A little Hilbert space problem

by Willie Wong

First let us consider the following question on a finite dimensional vector space. Let (V, \langle\rangle) be a k-dimensional Hermitian-product space. Let (e_i)_{1\leq i \leq k} be an orthonormal basis for V. Let T:V\to V be the linear operator defined by T(e_i) = e_{i+1} when i < k, and T(e_k) = 0. Does there exist any non-trivial vector v\in V such that \langle v,v\rangle = 1 and \langle v, T^jv\rangle = 0?

The answer, in this case, is no. Present v = \sum v_i e_i where v_i are complex numbers. Let a be the smallest index such that v_a \neq 0 and v_i = 0, i < a. Similarly let b be the largest non-vanishing index. If a = b, then v = v_a e_a is a multiple of a standard basis element, and so is trivial. So assume a < b. Now, by the requirement \langle v, T^{b-a}v\rangle = 0, we see that v_a v_b = 0, which contradicts our assumption that a,b are the minimum and maximum non-vanishing indices. In this proof, we used crucially that V is finite dimensional, so that a largest element b can exist.

Now, onto the real question

Take the complex Hilbert space \ell^2(\mathbb{N}), i.e. the set of all complex sequences (a_i)_{0\leq i < \infty} satisfying \sum_{i\in\mathbb{N}} |a_i|^2 < \infty. Let e = (1,0,0,\ldots), and let latex T$ be the right shift operator: (Ta)_{i+1} = a_i and (Ta)_0 = 0. Then T^ke is an orthonormal basis of \ell^2, and we have \langle e, T^ke\rangle = \delta_0^k. Does there exist non-trivial elements of \ell^2 for which \langle v, T^kv\rangle = \delta_0^k hold?

The answer is yes by the way.

It is easy to see that should such a v 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 v = v_a T^ae for some a.

Let me first give a construction which lists all such vectors. As is well known, \ell^2(\mathbb{N}) is the Fourier dual of H^2(\mathbb{T}), the twice-integrable Hardy space on the circle. So let v be one such vector, and take its inverse Fourier transform \check{v} = \sum_a v_a e^{2\pi i a x} where x runs from [0,1). Then we see that the right-shift operator T becomes \check{T}f(x) = e^{2\pi i x} f(x) the phase-shift operator. By Parsival’s Identity, \check{v} satisfies (up to a possible constant multiple that I always forget)

\displaystyle \int_\mathbb{T} \check{v}\check{v}^* e^{2\pi i a x} dx = \delta_0^a~,\quad a\geq 0

where f^* denotes the complex conjugation. Now, since \check{v} is in H^2, we have that \check{v}\check{v}^* is real-valued, so \int_\mathbb{T} \check{v}\check{v}^* e^{-2\pi i a x} dx = (\int_\mathbb{T} \check{v}\check{v}^* e^{2\pi i a x} dx)^* = \delta_0^a also. Therefore \check{v}\check{v}^* is a real constant, and hence is equal to 1. This implies that \check{v} only takes values in \mathbb{T}.

Now since \check{v} is in a Hardy space, it can be represented by an analytic function in the unit-disc of the complex plane. More precisely, let \tilde{v}(z) = \sum v_a z^a. We have that |\tilde{v}(e^{i\theta})| = 1 from before. Since \tilde{v} 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 \tilde{v}. Therefore each such vector v corresponds to an inner function \tilde{v}.

Now let us look at an example. Observe that the requirement \tilde{v} 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 \tilde{v}(z) = (2z - 1)/(2-z). A direct computation of its Taylor expansion at the origin shows that

\displaystyle \tilde{v}(z) = -\frac12 + \frac32 \sum_{k = 1}^{\infty}\left(\frac{z}{2}\right)^k

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 v to our question forms a semi-group, with e the identity. The group product is obtained via the following

\displaystyle (ab)_k = \sum_{j,l\geq 0; j + l = k} a_jb_l