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Homework 2 1. Let X and Y Be Hilbert Spaces Over C. Then a Sesquilinear Form H on X ×Y Is a Mapping H

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Homework 2 1. Let X and Y Be Hilbert Spaces Over C. Then a Sesquilinear Form H on X ×Y Is a Mapping H Homework 2 1. Let X and Y be Hilbert spaces over C. Then a sesquilinear form h on X ×Y is a mapping h : X × Y ! C such that for all x1; x2; x 2 X, y1; y2; y 2 Y and all scalars α; β 2 C we have (a) h(x1 + x2; y) = h(x1; y) + h(x2; y); (b) h(x; y1 + y2) = h(x; y1) + h(x; y2); (c) h(αx; y) = αh(x; y); (d) h(x; βy) = βh(x; y). A sesquilinear form is bounded if jh(x; y)j ≤ C kxk kyk and the norm of the form h is given by jh(x; y)j khk = sup = sup jh(x; y)j : X×Y !C x6=0;y6=0 kxk kyk kxk=1;kyk=1 Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y , then h has the representation h(x; y) = hSx; yiY where S : X ! Y is a bounded linear operator. Moreover, S is uniquely determined by h and has norm kSk = khk : X!Y X×Y !C Solution: The idea is to use the Riesz Representation to construct the linear operator S. Fix x 2 H and consider the functional on the space Y given by L(y) = h(x; y): By the properties of h being a sesquilinear form, we have that L is linear in y, and if h is a bounded bilinear form, we have that L is a bounded linear functional on Y . By the Riesz Representation Theorem, we have then that there exists a unique element z 2 Y such that L(y) = hy; ziY ; or in terms of h that h(x; y) = hz; yiY : While the point z is unique, it is depending on the point x that was fixed. We now define the map S : X ! Y by Sx = z. This map is clearly well-defined since the point z is well defined given the x. Substituting in this representation we find, h(x; y) = hSx; yiY as desired. It remains to show that S is linear, bounded and unique. To see that S is linear, we simply use the linearity of h in the first variable. Let α; β 2 C and x1; x2 2 X, then we have hS(αx1 + βx2); yiY = h(αx1 + βx2; y) = αh(x1; y) + βh(x2; y) = α hSx1; yiY + β hSx2; yiY = hαSx1; yiY + hβSx2; yiY = hαSx1 + βSx2; yiY : Since this holds for all y 2 Y we have that αSx1 + βSx2 = S(αx1 + βx2) and so S is linear. To see that S is bounded, first note kSxkY kSkX!Y = sup x6=0 kxkX jhSx; Sxi j = sup Y x6=0;Sx6=0 kxkX kSxkY jhSx; yi j ≤ sup Y x6=0;y6=0 kxkX kykY jh(x; y)j = sup = khk X×Y !C x6=0;y6=0 kxkX kykY In particular we have kSk ≤ khk , and so S is bounded. To see the other X!Y X×Y !C inequality, we have jh(x; y)j = jhSx; yiY j ≤ kSxkY kykY ≤ kSkX!Y kykY kxkX which gives khk ≤ kSk . X×Y !C X!Y For uniqueness, suppose that there are two linear operators S1 and S2 such that h(x; y) = hS1x; yiY = hS2x; yiY : This yields that S1x = S2x for all x 2 X, and so S1 = S2. 2. Suppose that H is Hilbert space that contains an orthonormal sequence fekg which is total in H. Show that H is separable (it has a countable dense subset). Solution: Let A denote the set of all linear combination of the form X γkj ekj J where J is a finite set, kj 2 J, and γkj = akj + ibkj , where akj ; bkj 2 Q. It is clear that A is countable. We now will show that A is the countable dense subset in H that we seek. We need to show that for every x 2 H and > 0 there is a v 2 A such that kx − vkH < Since the sequence fekg is total in H, there exists an integer n such that Yn = spanfe1; : : : ; eng contains a point y whose distance in x is less than 2 . In particular, by Pn Parseval, we can take y = k=1 hx; ekiH ek and have n X x − hx; e i e < : k H k 2 k=1 H Now for each coefficient hx; ekiH , we can find an element γk 2 C whose real and imaginary parts are rational such that n X (hx; e i − γ ) e < k H k k 2 k=1 H Pn Then define v = k=1 γkek 2 A, and note that n X kx − vk = x − γ e H k k k=1 H n n X X ≤ x − hx; e i e + (hx; e i − γ ) e k H k k H k k k=1 H k=1 H < . So we have that A is dense in H and A is countable, so H is separable. 