# Computing the Fourier transform (in the sense of tempered distributions) from a sequence of samples

Let $f\in L_{\operatorname{loc}}^2 (\mathbb{R})$ be a function of period $2\pi$. I'd like to get some help proving the following identity:

$$\hat{f}=\sum_{n=-\infty}^{\infty} \hat{f}(n)\delta_n$$

With some basic calculations, it seems that proving the following identity should suffice:

$$\int \int f(x)e^{-ix\xi} \phi(\xi)d\xi dx=\sum_{n=-\infty}^{\infty}\int e^{-inx}f(x)\phi(n)dx\mbox{ for every }\phi \in S(\mathbb{R}).$$

However, it seems that I can't prove this identity either, so perhaps this is not the right way to attack the problem, so I will appreciate any help proving the first identity (whether by proving the second identity or in any other way).

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I'm not exactly how to make sense of $\hat f(n)$ here; $\hat f$ most likely wont even be a function so how can you evaluate it at $n$? Or by $\hat f(n)$ do you mean the $n$th Fourier coefficient ${1\over 2\pi} \int_0^{2\pi} f(x)e^{-inx}\,dx$? The formula is true with $\hat f(n)$ equal to the $n$th Fourier coefficient (but then the notation is inconsistent). – Nick Strehlke Jun 2 '13 at 21:38

This first result may be known to you already, but I include it here for the benefit of other users.

Theorem 1. If $f:\mathbb{R}^{d}\rightarrow\mathbb{C}$ be a locally integrable $1$-periodic function, then $f$ defines a tempered distribution by integration.

Proof. Let $Q$ denote the unit half-open cube $[0,1)^{d}$, and let $Q_{k}:=Q+k$ for $k\in\mathbb{Z}^{d}$. Clearly, $\left\{Q_{k}\right\}_{k\in\mathbb{Z}^{d}}$ is a partition of $\mathbb{R}^{d}$. Since $f$ is locally integrable, it is in particular integrable on each $Q_{k}$. Whence, for any $\varphi\in\mathcal{S}(\mathbb{R}^{d})$, \begin{align*} \int_{\mathbb{R}^{d}}\left|\varphi(y)f(y)\right|dy&\leq\sup_{x\in\mathbb{R}^{d}}\left(1+\left|x\right|\right)^{m}\left|\varphi(x)\right|\int_{\mathbb{R}^{d}}\left(1+\left|y\right|\right)^{-m}\left|f(y)\right|dy\\ &=\sup_{x\in\mathbb{R}^{d}}(1+\left|x\right|)^{-m}\left|\varphi(x)\right|\sum_{k\in\mathbb{Z}^{d}}\int_{Q_{k}}(1+\left|y\right|)^{-m}\left|f(y)\right|dy\\ &=\sup_{x\in\mathbb{R}^{d}}(1+\left|x\right|)^{-m}\left|\varphi(x)\right|\sum_{k\in\mathbb{Z}^{d}}(1+\left|k\right|)^{-m}\left\|f\chi_{Q_{k}}\right\|_{L^{1}}\\ &\lesssim\left\|f\right\|_{L^{1}(\mathbb{T}^{d})}\sup_{x\in\mathbb{R}^{d}}(1+\left|x\right|)^{-m}\left|\varphi(x)\right|, \end{align*} where we use the convergence of $\sum_{k}(1+\left|k\right|)^{-m}$ for suitably large $m$. So $\left|\langle{f,\varphi}\rangle\right|$ is bounded by a constant multiple of a finite sum of Schwartz space seminorms. $\Box$

Given a locally integrable function $1$-periodic function $f\in L_{loc}^{1}(\mathbb{R}^{d})$, we write

$$\widehat{f}(n):=\int_{\mathbb{T}^{d}}f(x)e^{2\pi i n\cdot x}dx,$$ where we identify $\mathbb{T}^{d}\cong[0,1]^{d}$.

