Maximum value of function $\xi \mapsto |\hat{f}(\xi)|^2$ (Fourier Transformation) Let $f \in \mathcal{S(\mathbb{R})}$ a positive function. Find the point $\xi$ where the function $\xi \mapsto |\hat{f}(\xi)|^2$ reaches its maximum value.
Solution:
Defining the function $\varphi(\xi) = |\hat{f}(\xi)|^2$, we obtain that $\varphi '(\xi) = 2|\hat{f}(\xi)|(\hat{f}'(\xi))$ and using that $\hat{f}'(\xi) = -i\widehat{xf}(\xi) $, we obtain that
$\varphi '(\xi) = -2i|\hat{f}(\xi)|\widehat{xf}(\xi)$. We conclude that $\varphi '(\xi) = 0 \Leftrightarrow -2i|\widehat{f}(\xi)|\widehat{xf}(\xi) = 0$. Help.
 A: I think this is simpler than it looks: Recall $\hat{f}(\xi)= \int_{\mathbb{R}} e^{-2\pi ix\xi}f(x)\, dx$, and since $f$ is positive we have
$$
|\hat{f}(\xi)|\leq \int_{\mathbb{R}} |e^{-2\pi ix\xi}|f(x)\, dx  = \int_{\mathbb{R}} f(x)\, dx = \hat{f}(0).
$$
Put another way, since $f$ is positive, the complex exponential could be introducing cancellation, making things smaller. Therefore, we'll maximize the value of $\hat{f}$ when no such effects are in play (note though that I'm not claiming $\xi=0$ is the only such point; I'm not sure if that's true or not).
A: This is only a partial answer, but here is an approach to consider:
Analyze the expression $\varphi''(\xi) < 0 $. Note that $\varphi''$ exists because $f\in \mathcal S(\mathbb R)$, implying $\hat f\in \mathcal S(\mathbb R)$, which in particular implies $\varphi$ is smooth. If $\varphi''(\xi) < 0$ $\forall \xi$, then we have a single global maximum. We can of course use similar arguments for when the sign of $\varphi''$ changes to discuss local extrema, but that will depend on the particular $f$. The extrema of $\varphi(\xi)$ will vary depending on $f$. I am unsure if $f$ being positive gives us anything we can use to avoid case-by-case analysis.
