$f\in L^1\cap L^2$ implies $\hat f \in L^1$?

Question

Given $f\in L^1(\mathbb R^d)\cap L^2(\mathbb R^d)$. The Riemann-Lebesgue lemma and the unitarity of the Fourier transform on $L^2$ implies that $\hat f \in L^2\cap C_0$ where $C_0$ are continuous functions decaying at infinity. My professor claimed that $\hat f\in L^1$ which is not obvious in my opinion.

Is this correct?

Answer 1

This is not true.

Simply take $f = \chi_{[-1,1]}$ (the indicator function of the interval $[-1,1]$). Then $f \in L^p$ for all $p \in (0,\infty]$, but if $\widehat{f} \in L^1$ was true, then Fourier inversion would imply that

$$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$

would be (almost everywhere equal to) a continuous function. This is clearly not the case.

Answer 2

Your professor is wrong and Umberto P. is right. If we take $$ f(x)= K_0(|x|) $$ where $K_0$ is a modified Bessel function of the second kind, we have $f\geq 0$ and: $$ \int_{-\infty}^{+\infty}K_0(|x|)\,dx = \pi,\qquad \int_{0}^{+\infty}K_0(|x|)^2\,dx = \frac{\pi^2}{2},$$ so $f\in L^1\cap L^2(\mathbb{R})$, but: $$ \widehat{f}(s) = \sqrt{\frac{\pi}{2}}\frac{1}{\sqrt{1+s^2}} $$ does not belong to $L^1(\mathbb{R})$.