Fourier transform of a continuous function from $L^2$.

Question

Good day everyone, I have the following question. If $f$ from $L^2(\mathbb{R})$ and continuous, is it true, that its Fourier transform is also continuous? It is definitely true and easy to show if $f$ in $L^1$. But what is for $L^2$? And what if $f$ is even differentiable? Is $\hat f$ also differentiable in this case? Thanks in advance.

Answer 1

The intuition one should have is that if $f$ has a lot of decay (i.e. integrability), then $\hat{f}$ will have a lot of regularity (some $C^k$), respectively, if $f$ is very regular ($C^k$ for large $k$), then $\hat{f}$ will have a lot of decay. See for example here Smoothness and decay property of Fourier transformation

Unfortunately, regularity of a function does not imply regularity of its Fourier transform. For example take $f=\widehat{\chi_{[-1;1]}}$. As the Fourier transform maps $L^2(\mathbb{R})$ to $L^2(\mathbb{R})$, we get that $f$ is in $L^2(\mathbb{R})$. However, $\chi_{[-1;1]}$ is in $L^1(\mathbb{R})$, thus, $f$ is continuous (this is the Riemann-Lebesgue lemma, see Fourier transform is uniformly continuous). Now we can compute the Fourier transform of $f$. For this we note that $\chi_{[-1;1]}(x)=\chi_{[-1;1]}(-x)$ and therefore $\widehat{\chi_{[-1;1]}}=\check{\chi_{[-1;1]}}$ and hence

$$ \hat{f} = \widehat{\check{\chi_{[-1;1]}}} =\chi_{[-1;1]}.$$

Thus, the Fourier transform of $f$ is not even continuous.

In fact, it is much worse. Using that $\chi_{[-1;1]}$ has compact support, one can show (see Fourier transforms having compact support) that $f$ is in fact real-analytic, even though $\hat{f}=\chi_{[-1;1]}$ is discontinuous. This is telling us that there is no hope to get differentiability of the Fourier transform just by assuming differentiability of the function.