Given $ 2\pi $-periodic function $ f : [-\pi,\pi]\to\mathbb{R} $. Assume that $ f $ is bounded. If $\hat{f}(n)$, $ \hat{g}(n) $, and $ \hat{h}(n) $ are the Fourier coefficients of $ f $, $ g $, and convolution of $ f $ and $ g $ respectively, how to determine the function $ g $, such that:
- $ |\hat{h}(n)| = |\hat{f}(n)| $,
- $ |\hat{h}(n)| \leq |\hat{f}(n)| $.
Is there exist a function $ g $ such that $ |\hat{h}(n)| \geq |\hat{f}(n)| $?
$\hat {h} (n)$ is nothing but $\hat {f} (n) \hat {g} (n)$. So for 1) take $g(x)=1$ for all $x$. For 2) any function $g$ with $\|g\|_1 \leq 1$ will do because $|\hat {g} (n)| \leq \|g\|_1$. It is not always possible to find $g$ for reverse inequality. This inequality would require $\hat {f} (n)=0$ whenever $\hat {g} (n)=0$.