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.

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).
