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This might be a silly question. For $f$ an integrable, complex-valued function, its Fourier transform is

$$ \hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-isx}f(x)\, \mathrm{d}x $$

I want to show that if $\int xf(x) \, \mathrm{d}x$ exists then $\hat{f}$ is differentiable, with

$$ (\hat{f})'(s) = - \frac{i}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} xe^{-isx}f(x)\, \mathrm{d}x $$

for every $s$. I tried using the definition of derivative and got to

$$ (\hat{f})'(s) = \lim_{h \to 0}\frac{1}{h}\int_{-\infty}^{+\infty} f(x)e^{-isx}[e^{-ihx}-1] \, \mathrm{d}x $$

and I'm not really sure where to go. Can anyone point me in the right direction? Thanks.

Tom Offer
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1 Answers1

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You know, thanks to Riemann-Lebesgue theorem, that $\,\hat{f}$ and $\,\hat{g}$ are continuos and vanishing at $\infty$, where $g=\widehat{-ixf(x)}$. I prove that $$\hat{f}(y)-\hat{f}(0)=\int_0^y g(t)dt,$$ the result will follow from foundamental calculus theorem.

\begin{align*} \int_0^y g(t)dt&=\int_0^y\int_{\mathbb{R}}-ixe^{-ixt}dt\, f(x)dx=\\ &=\int_{\mathbb{R}} \left.e^{-ixt}\right|_{0}^y\,\, f(x)dx=\int_{\mathbb{R}}f(x)\left[e^{-ixy}-e^{ix0}\right]dx=\hat{f}(y)-\hat{f}(0). \end{align*}

The hypothesis $xf(x)\in L^1$, in addiction with $|e^{i \theta}|=1$ for all $\theta\in\mathbb{R}$, is used to apply Fubini's theorem.

Felice Iandoli
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