx; yi Hence, by continuity of the norm, it follows that hx; yi = hx; yi 8 R. Finally, if our eld is C; using that hix; yi = X i k x + i k y = i hx; yi k= We then have that hcx; yi = c hx; yi 8c = a + bi; with a; b Q. The result again follows by continuity of the norm. C6. Show that the scalar product depends continuously on its factors. Solution Let x n! x and y n! y. Then, using Cauchy Schwartz two times, jhx; yi hx n ; y n ij jhx x n ; yij + jhx n ; y y n ij kx x n k kyk + kx n k ky y n k Since x n is a bounded sequence, it follows that jhx; yi hx n ; y n ij! : C6. Prove Lemma 5: (i) The nullspace of a linear functional that is not is a linear subspace of codimension. (ii) If two linear functionals l and m have the same nullspace, they are constant multiples of each other. (iii) The nullspace of a linear functional that is bounded is a closed subspace. Solution 3 (i) Let l : V! R; N = N(l) null space. Then, we can de ne the linear functional b l : H=N! R s.t.: b l([x]h=n ) = l(x) Since l 6= ; it follows that b l([x] H=N ) = () x N () [x] H=N = [] H=N. Thus, b l is a bijective linear functional, and it follows that: dim(h=n) = dim(r) = It then follows that l has codimension ; since co dim(n) = dim(h=n): (ii) If l and m share the same nullspace N. Since co dim(n) = ; the complement N c = spanfeg; and any v V can be uniquely written as v = n + e; n N; F. Thus: l(v) = l(e) m(v) = m(e) If m(e) = ; it follows that m and the result follows trivially. Otherwise: l(v) = l(e) m(v) 8v V. m(e) (iii) Let u n N s.t. ku n uk! : Then, since l is bounded: Hence, u N; and so N is a closed subspace. jl(u)j = jl(u n u)j C ku n uk!

3 C7. Prove the Radon-Nikodym theorem for - nite measures. Solution In the case that and are - nite measures sharing the same -algebra B P (X); we want to prove that if << ; then 9f ; f L loc () such that: () = fd 8 B Since is - nite, 9X i X such that X = [ X i and (X i ) <. Now, lets consider the restrictions j i= and j of and to each X i ; with the corresponding subspace -algebra. It then follows that these measures are all nite, and that j << j 8j; since: j () = =) ( \ X i ) = =) ( \ X i ) = Thus, by the Radon Nikodym theorem for nite measures, 9f j ; f j L ( j ) such that: j () = f j d j 8 B j Finally, if we de ne f P j= f j; by Monotone Convergence, we have: () = ( j ) = fd j= C8. Let Y X closed subspace. Show (X=Y ) is isometrically isomorphic to Y?. Solution 5 We rst de ne a function ' : Y?! (X=Y ) such that '(l)[x] = l(x) and prove this is an isometric isomorphism between these two spaces. We check: (i) ' is well-de ned: [x] = [z] () x = z + y with y Y =) '(l)[x] = l(z + y) = l(z) = '(l)[z]: (ii) '(l)[x] = '( b l)[x] 8[x] X=Y =) l(x) = b l(x) 8x X =) l b l =) ' is injective. (iii) Let e l (X=Y ) : Then l X de ned s.t. l(x) = e l([x]) is in Y?, since l(y) = e l([y]) = 8y Y and '(l)[x] = l(x) = e l([x]) 8[x] X=Y. Thus, ' is surjective and Y? ' (X=Y ) : (iv) Finally, ' is an isometry. That is, we need to show that: k'(l)k (X=Y ) = sup fjl(x)jg = sup fjl(x)jg = klk Y? k[x]k= kxk= Since k[x]k = inf yy fkx + ykg kxk ; kxk =) k[x]k. Thus, fx X : kxk g fx X : k[x]k g; and so (by monotonicity of the supremum): k'(l)k (X=Y ) klk Y? Now, since l Y?, we know that l(x + y) = l(x) 8x X; y Y. Also, 8 k[x]k, 9y n y s.t. k[x]k kx + y n k k[x]k + n and l(x + y n) = l(x) 8n. Hence, sup fjl(x)jg sup fjl(x)jg = klk Y? k[x]k =n kxk 8n But fx X : k[x]k g = that: [ fx X : k[x]k n g and so, by monotonicity of the supremum, it follows n= n k'(l)k (X=Y ) klk Y? 3

