# Show that the Fourier Transform is differentiable [duplicate]

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.

• Show that if $\int xf(x)\,dx$ exists then $\hat{f}'$ exists for all $s$. – lemon Nov 17 '14 at 11:07

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.
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.
• Why have you assumed that $\hat{f}(0) = 0$? Also is $xf(x) \in L^1$ really sufficient to apply Fubini here? What about the $exp()$ term? – Tom Offer Nov 17 '14 at 18:53
• You don't have $\hat{f}(0)=0$, it was an error. By the way $|e^{ixy}|=1$, it is the complex exponential function. Anyway i'll edit in a few minutes. – Felice Iandoli Nov 18 '14 at 13:19