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