# A Little Hilbert Space Problem

First let us consider the following question on a finite dimensional vector space.

Let $(V, \langle\cdot\rangle)$ 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, giving a "trivial" solution. 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 $T$ 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 turns out to be yes.

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 Parseval's Identity, $\check{v}$ satisfies (up to a possible constant multiple that I always forget) $\int_\mathbb{T} \check{v}\check{v}^\star e^{2\pi i a x} dx = \delta_0^a~,\quad a\geq 0$ where $f^\star$ denotes the complex conjugation. Now, since $\check{v}$ is in $H^2$, we have that $\check{v}\check{v}^\star$ is real-valued, so $\int_\mathbb{T} \check{v}\check{v}^\star e^{-2\pi i a x} dx = \bigl(\int_\mathbb{T} \check{v}\check{v}^\star e^{2\pi i a x} dx\bigr)^\star = \delta_0^a$ also. Therefore $\check{v}\check{v}^\star$ 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 $\tilde{v}(z) = -\frac12 + \frac32 \sum_{k = 1}^{\infty}\left(\frac{z}{2}\right)^k .$ 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 convolution-type operation $(ab)_k = \sum_{j,l\geq 0; j + l = k} a_jb_l.$ ##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.