4 C8.3 Show that Y is isometrically isomorphic to X =Y?. Solution 6 Again, we de ne the function : X =Y?! Y such that ([ e l])(y) = e l(y) 8y Y (where it is implicitly understood that we use the restriction of e l to Y ). We check: (i) is well-de ned: [ b l] = [ e l] () b l = e l + l with l Y?. Hence, b l(y) = e l(y) + l (y) = e l(y) 8y Y; and so ([ b l]) ([ e l]) (ii) ([ e l])(y) = ([ b l])(y) 8y Y =) e l j Y b l j Y =) e l b l Y? =) [ e l] = [ b l]. Hence, is injective. (iii) Let l Y : Then e l X =Y? de ned s.t. e l(x) = [l](x), (where again it is implicitly understood that we use the class of the extension by of l). Then e l is in X =Y?, since e l = inf fkl + l kg klk X X =Y? l Y? = klk Y < And we have that ( e l)(y) = l(y) 8y Y. Thus, is surjective and Y ' X =Y? : (iv) Finally, is an isometry. That is, we need to show that: ( ) ([ e l]) = sup f e l j Y Y (y) g = f( e l + l )(x) g = [ e l] kyk= inf l Y? sup kxk= X =Y? Since [ e l] = inf X =Y? l Y?f e X l + l g, in particular we can choose b l Y? so that b l (x + y) = e l(x) 8x X; y Y. Thus,( e l + b l )(x + y) = e l(y) and: [ e l l] + b X l = sup f( e l + l X =Y? )(x) g = sup f e l j Y (y) g = ([ e l]) kxk= Now, by de nition of in mum, 9l n Y? such that [ e l] e X [ l + ln e X X =Y? l] + =Y? n. Also, we kyk= note that e l + l n j Y e l j Y 8n and that fx : kxk = ; x Y g fx : kxk = g, so: ([ e l]) = sup Y Hence, ([ e l]) [ e l]. Y X =Y? kyk= f e l j Y (y) g sup f( e l + l n )(x) g [ e l] + X =Y? n kxk= We note that, in this last step, we could have just used the result in Theorem 8.7 to show is an isometry. Y C8.7 Support functions have the following properties: (i) Subadditivity: 8l; m in X ; S M (l + m) S M (l) + S M (m) (ii) S M () = (iii) Positive Homogeneity: S M (al) = as M (l) for a >. (iv) Monotonicity: for M N; S M (l) S N (l): (v) Additivity: S M+N = S M + S N (vi) S M (l) = S M ( l) (vii) S M = S M (viii) S conv(m) = S M Solution 7 (i) 8y M; l(y) + m(y) S M (l) + S M (m) and so, sup ym fl(y) + m(y)g S M (l) + S M (m). (ii) l(y) = 8y; and so, S M () =. (iii) sup ym fal(y)g = a sup ym (l(y)g 8a >. (this is a basic property of the supremum. It follows from the fact that c an upper bound for A =) ac is an upper bound for aa). (iv) This is immediate, since S N (l) is a supremum over a larger set. (An upper bound for elements in N is an upper bound for elements in M). (v) sup ym+n fl(y)g = sup nn;mm fl(n) + l(m)g = S M (l) + S N (l) (vi) sup y M fl(y)g = sup ym fl( y)g = S M ( l).

5 (vii) By monotonicity, S M (l) S M (l). Now, by de nition of supremum, 9y n M such that S M (l) n l(y n ) S M (l). Also, by de nition of M and continuity of l; 9x n M s.t. S M (l) n l(x n) S M (l) But then, l(x n )! S M (l) =) S M (l) S M (l). (viii) Again, by monotonicity, S M S conv(m). However, for all conv(m); = P i im i, m i M we have l() = P i il(m i ) S M (l) P i i = S M (l). So, S conv(m) S M. Let C an open subset, the Lebesgue measure on C; denote: A () = ff : f is holomorphic on with jfj d < g Under the inner product hf; gi = R fgd; A () is a Hilbert space. (a) Fix z ; show that the linear functional l(f) = f(z) is continuous on A (): (b) By Riesz representation we may nd a function K(z; ) A () such that for every f A () such that for every f A (); f(z) = K(z; w)f(w)d(w) If (' j ) j= is an orthonormal base of A (); show that: K(z; w) = ' j (z)' j (w) In particular, K(z; w) = K(w; z): (c) For = fz C : jzj g; nd an explicit formula for K(z; w). Solution 8 (a) Using the results obtained on the last homework assignment, we know that, for a xed z ; B R (z ) : l(f) = f(z ) = R f(z)da j= B R (z ) Hence, for f n A (), kf n fk! =) jl(f n f)j = jf n (z ) f(z )j R jf n p kf n R fk! fj (z)da (b) Given an orthonormal basis, we can compute the Fourier coe cients of K(z; ) using the characterization of our kernel (this is called the reproducing property, and K is called a reproducing kernel for A ()): Hence, we have that: K(z; ); 'j K(z; w)' j (w)dw = ' j (z) j= K(z; w) = K(z; ); 'j 'j (w) = ' j (z)' j (w) (c) We begin by considering fz k g k=. We can see that: z k ; z m = z k z m da = 5 j= r k+m+ e i(k m) drd

6 q So, the family f z k ; z m = k+ ; k = m ; k 6= m k+ zk g k= is orthonormal. Finally, by Taylor s Theorem, we know that this is also a basis for A (). Applying (b), we have that: K(z; w) = j + (zw) j j= We immediately identify this as the derivative of a geometric series, with a = zw. Since jaj <, this series converges, and: K(z; w) = ( zw) And we obtain the identity: f(z) = j + j= f(w)w j dw z j = f(w)dw ( zw) Hence, we would like to con rm that this coincides with the Taylor expansion around. Let r (; ): Invoking Cauchy s formula for derivatives, we have: f k () k! = Multiplying both sides by r k+ and integrating yields: Hence, f k () (k + )k! f k () k! = = = k + f(re i ) r k d eik f(re i )r k e ik [rdrd] f(w)w k da f(w)w k da And we see that this kernel essentially encodes the Taylor expansion for a function f A (). 6

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