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.