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For a Dirichlet character $\chi$ modulo $N$, the Gauss sum attached to $\chi$ is given by $$G_\chi(m) = \sum_{k \in \mathbb{Z}_N} \chi(k) e^{2\pi i mk/N}.$$ Suppose one has an $N$-periodic function $f$ satisfying $$ G_\chi \ast f = G_\chi $$ for all Dirichlet characters $\chi$ modulo $N$. Here the convolution $\ast$ means $(g \ast h )(m) := \sum_{k \in \mathbb{Z}_N} g(k) h(m-k)$ for any two $N$-periodic functions $g,h$.

Question: Would you know what are (some) necessary and sufficient conditions that $f$ must satisfy for the equality to hold? It looks rather strange to me that this could happen. BTW this convolution equality is trivially satisfied if one takes $f = \delta$, the Kronecker delta function. But what can we say when $f \neq \delta$? Thanks a lot!

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We have that $f*G_\chi =G_\chi$ if and only if $f$ is of the form $$f(m)=\delta(m)+\sum_{r:\ (r,N)>1} c_r e^{2\pi m r /N},$$ where $\delta$ is the delta function and the $c_r$ can be equal to any complex numbers.

We can expand $G_\chi * f(m)$ as a double sum $$G_\chi * f(m)=\sum_{k\in\mathbb{Z}_{N}}\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi ikr/N}f(m-k).$$ Rearranging this we obtain $$G_\chi * f(m)=\sum_{r\in\mathbb{Z}_{N}}\chi(r)\sum_{k\in\mathbb{Z}_{N}}f(m-k)e^{2\pi ikr/N}$$ $$=\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi imr/N}\sum_{k\in\mathbb{Z}_{N}}f(k)e^{-2\pi ikr/N}=\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi imr/N}\hat{f}(r).$$ Now, you are asking when do we have the equality $$\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi imr/N}\hat{f}(r)=\sum_{r\in\mathbb{Z}_{N}}\chi(r)e^{2\pi imr/N}$$ for all $m$. This is automatically satisfied if $\hat{f}(r)=1$ for all $(r,N)=1$, and since I can isolate any particular coefficient by taking sums over many different values of $m$, this happens precisely when $\hat{f}(r)=1$ for all $(r,N)=1$. To obtain the stated result, we take the inverse Fourier transform.

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  • $\begingroup$ Unfortunately that $\chi(r) \hat{f}(r)= \chi(r)$ only implies that $\hat{f}(r) = 1$ whenever $(r, N) = 1$. When $(r, N) > 1$, we have $\chi(r) = 0$ and so comparing coefficients gives $0 = \hat{f}(r) \cdot 0$. So $\hat{f}$ may not be $1$ as you say. $\endgroup$ – user152169 Jul 15 '15 at 15:25
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    $\begingroup$ @user152169: I was thinking of $N$ being prime. In any case this solution completely characterizes all such functions $f$. We have $G_\chi*f=G_\chi$ if and only if $\hat{f}(r)=1$ for all $(r,N)=1$. It doesn't matter what the value of $\hat{f}(r)$ is for $(r,N)>1$. $\endgroup$ – Eric Naslund Jul 15 '15 at 18:34
  • $\begingroup$ So this is a characterization in terms of the Fourier transform of the function $f$. It would be very nice (and what I'd like to see) if something direct could be said purely in terms of $f$. What does your characterization say about how $f$ "looks", say in terms of its values? $\endgroup$ – user152169 Jul 15 '15 at 19:42
  • $\begingroup$ @user152169 Just take the inverse Fourier transform. It follows that $$f(m)=\delta(m) + \sum_{r:\ (r,N)>1} c_r e^{2\pi i m r/N}$$ where the $c_r$ can be any constants we like. $\endgroup$ – Eric Naslund Jul 15 '15 at 20:40

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