[Math] nonnegative Fourier Transform

Question

Let $\widehat{f}(\xi)$ be Fourier transform of $f$ given by
\begin{align}
\widehat{f}(\xi)=\int_{\mathbb{R}^n} e^{-ix\cdot\xi}f(x)dx.
\end{align}
Suppose that $\widehat{f}(\xi)$ is nonnegative and locally integrable function, easily seems (by inverse Fourier transform) that
\begin{align}
\Vert f\Vert_{L^{\infty}} \leq \Vert \widehat{f}\Vert_{L^1}.
\end{align}
How to show that there is a positive constant $c>0$ such that
\begin{align}
\Vert \widehat{f}\Vert_{L^1}\leq c \Vert f\Vert_{L^{\infty}}.
\end{align}

Answer 1

If $\hat f$ is nonnegative, then (up to a factor), $$f(0)=\int \hat f=\Vert \hat f \Vert_1 = \Vert f \Vert_\infty.$$