I try to understand the Poisson summation formula from the perspective of distribution theory. However, I got stuck at a problem on the way, namely proving that the distribution solution to $xT = 0$ in the Schwartz space (infinitely differentiable functions of rapid decay at infinity) $\mathcal{S}(\mathbb{R})$ is $C\sum_{k \in \mathbb{Z}} \delta_k$, where $\delta_k$ is the shifted Dirac delta distribution, $C$ a constant. (I hope) I managed to prove the related question, where the distribution need not be 'tempered', meaning instead of Schwartz functions we take all compactly supported infinitely differentiable functions. It basically amounts to using the Hadamard lemma. However, when trying to modify the proof for the Schwartz space, I do not know where should the sum $C\sum_{k \in \mathbb{Z}} \delta_k$ pop out? I would appreciate any hints!

  • $\begingroup$ Why do you think, any shifted delta plays a role? If $\Omega\subset \mathbb{R}$ is an open set with $0\notin\Omega$ it is obvious to see $T(\varphi)=0$ for every test function $\varphi$ with support$(\varphi)\subseteq\Omega$. This implies support$(T)\subseteq \{0\}$ and such distributions have a representation as a finite linear combination of derivatives of delta's. $\endgroup$ – Vobo Feb 16 '14 at 16:24
  • $\begingroup$ As I understand, PSF states that the Dirac comb is invariant under the Fourier transform. To prove it, I brought the Fourier transform of $\sum_{k \in \mathbb{Z}} \delta_k$ to the form $\sum_{k \in \mathbb{Z}} e^{-2\pi iks}$. If we call the last distribution T, then by periodicity it can be proven that $e^{-2\pi is}T = T$ so $(e^{-2\pi is}-1)T = 0$. Now I would like to know the solution to $xT = 0$ in Schwartz space, and I expect it to be exactly the Dirac comb. As for the part where you mention the support of distribution I'll try to read more about it and understand your comment. Thanks! $\endgroup$ – mateusz Feb 16 '14 at 18:45

Vobo's comment is correct: the space of solutions of $xT=0$ does not include the shifted Dirac functions. Indeed, for $T=\delta_k$ we have $xT=k\delta_k$ which is nonzero if $k\ne 0$.

To describe the space of solutions, suppose $xT=0$. Every test function $\varphi$ vanishing at $0$ can be written as $\varphi(x)=x\psi(x)$, see Quotient of two smooth functions is smooth. Thus, $T\varphi = T(x\psi) = xT(\psi)=0$. Fix any test function $\chi$ such that $\chi(0)=1$ and let $a=T\chi$. Then for every test function $\varphi$ we have $$T\varphi = T(\varphi-\varphi(0)\chi)+T(\varphi(0)\chi) = 0 +a\varphi(0)$$ which demonstrates that $T=a\delta_0$.

Thus, the space of solutions of $xT=0$ is one-dimensional: it contains only the multiples of the Dirac function at $0$.

In case of Poisson summation, you are looking at $gT=0$ where $g(x)=\exp(-2\pi ix)$. I claim that the solution space is spanned by $\delta$-functions at integers. Indeed, let $\chi$ be a test function supported on $[-1/2,1/2]$ such that $\chi(0)=1$. Let $a_n = T\chi_{ n}$ where the subscript indicates translate of $\chi$ by $ n$. Then for every test function $\varphi$ we have $$T\varphi = T\left(\varphi-\sum_n \varphi(n)\chi_n\right )+\sum_n T(\varphi(n)\chi_n) = 0 +\sum_n a_n \varphi(n)$$ which demonstrates that $T=\sum_n a_n\delta_n$.

To justify $T\left(\varphi-\sum_n \varphi(n)\chi_n\right )=0$, observe that $$\psi:= \frac{\varphi-\sum_n \varphi(n)\chi_n}{g}$$ is a test function: indeed, smoothness is a local property, and it is verified at each integer $n$ separately. The decay of derivatives may be a bit annoying to verify. My suggestion is to work with compactly supported $\varphi$; once you've shown that $T$ agrees with $\sum_n a_n\delta_n$ on compactly supported test functions, the equality extends to the Schwarz space by density.

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  • $\begingroup$ That's a nice and simple proof. $\endgroup$ – Vobo Feb 17 '14 at 7:50
  • $\begingroup$ Now I understand where I was confused.. But going back to the Poisson summation formula (my final goal), what 'trick' would I use to solve $(e^{-2\pi is}-1)T = 0$. Your solution relies on the Hadamard lemma, that every test function vanishing at 0 is of the form $x\psi(x)$. And it works perfectly for $xT = 0$, but do I need to come up with a different trick every time I change the function in front of $T$ or is there a more general approach? Sorry if the question seems naive, but I am a real beginner in distribution theory. $\endgroup$ – mateusz Feb 18 '14 at 20:17
  • $\begingroup$ @mateusz I expanded the answer to address your comment. $\endgroup$ – user127096 Feb 18 '14 at 20:46

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