The identity you are trying to prove holds in the sense of tempered distributions, meaning

$$\langle{f,\widehat{\varphi}}\rangle=\lim_{N\rightarrow\infty}\langle{\sum_{\left|n\right|\leq N}\widehat{f}(n)\delta_{n},\varphi}\rangle,\quad\forall \varphi\in\mathcal{S}(\mathbb{R}^{d})$$

We will obtain the identity as a consequence of the following more general result:

Theorem 2. Suppose $(c_{k})_{k\in\mathbb{Z}^{d}}$ is a sequence that satisfies $\left|c_{k}\right|\leq A(1+\left|k\right|)^{M}$ for all $k$, where $A,M>0$ are fixed. Then $$\sum_{\left|k\right|\leq N}c_{k}\delta_{k}$$ converges to a tempered distribution $u\in\mathcal{S}'(\mathbb{R}^{d})$ as $N\rightarrow\infty$. Moreover, $\widehat{u}$ is the limit in $\mathcal{S}'(\mathbb{R}^{d})$ of the sequence of functions $$h_{N}(\xi)=\sum_{\left|k\right|\leq > N}c_{k}e^{-2\pi i\xi\cdot k}$$

Proof. Let $\varphi\in\mathcal{S}(\mathbb{R}^{d})$.

$$\begin{array}{lcl}\displaystyle\left|\langle{\sum_{\left|k\right|\leq N}c_{k}\delta_{k},\varphi}\rangle\right|\leq\sum_{\left|k\right|\leq N}\left|c_{k}\right|\left|\langle{\delta_{k},\varphi}\rangle\right|&\lesssim&\displaystyle\sum_{\left|k\right|\leq N}(1+\left|k\right|)^{M}\left|\varphi(k)\right|\\[2 em]\displaystyle&\lesssim&\displaystyle\sup_{x\in\mathbb{R}^{d}}\left|(1+\left|x\right|)^{M+d+1}\varphi(x)\right|\sum_{k\in\mathbb{Z}^{d}}(1+\left|k\right|)^{-d-1}\end{array}$$

So the series $\sum_{k}c_{k}\langle{\delta_{k},\varphi}\rangle$ converges absolutely to $u(\varphi)$, and the above inequality shows that $\varphi\mapsto u(\varphi)$ defines a tempered distribution which we denote by $u$. From the continuity and linearity of the Fourier transform on $\mathcal{S}'(\mathbb{R}^{d})$ and recalling that $\widehat{\delta_{k}}=e^{-2\pi i\xi\cdot k}$, we obtain the second assertion. $\Box$

It is evident that the Fourier coefficients $\widehat{f}(k)$ are bounded. Since ${\delta_{k}}^{\vee}=\widehat{\delta}_{-k}=e^{2\pi i k\cdot()}$, we can apply the preceding theorem to obtain $$u^{\vee}=\lim_{N\rightarrow\infty}\sum_{\left|k\right|\leq N}(\widehat{f}(k)\delta_{k})^{\vee}=\lim_{N\rightarrow\infty}\sum_{\left|k\right|\leq N}\widehat{f}(k)e^{2\pi k\cdot ()}$$

But the partial sums on the RHS also converge to $f$ in $L^{2}(\mathbb{T}^{d})$. By density, it suffices to show that $\langle{u^{\vee},\phi}\rangle=\langle{f,\phi}\rangle$ for $\phi\in C_{c}^{\infty}(\mathbb{R}^{d})$.

$$\int_{\mathbb{R}^{d}}f(y)\varphi(y)dy=\sum_{m\in\mathbb{Z}^{d}}\lim_{N\rightarrow\infty}\int_{Q_{m}}\sum_{\left|k\right|\leq N}\widehat{f}(n)e^{2\pi i k\cdot y}\varphi(y)dy=\lim_{N\rightarrow\infty}\int_{\mathbb{R}^{d}}\sum_{\left|k\right|\leq N}\widehat{f}(k)e^{2\pi i k\cdot y}\varphi(y)dy,$$ where we can interchange the sum and limit without issue since $\text{supp}(\varphi)$ is contained in the union of finitely many $Q_{m}$.

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