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Let $a=(a_n)$ with $a_n\in\mathbb{C}$ be a vector indexed over all $n\in\mathbb{Z}$, and consider the system of equations $\sum\limits_{-\infty}^\infty\overline{a_n}a_{n+k}=\delta_{k0}$ for all $k\in\mathbb{Z}$. One may verify this has a family of trivial solutions given by $a_n=\delta_{nm}$ for some nonzero integer $m$. Assuming $a_0=0$, are there any other solutions?

This problem is inspired by (unsuccessful) attempts to find a tractable solution to an earlier question of mine. What I wanted was a closed curve $z(s)\in\mathbb{C}$ whose Fourier series was arc-length parametrized i.e.

$z(s)=\sum\limits_{n=-\infty}^\infty c_n e^{i n s}$ with $|z'(s)|^2=\sum\limits_{nm}(n\overline{c_n})(m c_m)e^{i (m-n)s}=1$ for all $s\in\mathbb{R}$. This requires $\sum\limits_{-\infty}^\infty nm\overline{c_n}c_{m}=\delta_{nm}$ for all $m$, which upon identifying $m=n+k$ and $a_n=n c_n$ yields the system of equations above.

Unfortunately, the only solutions which are obvious to me are the trivial ones given above (i.e. a single mode $e^{i m s}$). Any nontrivial solutions appear to involve all frequencies; a construction of such would clarify my questions greatly.

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Let $f(z)$ be holomorphic in a neighborhood of the unit circle such that $|f(z)|=1$ whenever $|z|=1$. There are plenty of such functions. Examples are $z^k$ for $k\in \mathbb{Z}$ and the Möbius functions $$\frac{z-a}{1-\overline{a}\,z}$$ for $|a|\neq 1$. Then you can take products of such functions or compositions etc. The key observation is the following. Let the Laurent expansion of $f$ on the unit circle be $$f(z)=\sum_{n\in \mathbb Z}a_nz^n.$$ Let $f^{\ast}$ be the holomorphic function defined by $f^{\ast}(z)=\overline{f(1/\overline{z})}$ then $$f(z)f^{\ast}(z)=\sum_{n\in\mathbb{Z}}\left(\sum_{k\in\mathbb{Z}}a_{k+n}\overline{a}_k\right) z^n.$$ If $|z|=1$ then $f(z)f^{\ast}(z)=|f(z)|^2=1$ and therefore (since it is holomorphic) $f(z)f^{\ast}(z)$ is constant on a neighborhood of the unit circle. In other words, all its Laurent coefficients except the constant term are equal to zero. So the Laurent coefficients of $f$ itself have the required property.

For $f(z)=z^k$ this gives the trivial examples. For $0<|a|<1$ the Möbius function $$f(z)=\frac{z-a}{1-\overline{a}\,z}=-a +(1-|a|^2)\sum_{n=0}^{\infty}\overline{a}^n z^{n+1}$$ gives non-trivial examples.

**Edit:** I overlooked the requirement $a_0=0$ but this can be fixed: Since the Möbius example above is a holomorphic function on the unit disc all its negative-index coefficients are $0$ so one can instead consider $z^kf(z)$ for some $k\geq 1$ to get $a_0=0$.

As $\sum_{k\in\mathbb Z}\lvert a_k\rvert^2<\infty$, then this corresponds to a $2\pi-$periodic $L^2-$function $f$, with

$$

f(x)=\sum_{k\in\mathbb Z}a_k\,\mathrm{e}^{ikx},

$$

and $\int_0^{2\pi}\lvert\,f(x)\rvert^2\,dx=2\pi\sum_{k\in\mathbb Z}\lvert a_k\rvert^2$.

Now the relation

$$

\sum_{n\in\mathbb Z}\overline{a}_na_{n+k}=\delta_{k0}, \qquad (\star)

$$

means that $\widehat{\lvert\,f^2\rvert}_k=\delta_{k0}$, or

$$

\int_0^{2\pi} \lvert \,f(x)\rvert^2\,\mathrm{e}^{ikx}\,dx=2\pi\,\delta_{k0},

$$

which means that $\lvert\, f(x)\rvert=1$, almost everywhere. So the general form of the infinite vector $\{a_k\}_{k\in\mathbb Z}$, satsfying $(\star)$ HAS TO BE of the form

$$

a_k = \frac{1}{2\pi}\int_0^{2\pi} f(x)\,\mathrm{e}^{ikx}\,dx,

$$

where $f :[0,2\pi]\to\mathbb C$, measurable, with $\lvert\,f(x)\rvert=1$.

In particular, if $f$ is real valued, then it has to be of the form

$$

f(x)=\left\{ \begin{array}{rcl}

1 & \text{if} & x\in E, \\

-1 & \text{if} & x\in [0,2\pi]\smallsetminus E,

\end{array}

\right.

$$

where $E$ is a measurable subset of $[0,2\pi]$.

Hence, the general $\{a_k\}_{k\in\mathbb Z}$ is of the form

$$

a_k=\frac{1}{2\pi}\int_0^{2\pi} f(x)\,\mathrm{e}^{ikx}\,dx=\frac{1}{2\pi}\left(\int_{E}\mathrm{e}^{ikx}\,dx-\int_{[0,2\pi]\smallsetminus E}\mathrm{e}^{ikx}\right).

$$

For example, if $E=[0,\pi]$, then

$$

a_k=\frac{1}{2\pi}\left(\int_{0}^\pi\mathrm{e}^{ikx}\,dx-\int_{\pi}^{2\pi}\mathrm{e}^{ikx}\right)=\frac{1}{2\pi}\left(\frac{e^{k\pi i}}{ik}-\frac{1}{ik}-\frac{e^{2k\pi i}}{ik}+\frac{e^{k\pi i}}{ik}\right) \\

=\left\{ \begin{array}{lll} 0 & \text{if} & k\,\, \text{even},\\

\dfrac{2i}{k\pi} & \text{if} & k\,\, \text{odd}.

\end{array}\right.

$$

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