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An exercise in Carlos ISNARD's Introdução à medida e integração:

Show that if $f$ and $f'$ $\in\mathscr{L}^1(\mathbb{R},\lambda,\mathbb{C})$ and $\lim_{x\to\pm\infty}f(x)=0$ then $\hat{(f')}(\zeta)=i\zeta\hat{f}(\zeta)$.

($\lambda$ is the Lebesgue measure on $\mathbb{R}$.)

I'm tempted to apply integration by parts on the integral from $-N$ to $N$ and then take limit as $N\to\infty$. But to obtain the result I seemingly need $f'e^{-i\zeta x}$ to be Riemann-integrable so as to use the fundamental theorem of Calculus.

What am I missing here?

Thank you.

Vincenzo Tibullo
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Weltschmerz
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2 Answers2

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We can find a sequence of $C^1$ functions with compact support $\{f_j\}$ such that $$\lVert f-f_j\rVert_{L^1}+\lVert f'-f_j'\rVert_{L^1}\leq \frac 1j.$$ This is a standard result about Sobolev spaces. Actually more is true: the smooth functions with compact support are dense in $W^{1,1}(\Bbb R)$, but that's not needed here. It's theorem 13 in Willie Wong's notes about Sobolev spaces.

Convergence in $L^1$ shows that for all $\xi$ we have $\widehat{f_j}(\xi)\to \widehat{f}(\xi)$ and $\widehat{f_j'}(\xi)\to \widehat{f'}(\xi)$. And the result for $C^1$ functions follows as stated in the OP by integration by parts.

Community
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Davide Giraudo
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There is a measure-theoretic version of the fundamental theorem of Calculus, see here.

Rasmus
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  • Thank you. The thing is, up to that point in the textbook no such theorem has been mentioned, so that's why I'm wondering. – Weltschmerz Jul 30 '12 at 13:45