I have read an equality about Fourier transforms which I can not proof. It is as following: Let $u\in C_0(\mathbb{R}^n)$ and \begin{equation} g(x_1,x_2,...,x_{n-1}):=u(x_1,x_2,...,x_{n-1},0). \end{equation} Then \begin{equation} \hat{g}(x_1,...,x_{n-1})=\frac{1}{2\pi}\int_{\mathbb{R}}\hat{u}(x_1,...,x_{n-1},x_n)dx_n. \end{equation} (Here the Fourier transform is defined by $\hat{u}(t)=\int_{\mathbb{R}^n} u(x)e^{-it\cdot x}dx$.)

How to prove this?

  • $\begingroup$ I'm pretty sure you just need to use the definition of Fourier transform and possibly interchanging some order of integration. EDIT: and inversion formula as well by the looks of it ! $\endgroup$ – Chee Han Jun 22 '15 at 18:01
  • $\begingroup$ Does $C_0$ mean "vanishes at infinity" or "compact support"? If the first there's no reason to think $u$ and $g$ even have Fourier transforms (in the classical sense, as you define it). And regardless there's no reason to think the integral in the second display exists. We need hypotheses $\endgroup$ – David C. Ullrich Jun 22 '15 at 18:08
  • $\begingroup$ @DavidC.Ullrich It is defined for compact support, since it is used to prove trace properties in Sobolev spaces. $\endgroup$ – Peter Jun 22 '15 at 18:16
  • $\begingroup$ @CheeHan Yeah perhaps, i know there are some relationships and i've tried, but I can still not obtain those equalities mentioned above -.-! $\endgroup$ – Peter Jun 22 '15 at 18:18
  • $\begingroup$ Note that this is really a one-variable question: $x_1,\dots,x_{n-1}$ only enter the problem as parameters, so you are really just asking one to prove that $f(0)$ is the integral of the Fourier transform of $f$, multiplied by a (universal) normalization constant. $\endgroup$ – Ian Jun 22 '15 at 18:38

I'm going to assume $u$ is in the Schwarz space, so we don't have to worry about various technicalities. I'm going to take $n=2$ for convenience. And I'm going to assume the Littlewood convention $2\pi=1$, because everyone puts the $2\pi$'s in different places (feel free to edit this, inserting the $2\pi$'s as needed for your version of the Fourier transform).

Hmm. Using the same letter for the variable on both sides is bad; I'm going to write $\hat u(s,t)$, where $u=u(x,y)$.

Ok. Let $$\phi(s)=\int_\mathbb R\hat u(s,t)\,dt.$$ By the inversion theorem (for $n=1$) it's enough to show that $$\int_{\mathbb R}\phi(s)e^{ixs}\,ds=u(x,0).$$ But this is clear from the definitions and the inversion theorem (for $n=2$): $$\int_{\mathbb R}\phi(s)e^{ixs}\,ds=\int_{{\mathbb R}^2}\hat u(s,t)e^{i(x,0)\cdot(s,t)}\,dsdt=u(x,0).$$

  • $\begingroup$ Nice proof! Thank you very much! $\endgroup$ – Peter Jun 22 '15 at 18:55

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