assume for a function $\varphi \in L^1$, that $\varphi \geq 0$ and $\int_\mathbb{R}\varphi(x) dx = 1$. Then the following should hold:
$$|\hat{\varphi}(\omega)| \leq \int_\mathbb{R} |\varphi(x)||e^{-ix\omega}|dx =\int_\mathbb{R}\varphi(x) dx = 1 = \hat{\varphi}(0) $$ which means, that $\hat{\varphi}$ has a maximum at $\omega =0$.
Therefore $$ |0|=|\hat{\varphi}'(0)| = |\int_\mathbb{R}x\varphi(x) dx | $$ Now assume $\varphi(x)= \begin{cases}1 & x \in [0,1],\\ 0 & otherwise \end{cases}$ which means all assumptions are fullfiled.
But now $|\hat{\varphi}'(0)| = \frac{1}{2}$
Where is my error?
Thanks for any help Matthias
It is $|\hat\varphi$| that attains its maximum at $\omega=0$, not $\hat\varphi$. In fact, since $\varphi$ is complex valued, it does not make sense to talk about its maximum. $$ \varphi(\omega)=\frac{1-e^{-i\omega}}{i\,\omega} $$ and $$ |\varphi(\omega)|^2=\frac{2(1-\cos\omega)}{\omega^2}=1-\frac{\omega^2}{12}+\dots $$