I have a comprehension problem regarding Fourier transforms. So far I know, the Fourier transform can be defined on the whole Schwartz space $\mathcal{S}(\mathbb{R})$ and is bijective on it. So I have a function $u(x) \in \mathcal{S}(\mathbb{R})$ and want to Fourier transform $u(x)\partial_x u(x)$ according to the prescription $\partial_x \rightarrow ik$. This gives:
$u(x)\frac{\partial u(x)}{\partial x} \rightarrow \widehat{u}*(ik\widehat{u}) = i \intop_{-\infty}^{+\infty}\widehat{u}(p)(k-p)\widehat{u}(k-p)dp = \intop_{-\infty}^{+\infty}p\widehat{u}(p)\widehat{u}(k-p)dp.$
But on the other hand it is also true that
$u(x)\frac{\partial u(x)}{\partial x} = \frac{1}{2}\frac{\partial}{\partial x}(u^2(x)) \rightarrow \frac{1}{2}ik\widehat{u^2} = \frac{1}{2}ik\intop_{-\infty}^{+\infty}\widehat{u}(p)\widehat{u}(k-p)dp.$
The two integrals in Fourier space on the very right should be the same but I cannot see how. For this one should show that
$\intop_{-\infty}^{+\infty}\widehat{u}(p)\widehat{u}(k-p)dp = \frac{2}{k}\intop_{-\infty}^{+\infty}p\widehat{u}(p)\widehat{u}(k-p)dp.$ Does anyone have an idea how to do that?