The Fourier transform of functions with compact support is differentiable.

Question

1) How can I prove that if $f(x)$ is a continous function with compact support (let's say $f(x)=0$ $\forall x\in B(0,R)^c$), then its Fourier transform $\hat{f}(\xi)$ is differentiable?

2) Is there any counter example that $\hat{f}(\xi)$ is differentiable if $f\in C^0 (\mathbb{R})$ (without necessarely having a compact support)?

Thank you!

Answer 1

If $f\in C_c(\Bbb R)$, then you can show that $\hat f$ is differentiable by definition and dominated convergence theorem.

If we want to build $f\in C(\Bbb R)$ such that $\hat f$ is not differentiable, we can use the fact that for even functions $f = \hat{\hat f}$ (up to a multiplicative constant). Now take, for example, $\hat f = \mathbf 1_{[-1,1]}(x)$. This function is not differentiable on $\Bbb R$. Its Fourier transform $\hat{\hat f}$ (and hence Fourier inverse $f$) is continuous (easy to check by definition).

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Suppose that $supp (f)\subset [-R,R]$ and $\sup_{x\in[-R,R]} |f(x)|=M$. $$\frac{\hat f(\xi+h)-\hat f(\xi)}{h} -\int_{-R}^Rixf(x) \exp(ix\xi) dx =\int_{-R}^Rf(x) \exp(ix\xi)\left(\frac{\exp(ixh)-1}{h}-ix\right)dx.$$

The factor $\frac{\exp(ixh)-1}{h}-ix$ converges uniformly to zero on $[-R,R]$ as $h\to 0$, hence

$$\left|\frac{\hat f(\xi+h)-\hat f(\xi)}{h} -\int_{-R}^Rixf(x) \exp(ix\xi) dx\right| \le \left|\int_{-R}^Rf(x) \exp(ix\xi)\left(\frac{\exp(ixh)-1}{h}-ix\right)dx \right|$$ $$\le 2RM\sup_{x\in[-R,R]}\left |\frac{\exp(ixh)-1}{h}-ix\right| \to 0\mbox { as } h\to 0.$$

Answer 2

If $f$ is continuous and has compact support then $xf$ and $f$ are in $L^1$. Then $x\mapsto f(y)e^{-ixy}$ is differentiable for each $y\in\mathbb R$ and $|\frac{\partial}{\partial x} f(y)e^{-ixy}| = |-iyf(y)e^{-ixy}| \leq |yf(y)|\in L^1$.

Now we have a theorem that states that $x\mapsto \hat f(x) = \int f(y)e^{-ixy}dy$ is differentiable in this case with derivative $\frac{d}{dx}f(x) = \int \frac{\partial}{\partial x}f(y)e^{-ixy}dy = -i (yf)^\hat{}$.

If you consider the plancherel theorem, then the example of TZakreveskiy works fine of course.

However, if you want $f\in L^1(\mathbb R)\cap C^0(\mathbb R)$ with $\hat f\not\in C^1(\mathbb R)$, consider $f = \mathbb 1_{[-1,1]}\ast 1_{[-1,1]}$. Then $\hat f(x) = \frac{\sin^2(x)}{x^2}$ up to some constant. $\hat f$ is in $L^1\cap C^0$ but $\hat{\hat f}$ is not differentiable as you can check easily.