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Let $f(x)=\sum_{n\in\mathbb{Z}}\delta(x-n).$

(a) Show $f$ is a tempered distribution.

(b) Compute $\hat{f}$ using the convention $\int_{\mathbb{R}}f(x)e^{-ix\xi}\;dx$ convention for $\mathcal{F}$.

(c) Let $\{e_{k}\}_{k\leq d}$ is a basis for $\mathbb{R}^{d}$ and compute $\hat{f}\in\mathscr{S}'(\mathbb{R}^{d})$ where $$f(x):=\sum_{n\in\mathbb{Z}^{d}}\delta(x-[n_{1}e_{1}+\ldots+n_{d}e_{d}]),$$ and formulate the result in terms of the dual basis.


Part (a) is straightforward:

Let $f_{n}$ denote the nth symmetric partial sum of the series defining $f$. Then $\{f_{n}\}_{n\in\mathbb{N}}$ defines a sequence of tempered distributions (as they are compactly supported) whereby for each $\phi\in\mathscr{D}$ we have $(f_{n},\phi)\to\sum_{n\in\mathbb{Z}}\phi(n)$ as $n\to\infty$. The righthand sum is finite since $\phi$ is rapidly decreasing (in particular, for large $n$, $\phi(n)$ is majorized by $|n|^{-k}$ for every $k>0$) and thus $f_{n}\to f$ in the distributional sense. Since $\mathscr{S}'$ is closed in $\mathscr{D}'$, the claim follows.

Part (b) is also relatively straight forward if we proceed formally:

$f$ is periodic with period $1$ so it can be expressed as a Fourier series $$f(x)=\sum_{n\in\mathbb{Z}}a_{n}e^{2\pi inx}$$ where $$a_{n}=\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x)e^{-2\pi inx}\;dx=\int_{-\frac{1}{2}}^{\frac{1}{2}}\delta(x)e^{-2\pi inx}\;dx=1.$$ On the other hand, the inversion theorem gives \begin{align*} \hat{f}(x)&=\int_{-\infty}^{\infty}f(x)e^{ix\xi}\;dx\\ &=\int_{-\infty}^{\infty}\left(\sum_{n=-\infty}^{\infty}e^{2\pi i nx}\right)e^{ix\xi}\;d\xi\\ &=\sum_{n=-\infty}^{\infty}\mathcal{F}\left\{e^{2\pi inx}\right\}\\ &=2\pi\sum_{n\in\mathbb{Z}}\delta(\xi-2\pi n) \end{align*} as desired.

Part (c) I am having difficulty proceeding. I know that it will be similar to (b), but the sum over $\mathbb{Z}^{d}$ is giving me a headache; furthermore, I am questioning the rigor in (b). I think that I need to express everything in distributional form (i.e. as linear functionals) and then apply the operations of Fourier transforms, inversion, etc. as they are so-defined for distributions, no?

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1 Answer 1

The formula(s) you want to prove are examples of "the Poisson summation formula", eminently googleable...

There are at least two proofs. The more elementary one is to observe that, given Schwartz function $f$, the function $F(x)=\sum_{\lambda\in\Lambda} f(x+\lambda)$ for any lattice $\Lambda$ is a $\Lambda$-periodic function, so descends to $\mathbb R^n/\Lambda$ and is smooth. Thus, its Fourier series on that quotient converges pointwise (and more). One observes that the $\xi$th Fourier coefficient (for $\xi$ in the dual lattice to $\Lambda$) of $F$ is the $\xi$th Fourier-transform component of $f$ ...

Another proof, more distributional, observes that Fourier transform converts annihilation by multiplication by $e^{i\langle x,\xi\rangle}-1$ to annihilation by translations by the lattice. Then one should prove that the space of such tempered distributions is just one dimensional (by devices depending on one's grounding...), so the Fourier transform of this Poisson-summation distribution is the corresponding distribution on the dual lattice.

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