Homework 2

1. Let X and Y be Hilbert spaces over C. Then a sesquilinear form h on XY is a mappingh : X Y C such that for all x1, x2, x X, y1, y2, y Y and all scalars , 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(x, y) = h(x, y);

(d) h(x, y) = h(x, y).

A sesquilinear form is bounded if |h(x, y)| C x y and the norm of the form h is givenby

hXYC = supx 6=0,y 6=0

|h(x, y)|x y = supx=1,y=1 |h(x, y)| .

Show that if h is a bounded sesquilinear form on the Hilbert spaces X and Y , then h hasthe representation

h(x, y) = Sx, yYwhere S : X Y is a bounded linear operator. Moreover, S is uniquely determined by hand has norm

SXY = hXYC .

Solution: The idea is to use the Riesz Representation to construct the linear operatorS. 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 his a bounded bilinear form, we have that L is a bounded linear functional on Y . By theRiesz Representation Theorem, we have then that there exists a unique element z Ysuch that

L(y) = y, zY ,or in terms of h that

h(x, y) = z, yY .While the point z is unique, it is depending on the point x that was fixed. We now definethe map S : X Y by Sx = z. This map is clearly well-defined since the point z is welldefined given the x. Substituting in this representation we find,

h(x, y) = Sx, yYas 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 , Cand x1, x2 X, then we have

S(x1 + x2), yY = h(x1 + x2, y)= h(x1, y) + h(x2, y)

= Sx1, yY + Sx2, yY= Sx1, yY + Sx2, yY= Sx1 + Sx2, yY .

Since this holds for all y Y we have that

Sx1 + Sx2 = S(x1 + x2)

and so S is linear.

To see that S is bounded, first note

SXY = supx 6=0

SxYxX

= supx 6=0,Sx6=0

|Sx, SxY |xX SxY

supx 6=0,y 6=0

|Sx, yY |xX yY

= supx 6=0,y 6=0

|h(x, y)|xX yY

= hXYC

In particular we have SXY hXYC, and so S is bounded. To see the otherinequality, we have

|h(x, y)| = |Sx, yY | SxY yY SXY yY xXwhich gives hXYC SXY .

For uniqueness, suppose that there are two linear operators S1 and S2 such that

h(x, y) = S1x, yY = S2x, yY .

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 istotal 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 formJ

kjekj

where J is a finite set, kj J , and kj = akj + ibkj , where akj , bkj Q. It is clear thatA 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

x vH <

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, byParseval, we can take y =

nk=1 x, ekH ek and havex

nk=1

x, ekH ekH