$h(\xi)=\xi \hat{f}(\xi)$ Show that $f$ is continuous

Question

Let $f\in L^2(\mathbb R)$ and $\hat{f}$ is the Fourier transform of $f$. Let $h:\mathbb R\rightarrow \mathbb R, \xi \mapsto \xi \hat{f}(\xi)$.
Show: $h\in L^2(\mathbb R)\Rightarrow f\in C(\mathbb R)$. I dont know how to show this. If I want to show that $f$ is continuous I have to show that for every series $x_n\rightarrow x$, $|f(x_n)-f(x)|\rightarrow 0$. But the only thing I know about $f$ is $h(\xi)=\xi \hat{f}(\xi)$.

Answer 1

Hints: $\int_{|\xi|\leq 1} |\hat f (\xi)|d\xi$ is finite and $\int_{|\xi|> 1} |\hat f (\xi)|d\xi\leq \sqrt {\int |h|^{2}}\sqrt {\int_{|\xi|> 1} \frac 1 {|\xi|^{2}}}<\infty$. So $\hat f \in L^{1}$. Apply Inversion Formula (Theorem 9.14 in Rudin's RCA) to see that $f$ is continuous. [F.T. of an $L^{1}$ function is continuous].