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 ∈ X, y1, y2, y ∈ Y and all scalars α, β ∈ 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 |h(x, y)| ≤ C kxk kyk and the norm of the form h is given by |h(x, y)| khk = sup = sup |h(x, y)| . 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 ∈ 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 ∈ 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 ∈ C and x1, x2 ∈ 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 ∈ Y we have that

αSx1 + βSx2 = S(αx1 + βx2)

and so S is linear.

To see that S is bounded, ﬁrst note

In particular we have kSk ≤ khk , and so S is bounded. To see the other X→Y X×Y →C inequality, we have

|h(x, y)| = |hSx, yiY | ≤ 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 ∈ X, and so S1 = S2.

2. Suppose that H is Hilbert space that contains an orthonormal sequence {ek} 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 ∈ J, and γkj = akj + ibkj , where akj , bkj ∈ 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 ∈ H and > 0 there is a v ∈ A such that

kx − vkH <

Since the sequence {ek} is total in H, there exists an integer n such that Yn = span{e1, . . . , en} 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 ∈ 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 ∈ 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 |f(x)| ≤ c for all f ∈ X∗ of norm at most 1. Show that kxk ≤ c.

Solution: We have the following fact at our disposal:

|f(x)| kxkX = sup f∈X∗,f6=0 kfk X→C . It is easy to show that

|f(x)| sup = sup |f(x)| f∈X∗,f6=0 kfk f∈X∗,kfk =1 X→C X→C The result then follows easily, since

kxkX = sup |f(x)| ≤ c f∈X∗,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 ∈ 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 ∈ 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. S∞ Let q1, q2,... be an enumeration of Q, i.e., we can write Q = k=1 qk. Then it is clear that each set {qk} 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

∞ ∞ c [ [ M = Ak M = Bk k=1 k=1

where each Ak and Bk is nowhere dense in X. S c S∞ S∞ S∞ 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 ∈ B(X,Y ) is such that supn kTnk = +∞. Show that there is a point x0 ∈ X such that supn kTnx0kY = +∞. The point x0 is called a point of resonance.

Solution: Suppose that there does not exist a point x0 ∈ X with the desired conclusion. Then for all x ∈ X we have that

sup kTnxkY < +∞ 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 < ∞. n

However, this clearly contradicts the hypotheses of the operators Tn, and so we must have that there exists x0 ∈ X such that

sup kTnx0kY = +∞ n

9. If X and Y are Banach spaces and Tn ∈ B(X,Y ), show that the following are equivalent:

(a) {kTnk} is bounded;

(b) {kTnxk} is bounded for all x ∈ X;

∗ (c) {|g(Tnx)|} is bounded for all x ∈ X and for all g ∈ Y . Solution: It is immediate that (a) implies (b) and (b) implies (c). So it suﬃces to prove that (c) implies (a). But, two applications of the Uniform Boundedness Principle gives this result. The ﬁrst application of UBP implies that the sequence {Tnx} is bounded for all x. This is problem 10 below. Then we can simply apply the standard version of the UBP to get that {kTnk} is bounded.

∗ 10. If {xn} is a sequence in a Banach space is such that for all f ∈ X the sequence {f(xn)} is bounded, show that {kxnk} is bounded.

∗ Solution: Deﬁne an operator Tn : X → C by Tn(f) = f(xn), then we have that Tn is a linear operator on the space X∗. By the hypothesis, we have that

sup |Tn(f)| = sup |f(xn)| n n is bounded for all f ∈ X∗. Then, by the Uniform Boundedness Principle, we have there there exists some universal constant C such that

kTnk ∗ ≤ C. X →C This implies that for all f ∈ X∗ of norm at most 1 that

sup |f(xn)| = sup |Tn(f)| ≤ C ∗ ∗ f∈X ,kfkX∗ ≤1 f∈X ,kfkX∗ ≤1 But, by the Theorem in class, we have that

kxnkX = sup |f(xn)| ≤ C ∗ f∈X ,kfkX∗ ≤1

which gives that {kxnkX } is bounded.