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?
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 $n=2$): $$\int_{\mathbb R}\phi(s)e^{ixs}\,ds=\int_{{\mathbb R}^2}\hat u(s,t)e^{i(x,0)\cdot(s,t)}\,dsdt=u(x,0).$$