2
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Call $I(k)$ the integral of the RHS. Using the substitution $t=k-p$, we get that $$I(k)=k\int_{-\infty}^{+\infty}\widehat u(p)\widehat u(k-p)dp-I(k),$$ hence $$2I(k)=k\int_{-\infty}^{+\infty}\widehat u(p)\widehat u(k-p)dp,$$ which is the wanted formula.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .