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If $u$ is a compactly-supported distribution on $\mathbb{R}^n$, how can we prove that its Fourier transform $\mathcal{F}u$ is the tempered distribution given by the function $\xi\mapsto u(e^{-ix\xi})$?

Here, the Fourier transform is defined on Schwartz functions as $\mathcal{F}\phi(\xi)=\int e^{-ix\xi}\phi(x) dx $, and on distributions as $\mathcal{F}u(\phi)=u(\mathcal{F}\phi)$. If $u$ is compactly-supported, then $u=\chi u$ for some compactly-supported smooth $\chi$, and so $u(e^{-ix\xi}) = u(\chi(x)e^{-ix\xi})$ is well-defined for any $\xi$. Intuitively, one has

$$ \int_{\xi\in\mathbb{R}^n} u(\chi(x)e^{-ix\xi}) \phi(\xi) d\xi = u\left(\int_{\xi\in\mathbb{R}^n}\chi(x)e^{-ix\xi}\phi(\xi) d\xi \right) $$

which is what we want to show (the LHS is $u$ applied to that function on $x$), but I'm not sure how rigorous it is to pull the integral sign inside the distribution.

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    $\begingroup$ This fact is part of the Paley-Wiener theorem (en.m.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem), a proof of which can be found in many books on Fourier analysis (though the exact equality that you have an issue with is not always proven formally). $\endgroup$
    – PhoemueX
    Commented Mar 21, 2020 at 18:16
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    $\begingroup$ Not that it helps, but it's not just you. Once some years ago I was trying to reconstruct that stuff on my own. This is the one point I got stuck on; looked it up in Folland and saw it's more or less left to the reader... $\endgroup$ Commented Sep 16, 2020 at 16:35

3 Answers 3

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There's a nice proof of this in Hörmander's book "The Analysis of Linear Partial Differential Operators I" (Theorem 7.1.14). The main tool needed is a distributional version of Fubini's theorem (Theorem 5.1.1 in Hörmander's book). I'll summarize the main points:

Preliminary step 1: Fubini's theorem

If $u$ and $v$ are distributions on ${\mathbb R}^m$ and ${\mathbb R}^n$ respectively, then there is a unique product distribution $u \otimes v$ on ${\mathbb R}^{m+n}$ characterized by the condition $$ (u \otimes v)(\varphi \otimes \psi) = u(\varphi) v(\psi) $$ for all $\varphi \in {\mathcal D}({\mathbb R}^m)$ and $\psi \in {\mathcal D}({\mathbb R}^n)$, where we write $(\varphi \otimes \psi)(x,y) := \varphi(x) \psi(y)$. Moreover, $u \otimes v$ can be evaluated on an arbitrary test function $\varphi \in {\mathcal D}({\mathbb R}^{m+n})$ by $$ (u \otimes v)(\varphi) = u(x \mapsto v(\varphi(x,\cdot))) = v(y \mapsto u(\varphi(\cdot,y))). $$ (Note that there is a slightly nontrivial exercise required to show that both expressions on the right make sense, e.g. that $x \mapsto v(\varphi(x,\cdot))$ defines a smooth compactly supported function on ${\mathbb R}^m$. This rests mainly on the fact that by uniform continuity, $x \mapsto \varphi(x,\cdot)$ is a continuous map to the space of test functions, and other arguments of this sort.) If you write down this formula in the case where $u$ and $v$ are given by locally integrable functions, you'll find that it follows easily from the classical Fubini's theorem.

If you know that the product distribution is unique, then the formula follows by verifying directly that both expressions on the right hand side define distributions that satisfy the defining property of a product distribution. Uniqueness can be proved via mollification: if $u \otimes v$ were not unique, then there would exist a nontrivial distribution $w$ on ${\mathbb R}^{m+n}$ such that $w(\varphi \otimes \psi) = 0$ for all $\varphi \in {\mathcal D}({\mathbb R}^m)$ and $\psi \in {\mathcal D}({\mathbb R}^n)$. Choose approximate identities, i.e. sequences of smooth functions $\rho_j : {\mathbb R}^m \to [0,\infty)$ and $\sigma_j : {\mathbb R}^n \to [0,\infty)$ with shrinking compact supports near $\{0\}$ that converge in the space of distributions to $\delta$-functions. Then the classical Fubini theorem implies that the sequence $\rho_j \otimes \sigma_j : {\mathbb R}^{m+n} \to [0,\infty)$ also defines an approximate identity in the same sense, and it follows that the sequence of smooth functions $(\rho_j \otimes \sigma_j) * w$ converges to $w$ in the space of distributions. But those functions are all $0$ due to the defining property of $w$, thus $w=0$.

