Suppose that $f \in L^1 $ and $\hat{f} \geq 0$. If $f$ is continuous in $0$ then $\hat{f} \in L^1$

Question

Suppose that $f \in L^1 $ and $\hat{f} \geq 0$ be the the Fourier transform of $f$. If $f$ is continuous in $0$ then $\hat{f} \in L^1$.

I know that $\hat{f}(\xi) \leq ||f||_1 < \infty$. But I don't know how to use the fact that $f$ is continuous.

Could someone help me?

Answer 1

Hints: Let $g_n(x)=\frac n {\sqrt {2\pi}}e^{-n^{2}x^{2}/2}$. Then $\hat {f*g_n}=\hat f \hat {g_n}\in L^{1}$. Apply Inversion Formula for $f*g_n$ and take limit as $n \to \infty$. Continuity of $f$ at $0$ implies that $f*g_n(0) \to f(0)$ as $ n \to \infty$ and this gives $f(0)=\int \hat f(x)dx$.