1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.