Preliminary step 2: polynomial growth

Before we can view the function $g(\xi) := u(\chi(x) e^{-i x \xi})$ as a plausible candidate for the Fourier transform of $\chi u$, we need to know that it behaves reasonably enough at infinity to define a tempered distribution. As I indicated in my comments on the previous answer, $g$ is definitely not a Schwartz function in general, but one can show that it has polynomial growth. Perhaps the quickest way is to use standard properties of the Fourier transform and rewrite $g$ as $$ g(\xi) = \left( ( {\mathcal F}\chi)^- * {\mathcal F}^*u\right)(-\xi), $$ where I'm using the notation $f^-(x) := f(-x)$. As the convolution of a Schwartz function with a tempered distribution, it follows from standard results about the convolution that this function has polynomial growth.

The main argument

As stated in the question, we need to prove that the relation $$ \int_{{\mathbb R}^n} u(\chi(x) e^{-i x \xi}) \phi(\xi) \, d\xi = u\left( \int_{{\mathbb R}^n} \chi(x) e^{- i x \xi} \phi(\xi)\, d\xi \right) $$ holds for every $u \in {\mathcal D}'({\mathbb R}^n)$, $\chi \in {\mathcal D}({\mathbb R}^n)$ and $\phi \in {\mathcal S}({\mathbb R}^n)$. By step 2, we already know that both sides give well-defined tempered distributions when regarded as functionals of $\phi$, so by density, it will suffice to assume $\phi \in {\mathcal D}({\mathbb R}^n)$. The key observation now is that by the theorem in step 1, both sides can be identified with $(u \otimes 1)(f)$, where $1 \in {\mathcal D}'({\mathbb R}^n)$ is the distribution $1(\varphi) := \int_{{\mathbb R}^n} \varphi(x)\, dx$ and $f \in {\mathcal D}({\mathbb R}^{m+n})$ is given by $$ f(x,\xi) := \chi(x) \phi(\xi) e^{-i x \xi}. $$

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To prove the equality, we compare the two mappings from the space of tempered distributions to the complex plane: $$ \mathcal{S}' \to \mathbb{C} \\ u\mapsto \int u(\chi(x)e^{-ix\xi}) \phi(\xi)d\xi $$ and $$ \mathcal{S}' \to \mathbb{C} \\ u\mapsto u\left(\int \chi(x)e^{-ix\xi} \phi(\xi)d\xi\right) $$ The second one is continuous since it just evaluates at a compactly supported smooth function. The first one is continuous since it's the composition $\mathcal{S}'\to\mathcal{S}\to\mathbb{C}$ where the first map is $u\mapsto (\xi\mapsto u(\chi(x)e^{-ix\xi}))$ and the second map just integrates against $\phi(\xi)$. Hence, to prove the two maps coincide, it suffices to prove they coincide on a dense subset, say $\mathcal{S}\subset\mathcal{S}'$. So, say $u$ corresponds to Schwartz function $\psi(x)$. Then we must prove the equality $$ \int \left(\int \psi(x)\chi(x)e^{-ix\xi}dx\right) \phi(\xi)d\xi = \int \psi(x)\left(\int \chi(x)e^{-ix\xi}\phi(\xi)d\xi\right) dx $$ which is clear since they both integrate the function $$ (x,\xi) \mapsto \psi(x)\chi(x)e^{-ix\xi}\phi(\xi) $$ on $\mathbb{R}^{2n}$, and Fubini-Tonelli applies since this function is in $L^1(\mathbb{R}^{2n})$ since its norm is bounded by $|\psi(x)\chi(x)|\cdot|\phi(\xi)|$ and $\psi,\chi,\phi$ are all Schwartz.

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  • $\begingroup$ I'm stuck on one detail in this answer: it seems not so obvious that $\xi \mapsto u(\chi(x) e^{-i x \xi})$ is a Schwartz function, or that the resulting map ${\mathcal S}' \to {\mathcal S}$ is continuous, especially since the topologies on ${\mathcal S}'$ and ${\mathcal S}$ are very weak and strong respectively. Is this a standard fact? The most I've been able to convince myself is that the function has polynomial growth. $\endgroup$ Commented Sep 13, 2020 at 8:05
  • $\begingroup$ Addendum to my previous comment: $\xi \mapsto u(\chi(x) e^{-ix\xi})$ is definitely not a Schwartz function in general, so this is an error in the above proof. For example, take $u$ to be the Dirac $\delta$-function, then $u(\chi(x) e^{- i x \xi}) = \chi(0)$ is a constant (and generally nonzero) function of $\xi$. $\endgroup$ Commented Sep 15, 2020 at 14:52
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Take a sequence $u_k \in C_c^\infty \subset \mathcal{S}'$ such that $u_k \to u$ in $\mathcal{S}'$ as $k \to \infty.$ For these the Fourier transform, distributional as well as classical, is given by $\widehat{u_k}(\xi) = \langle u_k(x), e^{-i\xi x} \rangle.$

The Fourier transform is continuous on $\mathcal{S}'$, i.e. if $u_k \to u$ in $\mathcal{S}'$ then $\widehat{u_k} \to \widehat{u}$ in $\mathcal{S}'$. Therefore, taking limits of the previous result, we get $\widehat{u}(\xi) = \langle u(x), e^{-i\xi x} \rangle.$

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