This is a problem from Katznelson's Harmonic Analysis book(Page 143, Problem 8):
Suppose $f\in L^1(\mathbb{R})$ and is continuous at $0$. Also suppose $\hat{f}\geq 0$, then show that $\hat{f}$ is in $L^1(\mathbb{R})$ too.
All hints and suggestions are welcome.
You could use the heat kernel. For example, the following holds for all $f\in L^{1}(\mathbb{R})$: $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{-s^2t}e^{isx}ds = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(y)e^{-(x-y)^{2}/4t}dy. $$ Under your conditions, $\hat{f} \ge 0$ and $f$ is continuous at $0$. Setting $x=0$ gives $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{-s^{2}t}ds =\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(y)e^{-y^{2}/4t}dy $$ Now, letting $t\downarrow 0$, the right side converges to $f(0)$ because $f$ is continuous at $0$. By the monotone convergence theorem, $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)ds = f(0). $$ The left side is finite because the limit on the right exists.