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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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    $\begingroup$ You require a sequence $a_n$ that satisfies $|\sum_n a_n e^{-iun}| = 1$. A "square wave" in $u$ satisfies this condition and would have Fourier coefficients $a_n = 1/n$ for odd $n$, and $a_n = 0$ for even $n$. If you normalize so that $\sum_n |a_n|^2 = 1$ then I think this sequence would work. Posting as a comment instead of an answer since I haven't verified it yet. $\endgroup$ – Bungo Aug 25 '14 at 2:49
  • $\begingroup$ @Bungo: The problem with the square-wave as a particular example is that it's not smooth, and therefore z(s) isn't a nice plane curve. But that's an objection at the level of the underlying motivation, rather than the problem as stated. (I probably should've asked the problem directly at the level of finding a suitable function, rather than focusing on the algebraic constraints in particular.) $\endgroup$ – Semiclassical Aug 25 '14 at 4:23
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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 $a_0=0$.

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  • $\begingroup$ Nice job. These definitely fulfill the conditions I set out. I think the example of products of such transforms has the most interesting possibilities. $\endgroup$ – Semiclassical Aug 26 '14 at 22:02
  • $\begingroup$ 1) Do you know any special functions which are of the form you state? (Perhaps with some complicated dependence of parameters for the mobius transforms.) 2) You might find my most recent question of interest as well; it's related to the motivation of this one, but without an explicit appeal to Fourier coefficients. $\endgroup$ – Semiclassical Aug 26 '14 at 22:20
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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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