Fourier Transform of continuous/smooth $L^2$ function

Question

Given the following two propositions:
(i) $f \in L^2(\mathbb{R}^d)$ continuous $\Rightarrow$ $\hat{f} \in L^2(\mathbb{R}^d)$ continuous
(ii) $f \in L^2(\mathbb{R}^d)$ smooth$\Rightarrow$ $\hat{f} \in L^2(\mathbb{R}^d)$ smooth
where $\hat{f}$ is the fourier transform of $f$.

I feel like both statements are true because of Plencherel but i can't quite show why. I tried to show it by going through the definition of the fourier transform in $L^2$ but couldn't really get an idea, especially for the 2nd statement.

Answer 1

Both statements are false.

Take $f(x)=\frac {\sin x}{x}$. It's $\mathcal C^{\infty}$. In fact it is analytic on the line.

It's also square integrable.

But its Fourier transform is the indicator function of an interval. So not even continuous.

The problem is not the square integrability, that's ensured by Plancherel, as you noted.

The smoothness of a Fourier transform $\hat f$ depends on the decay rate of the original function $f$ at infinity. It's got nothing to do with the smoothness of $f$.

Answer 2

You should review the statement of the theorem of Plancherel. Think carefully about the function $$ f(x)=2\frac{\sin(x)}{x}, $$ whose Fourier transform is...