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}



nonnegative Fourier Transform
If $\hat f$ is nonnegative, then (up to a factor), 
$$f(0)=\int \hat f=\Vert \hat f \Vert\_1 = \Vert f \Vert\_\infty.$$


