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Assume that a function $f$ is defined such that $xf\left(x\right)$ is integrable: \begin{equation} \int_{-\infty}^{\infty}\lvert xf\left(x\right)\rvert \mathrm{d}x < \infty. \end{equation} Then the function $xf\left(x\right)$ is of a Fourier transform, and so does $-2\pi i x f\left(x\right)$. The derivation is given by \begin{equation} \begin{aligned} \mathcal{F}\left(-2\pi i x f\left(x\right)\right) &= \int_{-\infty}^{\infty}\left(-2\pi i x\right)e^{-2\pi i s x}f\left(x\right)\mathrm{d}x\\ &= \int_{-\infty}^{\infty}\left(\frac{\mathrm{d}}{\mathrm{d}s}e^{-2\pi i s x}\right)f\left(x\right)\mathrm{d}x\\ &\stackrel{1}{=}\frac{\mathrm{d}}{\mathrm{d}s}\int_{-\infty}^{\infty}e^{-2\pi i s x}f\left(x\right)\mathrm{d}x\\ &= \frac{\mathrm{d}}{\mathrm{d}s}\left(\mathcal{F}f\right)\left(s\right). \end{aligned} \end{equation}

I am not sure why in step 1, the differential can be moved out. The reason given by the author is that $xf\left(x\right)$ is integrable. Can someone let me know more details about the argument? Also, it was mentioned that the derivative is continuous. Can someone also explain why the derivative is continuous?

Ziqi Fan
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  • Either by the dominated convergence theorem as @CalvinKhor pointed out or by Fubini's theorem (note that $xf(x)\in L(\mathbb R)$ implies that $x\exp(-2π\mathrm isx)f(x)\in L(\mathbb R×[s_0-1,s_0+1])$ for any $s_0\in\mathbb R$). – Ѕᴀᴀᴅ May 22 '21 at 03:56

1 Answers1

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You need to know under what conditions $$ \lim_{h\to 0} \int \frac{ f(x,s+h)-f(x,s)}h dx$$ exists. Suppose that $\partial_s f$ exists for a.e. $x$ and all $s$. Then Mean value theorem gives that there exists $t$ in between $s$ and $s+h$ such that $$\frac{ f(x,s+h)-f(x,s)}h = \partial_s f(x,t(x,s))$$ So one way to proceed would be to ask for this derivative to satisfy a dominated convergence type estimate, uniform in the second parameter: that is, there exists $g\in L^1$ such that $$|\partial_s f(x,s)|\le g(x)$$ for any $s$.

You should be able to check that the condition on $xf(x)$ allows you to apply this result

(In summary: $f(x,s)\in L^1(dx)$ for all $s$, $\partial_s f(x,s)$ exists for a.e. $x$ and all $s$ with a dominating function lets you interchange $d/ds$ with $\int dx$.)

I did some more searching. Heres what I found:

I did search Royden and Axler but did not find this theorem. Wikipedia also has this theorem and proof but oddly, does not source it.

Calvin Khor
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    Anything to read to understand this systematically? I guess real analysis and complex analysis? – Ziqi Fan May 22 '21 at 03:44
  • The ‘only’ tool used is Dominated Convergence Theorem, which is one of the main results of a typical measure theory course or book. Royden’s book is good for this IMO. – Calvin Khor May 22 '21 at 03:48
  • This specific theorem is so useful, it'd be nice to know a reference where the theorem and proof are given in detail. – littleO May 22 '21 at 03:58
  • @littleO Differentiation under the integral sign? Or DCT? For the first I wrote the answer because I recall trying to find a source and failed. Perhaps my googlefu is bad. For the second it’s normally clearly named as ‘Lebesgue Dominated Convergence’ or other names with many signposts telling the reader that it’s one of the big payoffs for sitting through sigma algebras. But I also don’t have my copy of Royden handy (sorry) – Calvin Khor May 22 '21 at 04:01
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    @CalvinKhor I meant differentiation under the integral sign. Yeah, the dominated convergence theorem is covered very explicitly in every measure theory book; but differentiation under the integral sign is a handy result that is often not spelled out explicitly. Once when I was reading the PDE book by Evans, long ago, I emailed Evans and asked him how to justify differentiation under the integral sign and he responded explaining that it can be justified using the DCT. – littleO May 22 '21 at 04:23
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    @littleO I was unable to source a textbook, but I did find a number of other links (have edited answer) – Calvin Khor May 25 '21 at 01:42
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    Folland's real analysis textbook (theorem 2.27) deals with this. Part (a) is about continuity of such parametr-dependent integrals, and part (b) deal with their differentiability. Also, Amann and Escher's Analysis III pages 107-108 discusses this theorem. They even discuss the holomorphic version of differentiation under the integral (for which by Cauchy's inequalities, one only needs a bound on the function itself rather than the derivative). – peek-a-boo Jul 27 '21 at 04:50