Let $(a_n)$ be a (double-sided) sequence of complex numbers satisfying $$\sum_{\mathbb{Z}}\vert n\vert\,\vert a_n\vert^2<\infty, \tag1$$ $$\sum_{\mathbb{Z}}a_n\bar{a}_{n+k}=\delta_0(k), \qquad \forall k\in\mathbb{Z}. \tag2$$

EDIT (revealing previously withheld information). A theorem of Boutet de Monvel and Gabber states that under these conditions, $\sum_{\mathbb Z}n|a_n|^2$ converges to an integer. This proof is high-tech and goes through an analysis of circle-valued functions of the circle to the circle that belong to the Sobolev space $H^{1/2}(\mathbb S^1)$. (The $a_n$ are the Fourier coefficients of the function and the integer is the "degree of the function" - a concept generalizing the classical "winding number" of a function around the unit circle).

QUESTION. Can you give a proof based on basic complex analysis?

REMARK. Condition (1) is an alternative Sobolev space qualification for $f\in H^{1/2}(\mathbb{S}^1)$ in terms of the Fourier coefficients $a_n$ of $f\in L^2(\mathbb{S}^1)$. Condition (2) ensures that $f$ is circle-valued. The quantity $\sum_{\mathbb Z}n|a_n|^2$ is $(1/2\pi i)\int \bar f(z)\,f'(z)\,dz=(1/2\pi i)\int f'(z)/f(z)\,dz$, which is the winding number in the case that $f$ is differentiable.

NOTATION. Here $\mathbb{S}^1$ is the unit circle, $\bar{a}$ is complex conjugation and $\delta_0(k)$ is the Dirac-delta function $\delta_0(0)=1$ and $\delta_0(k)=0$ otherwise.

  • $\begingroup$ When you say "$a_n$ is a sequence of infinite series", do you mean that each $a_n$ is an infinite series? Or did you just mean to say that $(a_n)_{n\in\mathbb Z}$ is a (double-sided) sequence? $\endgroup$ – Joe Silverman Nov 22 '16 at 19:13
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    $\begingroup$ So this is quite an interesting question. Your conditions basically say that $f$ satisfies $|f|=1$ and the quantity you are asking about is $(1/2\pi)\int f'(z)\bar f(z)\,dz$, which at least for differentiable functions is the number of times that the argument of $f(z)$ goes around the circle. I'm not familiar enough with Sobolev spaces to see quickly if this extends to $W^{1/2,2}$. In particular, is there a continuous (in the Sobolev space) path of functions from $f(z)=1$ to $f(z)=z$ with range in the circle? $\endgroup$ – Anthony Quas Nov 22 '16 at 20:09
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    $\begingroup$ If $f(x) = \sum_{n=-\infty}^{\infty }a_n e^{2\pi i n x}$ then your second condition is troubling me. Essentially it reads $\int_0^1 f(x)\overline{f(x)}e^{-2\pi i k x} \,dx = \delta_0(k)$. But then if $g(x) = |f(x)|^2$ it satisfies $\int_0^1 g(x)e^{2\pi i k x}\,dx = 0$ for all $k \neq 0$. I'm no pro at this but doesn't this imply $g$ and therefore $f$ is constant? particularly $f = 1$ by your normalization conditions? $\endgroup$ – user78249 Nov 22 '16 at 20:14
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    $\begingroup$ It is disingenuous to say that no proof has been furnished yet $\endgroup$ – Anthony Quas Nov 24 '16 at 2:37
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    $\begingroup$ I have edited the post to include the information that you withheld in the original posting. I believe this fixes the mistake that you have acknowledged making (I would have expected you to do this yourself). $\endgroup$ – Anthony Quas Nov 24 '16 at 7:42

As correctly pointed by Anthony in the comments below, the argument I attempted here originally did not hold water.

These issues apparently have been studied quite extensively recently. See for example J. Bourgain, One cannot hear the winding number. In particular, in the introduction to this paper, Bourgain mentions that the answer to your question is yes; the result is attributed to Boutet de Monvel and Gabber.

