Fourier transform of continuous and integrable function

Question

$f\in L^1(R^n)$,and $f$ is continuous at 0,the Fourier transform of $f$ noted $\widehat{f}$ is non-negative,prove that$ \widehat{f}\in L^1(R^n)$

I have no idea of how to use the condition of continuous at $0$. Any idea will be helpful.

Answer 1

I'll do with $n=1$ it is the same for other $n$.

Since $f$ is $L^1$ then $\hat{f}$ is bounded and continuous and so let $f_k = f \ast k e^{-\pi k^2 x^2}$ (convolution) then we can apply the Fourier inversion theorem without fear to

$$\int_{-\infty}^\infty \hat{f}(\xi)e^{-\pi \xi^2/k^2}d\xi= \int_{-\infty}^\infty \hat{f_k}(\xi)d\xi =f_k(0)$$

So for $\hat{f}$ non-negative we have $$f_k(0) = \|\hat{f} e^{-\pi \xi^2/k^2}\|_{L^1}$$ And hence $$\|\hat{f} \|_{L^1} = \lim_{k \to \infty}f_k(0) $$

Do you see now where the continuity of $f$ at $0$ is involved ?