Show: $h \in L^2(\mathbb{R})$ implies $ f \in C(\mathbb{R})$

Question

Is $ f \in L^2(\mathbb{R})$, $\hat{f}$ the Fourier transform from $f$ and $h: \mathbb{R} \to \mathbb{R}, h(\xi) = \xi \hat{f}(\xi))$

Show: $h \in L^2(\mathbb{R})$ implies $ f \in C(\mathbb{R})$

Does someone know where to begin? I can't find a stepping stone for this one. I tried to get something out of the integral of the Fourier transform but i can't find an argument why it should be $ f \in C(\mathbb{R})$.

Answer 1

Since $\int_{-1}^1|\hat f|\le\sqrt 2\left(\int_{-1}^1|\hat f|^2\right)^{1/2}$ it follows easily that $\hat f\in L^1$. Now the Inversion Theorem and the Riemann-Lebesgue Lemma show that $f\in C_0(\Bbb R)$.