# Equivalence of Fourier Transform Regularizations

Let $f:\mathbb R\to\mathbb C$ be a function that "isn't too discontinuous" and "doesn't grow too rapidly" at $\pm\infty$. Suppose we define the following two "regularized" Fourier transforms of $f$: \begin{align} F_\ell(k) &= \int_{-\ell}^\ell f(x)\, e^{-ikx}\, \frac{dx}{\sqrt{2\pi}} , \qquad \ell>0\\ G_\epsilon(k) &= \int_{-\infty}^\infty f(x)e^{-\epsilon |x|} \,e^{-ikx} \frac{dx}{\sqrt{2\pi}}, \qquad \epsilon>0. \end{align} Each of $F_\ell$ and $G_\epsilon$ is a one-parameter family of functions concocted to make the integrals converge even in $f$ isn't super well-behaved. The first accomplishes this by hard cutoffs on the integrals, and the second does so by taming $f$ through multiplication by a rapidly decaying exponential. When $\ell\to\infty$ and $\epsilon\to 0$, it seems that morally speaking one should obtain the Fourier transform of $f$.

More precisely, I want to say that if $f$ is suitably nice (perhaps something like piecewise smooth and of at most polynomial growth at infinity), then these families are equivalent "regularizations" of the Fourier transform $\hat f$ of $f$ in the sense that for any smooth, rapidly-decaying test function $\varphi$ (i.e. $\varphi$ in Schwartz space $S(\mathbb R)$), we have \begin{align} \lim_{\ell\to\infty}\int_{-\infty}^\infty F_\ell(k)\,\varphi(k)\, dk = \lim_{\epsilon\to 0}\int_{-\infty}^\infty G_\ell(k)\,\varphi(k)\, dk = \int_{-\infty}^\infty \hat f(k)\, \varphi(k)\, dk \end{align} Is this true?

This question is motivated by thinking of $F_\ell$ and $G_\ell$ as families of distributions in which case the equivalence condition above says that these families converge to the same desired tempered distribution -- the Fourier transform of $f$.

• For $F_l$ you could use a suitably scaled boxcar function as part of the integrand, instead of hard cutoffs. In the transform domain, that would give you the fourier transform of $f$ convolved with an ever narrowing $sinc()$ as $l$ approaches $\infty$ – Andy Walls Feb 18 '18 at 13:04
• Not sure if helpful, but if your function is in $L^2$ then the Plancherel theorem implies that the Fourier transform is independent of the regularisation. – UtilityMaximiser Feb 10 at 22:58

## 1 Answer

What you want is to prove that $$\chi_{[-l,+l]} f\xrightarrow{l\to\infty} f \tag{1}$$ and $$e^{-\epsilon|x|} f \xrightarrow{\epsilon\searrow 0} f \tag{2}$$ in $\mathcal{S}'$, because then your claim follows from continuity of the FT as an operator $\mathcal{S}'\to\mathcal{S}'$.

There are a numbers of conditions under which one can prove these assertions because a whole variety of different notions of convergence imply $\mathcal{S}'$-convergence. For example, if $f\in L^p$, then both (1) and (2) hold in the $L^p$-sense by an easy application of dominated convergence. More generally one can use convergence in a weighted $L^p$-space $L^p(\mathbb{R},w\cdot dx)$ for weight functions $w$ with $1/w\in O(x^N)$.

That would also answer the implicit question whether the integrals you used to define $F_l$ and $G_\epsilon$ actually exist.

• Thanks Johannes. This is helpful. Do you know of a nice, preferably introductory reference in which (i) continuity of the FT on $\mathcal S'$ is discussed and (ii) this idea that many notions of convergence imply $\mathcal S'$ convergence are discussed in some detail? – joshphysics Feb 21 '18 at 1:07
• Any introductory treatment of tempered distributions should cover that. – Johannes Hahn Feb 21 '18 at 1:47