### A little Hilbert space problem

#### by Willie Wong

First let us consider the following question on a finite dimensional vector space. Let be a -dimensional Hermitian-product space. Let be an orthonormal basis for . Let be the linear operator defined by when , and . Does there exist any non-trivial vector such that and ?

The answer, in this case, is no. Present where are complex numbers. Let be the smallest index such that and . Similarly let be the largest non-vanishing index. If , then is a multiple of a standard basis element, and so is trivial. So assume . Now, by the requirement , we see that , which contradicts our assumption that are the minimum and maximum non-vanishing indices. In this proof, we used crucially that is finite dimensional, so that a largest element can exist.

Now, onto the real question

Question

Take the complex Hilbert space , i.e. the set of all complex sequences satisfying . Let latex T$ be the right shift operator: and . Then is an orthonormal basis of , and we have . Does there exist non-trivial elements of for which hold?

The answer is *yes* by the way.

It is easy to see that should such a exist, it must not be a finite sequence. Else the argument given in the finite dimensional case can easily be adapted again to show that for some .

Let me first give a construction which lists all such vectors. As is well known, is the Fourier dual of , the twice-integrable Hardy space on the circle. So let be one such vector, and take its inverse Fourier transform where runs from . Then we see that the right-shift operator becomes the phase-shift operator. By Parsival’s Identity, satisfies (up to a possible constant multiple that I always forget)

where denotes the complex conjugation. Now, since is in , we have that is real-valued, so also. Therefore is a real constant, and hence is equal to 1. This implies that only takes values in .

Now since is in a Hardy space, it can be represented by an analytic function in the unit-disc of the complex plane. More precisely, let . We have that from before. Since is an analytic function in the unit disc that extends continuously to the boundary, and that maps the boundary to itself, we can take the Blaschke product representation of . Therefore each such vector corresponds to an inner function .

Now let us look at an example. Observe that the requirement is analytic in the unit disc and maps the boundary to itself can be satisfied by any fractional linear transformation. So for our example we will choose . A direct computation of its Taylor expansion at the origin shows that

And it is now simple to check that the coefficients indeed answer our question above!

Using the Blaschke product representation also shows the following: the set of solutions to our question forms a semi-group, with the identity. The group product is obtained via the following