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 $f$.

  • $\begingroup$ 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 $\infty$ $\endgroup$ – Andy Walls Feb 18 '18 at 13:04
  • $\begingroup$ 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. $\endgroup$ – UtilityMaximiser Feb 10 at 22:58

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.

  • $\begingroup$ 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? $\endgroup$ – joshphysics Feb 21 '18 at 1:07
  • $\begingroup$ Any introductory treatment of tempered distributions should cover that. $\endgroup$ – Johannes Hahn Feb 21 '18 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.