Suppose that $f$ is a Schwartz function, and suppose that its periodization vanishes, i.e. $$ \sum_{k\in \mathbb{Z}}f(x+k)=0$$

Next, suppose that $\pi$ is a periodic tempered distribution, i.e. $\pi(T_k g)=\pi(g)$, for all $k\in \mathbb{Z}$ and all Schwartz functions $g$, where $T_k g(x)=g(x+k)$ is the translation operator.

Prove that $\langle\pi,f\rangle=0$.

Ideas: (1) If we can show that $(2N)^{-1}\sum_{k=-N}^{N}f(x+k)$ converges weakly to zero (under the assumption that $f$'s periodization vanishes) then we reach a solution.

(2) The fact that the periodization of $f$ vanishes implies that its (inverse) Fourier transform $f^{\vee}(\xi)$ vanishes on the integers. The fact that $\pi$ is periodic implies that $(1-e^{2\pi i \xi})\hat{\pi}=0$, where $\hat{\pi}$ is the Fourier transform of $\pi$. If we can show that $(1-e^{2\pi i\xi})^{-1}f^\vee(\xi)$ is a Schwartz function, then we can establish that $$\langle \pi,f\rangle=\langle\hat{\pi},f^\vee\rangle=\langle(1-e^{2\pi i \xi})\hat{\pi},(1-e^{2\pi i \xi})^{-1}f^\vee\rangle=0,$$ as desired.


Let $T$ be a $1$-periodic (tempered) distribution.

Take $\phi \in C^\infty_c([0,1/2])$ constant $=1$ on $[1/5,2/5]$. Let $$T_1 = T\ \sum_n \phi(x+n), \qquad T_2 = T\ (1-\sum_n \phi(x+n))$$ so that $T = T_1+T_2$ where $T_1 = 0$ on $\mathbb{Z}+(1/2,1)$, $T_2 = 0$ on $\mathbb{Z}+(1/5,2/5)$. Then

$$\forall \varphi \in S(\mathbb{R}),\qquad P_\varphi(x) = \sum_n \varphi(x+n) \in C^\infty_{1-\text{periodic}} \\ \langle T, \varphi \rangle = \langle T_1,\varphi \rangle+\langle T_2,\varphi \rangle =\int_{\frac{3}{4}}^{\frac{3}{4}+1}T_1(x)P_\varphi(x)dx+\int_{\frac{1}{4}}^{\frac{1}{4}+1}T_2(x) P_\varphi(x)dx$$ and hence $T$ can be seen as a distribution acting on $C^\infty_{1-\text{periodic}}$ functions.

  • $\begingroup$ Thanks for the answer, but I'm still confused. $\endgroup$ – Dylan Sep 6 '17 at 4:06
  • $\begingroup$ @Dylan With what ? Did you realize I proved periodic distributions have a Fourier series, since with the decomposition $T = T_1+T_2$ then $\int_0^1 T(x) e^{-2i \pi nx}dx \overset{def} = \int_{3/4}^{3/4+1} T_1(x)e^{-2i \pi nx}dx+\int_{1/4}^{1/4+1} T_2(x)e^{-2i \pi nx}dx$ is well-defined $\endgroup$ – reuns Sep 6 '17 at 4:11
  • $\begingroup$ I don't get how $\int_{3/4}^{3/4+1}T_{1}(x)e^{-2i\pi nx}\,dx$ makes sense when $T_1$ is merely a distribution and not a function. $\endgroup$ – Dylan Sep 6 '17 at 4:14
  • $\begingroup$ @Dylan $\langle \delta, e^{2i \pi n x} 1_{|x| < 1/2} \rangle = \int_{-1/2}^{1/2} \delta(x) e^{-2i \pi nx}dx$ makes sense because $\delta$ is the zero distribution around $\pm 1/2$. What clearly doesn't make sense is $\int_0^1 \delta'(x)dx$. $\endgroup$ – reuns Sep 6 '17 at 4:22
  • $\begingroup$ yes but $T_1$ isn't $\delta$. What if $T_1$ happened to be $\delta(x-3/4)\,dx$, what would $\int_{3/4}^{3/4+1}T_1(x)e^{-2i\pi nx}dx$ be? $\endgroup$ – Dylan Sep 6 '17 at 4:26

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