3. If p is a sublinear functional on a real vector space X, show that there exists a linear functional f~ on X such that −p(−x) ≤ f~(x) ≤ p(x). Solution: Note that by Hahn-Banach we have that for any linear functional f with f(x) ≤ p(x) we have a corresponding extension f~(x) that satisfies f~(x) ≤ p(x). Now simply note that if we evaluate this expression at −x then we have −f~(x) = f~(−x) ≤ p(−x) and so we have f~(x) ≥ −p(−x). Combining these inequalities gives the result. 4. If x in a normed space X such that jf(x)j ≤ c for all f 2 X∗ of norm at most 1. Show that kxk ≤ c. Solution: We have the following fact at our disposal: jf(x)j kxkX = sup f2X∗;f6=0 kfk X!C . It is easy to show that jf(x)j sup = sup jf(x)j f2X∗;f6=0 kfk f2X∗;kfk =1 X!C X!C The result then follows easily, since kxkX = sup jf(x)j ≤ c f2X∗;kfk =1 X!C 5. Show that for any sphere centered at the origin with radius r, Sr(0), in a normed space X and any point y 2 Sr(0) there is a hyperplane Hy 3 y such that Br(0) lies entirely in one of the two half spaces determined by the hyperplane Hy. Solution: Let y 2 Sr(0), and let Hy be the hyperplane containing y. As X is a normed space, it is possible to write X = Hy ⊕Ry (here we view things as a real vector space, the complex case is not any different). The space Ry is the span of the vector y. Note that the functional f : X ! R given by fy(h+ry) = r has kernel Hy. This algebraic reasoning can be turned around. From this decomposition of the space X, we can then see that a −1 hyperplane is given by the level sets of a linear functional, i.e., Hy = fy (0) = ker fy. Now consider the bounded linear functional f : Ry ! R given by f(ry) = r. Then, we can extend this linear functional to all of X by Hahn-Banach. Then we have that −1 Hy = f (0), and it is easy to see that the set Br(0) must lie in one of the half-spaces. It isn't too much more work to show that the same conclusion in fact holds when one is given a non-empty convex subset C of X. The interested student should attempt to work this out. 6. Of what category is the set of all rational numbers Q in R? Solution: The rationals Q are meagre in R. Note that the rationals Q are countable. S1 Let q1; q2;::: be an enumeration of Q, i.e., we can write Q = k=1 qk. Then it is clear that each set fqkg is nowhere dense. Thus, we have written Q as a countable union of nowhere dense sets in R. 7. Show that the complement M c of a meager subset M of a complete metric space X is non-meager. Solution: Let M be a meager subset of a complete metric space X. Suppose, for a contradiction, that M c is also meager. Then we can write 1 1 c [ [ M = Ak M = Bk k=1 k=1 where each Ak and Bk is nowhere dense in X. S c S1 S1 S1 But we have that X = M M = k=1 Ak [ k=1 Bk = k=1 Ck. So we have written X as a countable union of nowhere dense sets, and so X would be of the first category. But since X is complete, it must be of the second category, and so we have a contradiction. This then implies that M c is non-meager. 8. Let X and Y be Banach spaces. Suppose that Tn 2 B(X; Y ) is such that supn kTnk = +1. Show that there is a point x0 2 X such that supn kTnx0kY = +1. The point x0 is called a point of resonance. Solution: Suppose that there does not exist a point x0 2 X with the desired conclusion. Then for all x 2 X we have that sup kTnxkY < +1 n But, then the operators Tn satisfy the hypotheses of the Uniform Boundedness Principle, so we have that the sequence of norms kTnk is finite, i.e., sup kTnk < 1: n However, this clearly contradicts the hypotheses of the operators Tn, and so we must have that there exists x0 2 X such that sup kTnx0kY = +1 n 9.
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