  • $\begingroup$ I don't seem to follow where property (3) above has been shown for $h_k$ and they lead to property (3) for $f$. $\endgroup$ – T. Amdeberhan Nov 23 '16 at 4:32
  • $\begingroup$ Hang on a minute! I agree that $|u(y)-u(x)|$ is comparable to $|g(x)-g(y)|$ on every $A_n$, but the comparison gets worse and worse as $n$ increases. I don't think I believe that there is a $g$ with $|g(x)|<\pi$ that lies in $H^{1/2}$. For a specific example, what if $f(z)=z^2$. Then $g(x)$ should be $g(x)=4\pi x$, except this doesn't take values in $(-\pi,\pi)$. If you force it to take principal values, I think you don't satisfy the integrability condition any more. $\endgroup$ – Anthony Quas Nov 23 '16 at 6:25
  • $\begingroup$ @AnthonyQuas: You are right of course. $\endgroup$ – Christian Remling Nov 23 '16 at 20:58
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    $\begingroup$ So, wait: you knew about the Boutet de Monvel & Gabber work, Amdeberhan, and deliberatley withheld this information when posting the question? $\endgroup$ – Gerry Myerson Nov 23 '16 at 23:23
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    $\begingroup$ Yes! This does bother us! Why not say: here is a fact about sequences of complex numbers where the only proof I know goes through high level harmonic analysis (give reference). Does anyone see how to give a more elementary proof? $\endgroup$ – Anthony Quas Nov 24 '16 at 2:35

This is probably true. Here's a long way to try to prove a special case that was already completely proven in the comments to the OP.

Let's wonder what could be said if $a_{j} = 0$ for $j < 0$. Consider the formal power series $f(z) = \sum_{n\geq 0} a_n z^n$. Since $|a_n|$ is summable, $f(z)$ is analytic in $|z| < 1$, and in fact it can be extended to a continuous function on $|z|\leq 1$. Consider $f(u) \overline{f(v)}$, which has the Taylor series $$ f(u) \overline{f(v)} = \left(\sum_{n \geq 0} a_n u^n \right) \overline{\left(\sum_{m \geq 0} a_m v^m \right) } = \sum_{n,m} a_n u^{n} \overline{a_m v^m} = \sum_{k \in \mathbb{Z}} \sum_{n\geq 0} a_n u^{n} \overline{a_{n+k} v^{n+k}}, $$ where we interpret $a_{j} = 0$ if $j < 0$. This in turn is equal to $$ f(u) \overline{f(v)} = \sum_{k \in \mathbb{Z}} \overline{v}^k \sum_{n\geq 0} a_n \overline{a_{n+k}} (u\overline{v})^{n}. $$ Specializing $u=v$ for $|u| = 1$ (or taking limits as $|u| \to 1$) shows $$ f(u) \overline{f(u)} = \sum_{k \in \mathbb{Z}} \overline{u}^k \sum_{n\geq 0} a_n \overline{a_{n+k}} = 1. $$ Thus, we have that $|f(u)| = 1$ for all $|u|=1$ (or rather, view this as a statement about limits). This implies that $f(u)$ is a finite Blaschke product, meaning it is of the form $$ f(u) = c \prod_{i=1} ^{N} \left( \frac{u - r_i}{1-\overline{r_i}u}\right)^{m_i}, $$ where $m_i$ are non-negative integers, $c$ is a constant of modulus $1$ and the $r_i$ are constants of modulus less than $1$.

From here, I suspect it might be possible to finish the proof by pushing pretty hard. Let's see what happens for the first non-trivial case, which is $f(z) = \frac{z-r}{1-rz}$, where $|r|<1$ is a real number parameter. This has Taylor series $$ f(z) = -r + \sum_{n=0} ^{\infty} z^{n+1} r^n (1-r^2). $$

So our corresponding sequence will be $$a_{j} = \begin{cases}-r, \qquad &\text{if $j=0$}\\ r^{j-1} (1-r^2), \qquad &\text{if $j> 0$}\\ 0, \qquad &\text{if $j<0$} \end{cases}$$

This sequence satisfies $\sum_{n} n |a_n|^2 < \infty$ (clearly since $|r| < 1$), and $\sum_{n} a_{n}^2 = r^2 + (1-r^2)^2 \sum_{n=0}^{\infty}r^{2n} = 1,$ and for $k > 0$ (which is all we need to check by symmetry) we have $$ \sum_{n} a_{n} \overline{a_{n+k}} = -r \cdot r^{k-1}(1-r^2) + \sum_{n\geq 0} r^{n}(1-r^2)r^{n+k}(1-r^2) = 0.$$

And we have $$ \sum_{n} n |a_n|^2 = \sum_{n \geq 1} n (1-r^2)^2 r^{2(n-1)} = 1, $$ as we wanted.

The fact that it works out for this first case is a very positive sign, but not yet a proof.

  • $\begingroup$ No. Rather, in sentence 3 of my answer, I asked what can be said if $a_j = 0$ for $j< 0$. $\endgroup$ – Pat Devlin Nov 22 '16 at 22:39
  • $\begingroup$ Oh! So it is. Nevermind then! $\endgroup$ – Pat Devlin Nov 22 '16 at 22:50

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