Orthonormal bases formed by translation of sequences

The standard orthonormal basis (ONB) in the Hilbert space ${\ell^2(\mathbb N_0)}$ consists of the vectors
(1, 0, 0, 0, …)
(0, 1, 0, 0, …)
(0, 0, 1, 0, …)
…
Let S be the forward shift operator: ${S(x_0, x_1, \dots) = (0, x_0, x_1, \dots)}$ . The aforementioned ONB is precisely the orbit of the first basis vector ${e_0}$ under the iteration of S. Are there any other vectors x whose orbit under S is an ONB?

If one tries to construct such x by hand, taking some finite linear combination of ${e_k}$ , this is not going to work. Indeed, if the coefficient sequence has finitely many nonzero terms, then one of them, say, ${c_m}$ , is the last one. Then ${S^m x}$ is not orthogonal to ${x}$ because the inner product is ${\overline{c_0} c_m}$ and that is not zero.

However, such vectors x exist, and arise naturally in complex analysis. Indeed, to a sequence ${(c_n)\in\ell^2(\mathbb N_0)}$ we can associate a function ${f(z) = \sum_{n=1}^\infty c_n z^n}$ . The series converges in the open unit disk to a holomorphic function which, being in the Hardy space ${H^2}$ , has boundary values represented by an square-integrable function on the unit circle ${\mathbb T}$ . Forward shift corresponds to multiplication by ${z}$ . Thus, the orthogonality requires that for every ${k\in \mathbb N}$ the function ${z^kf}$ be orthogonal to ${f}$ in ${L^2(\mathbb T)}$ . This means that ${|f|^2}$ is orthogonal to all such ${z^k}$ ; and since it’s real, it is orthogonal to ${z^k}$ for all ${k\in \mathbb Z\setminus \{0\}}$ by virtue of conjugation. Conclusion: |f| has to be constant on the boundary; specifically we need |f|=1 a.e. to have a normalized basis. All the steps can be reversed: |f|=1 is also sufficient for orthogonality.

So, all we need is a holomorphic function f on the unit disk such that almost all boundary values are unimodular and f(0) is nonzero; the latter requirement comes from having to span the entire space. In addition to the constant 1, which yields the canonical ONB, one can use

A Möbius transformation ${f(z)=(z-a)/(1-\bar a z)}$ where ${a\ne 0}$ .
A product of those (a Blaschke product), which can be infinite if the numbers ${a_k}$ converge to the boundary at a sufficient rate to ensure the convergence of the series.
The function ${f(z) = \exp((z+1)/(z-1))}$ which is not a Blaschke product (indeed, it has no zeros) yet satisfies ${|f(z)|=1}$ for all ${z\in \mathbb T\setminus \{-1\}}$ .
Most generally, an inner function which is a Blaschke product multiplied by an integral of rotated versions of the aforementioned exponential function.

Arguably the simplest of these is the Möbius transformation with ${a=1/2}$ ; expanding it into the Taylor series we get
$-\frac12 +\sum_{n=1}^\infty \frac{3}{2^{n+1}} z^n$
Thus, the second simplest ONB-by-translations after the canonical one consists of
(-1/2, 3/4, 3/8, 3/16, 3/32, 3/64, …)
(0, -1/2, 3/4, 3/8, 3/16, 3/32, …)
(0, 0, -1/2, 3/4, 3/8, 3/16, …)
and so on. Direct verification of the ONB properties is an exercise in summing geometric series.

What about the exponential one? The Taylor series of ${\exp((z+1)/(z-1))}$ begins with
$\frac{1}{e}\left(1 - 2z +\frac23 z^3 + \frac23 z^4 +\frac25 z^5 + \frac{4}{45} z^6 - \frac{10}{63} z^7 - \frac{32}{105}z^8 - \cdots \right)$
I don’t know if these coefficients in parentheses have any significance. Well perhaps they do because the sum of their squares is ${e^2}$ . But I don’t know anything else about them. For example, are there infinitely many terms of either sign?

Geometrically, a Möbius transform corresponds to traversing the boundary circle once, a Blaschke product of degree n means doing it n times, while the exponential function, as well as infinite Blaschke products, manage to map a circle onto itself so that it winds around infinitely many times.

Finally, is there anything like that for the backward shift ${S^*(x_0, x_1, x_2, \dots) = (x_1, x_2, \dots)}$ ? The vector ${(S^*)^k x}$ is orthogonal to ${x}$ if and only if ${x}$ is orthogonal to ${S^kx}$ , so the condition for orthogonality is the same as above. But the orbit of any ${\ell^2(\mathbb N_0)}$ vector under ${S^*}$ tends to zero, thus cannot be an orthonormal basis.

Orthonormal bases formed by translation of sequences

Related

Leave a Reply Cancel reply