Suppose we are given a Schwartz function $f\in\mathcal{S}(\mathbb{R}^2)$ and we define $\varphi\in\mathcal{S}(\mathbb{R})$ by $\varphi(x):=f(x,0)$. I am wondering if there is a nice relationship between the Fourier transforms $\hat{\varphi}$ and $\hat{f}$.
I deduced several equations involving $\hat{\varphi}$ and $\hat{f}$, but my equations are not very satisfying. They look very artificial (e.g. one of them uses the Heaviside function). Is it possible to relate both Fourier transforms more naturally? I would appreciate your help.
So first remark that the Fourier transform of a function of two variables is just the composition of the Fourier transform for each variables, i.e. $$ \hat{g}(x,y) = \mathcal{F}_{2}\mathcal{F}_{1}g(x,y) $$ where $\mathcal{F}_{1}$ is the Fourier transform with respect to the first variable, and $\mathcal{F}_{2}$ to the second. Now, taking $y=0$, $f=\hat{g}$ (so that $g(z,w) = \hat{f}(-z,-w)$ perhaps with a multiplication by a constant depending on your definition of the Fourier transform) and using the fact that $\hat{u}(0) = ∫_{\mathbb{R}}u$, you get $$ \begin{align*} \varphi(x) &= f(x,0) = \hat{g}(x,0) = ∫_{\mathbb{R}}\mathcal{F}_{1}g(x,y)\,\mathrm{d} y \\ &= ∫_{\mathbb{R}}\mathcal{F}_{1}\hat{f}(-x,y)\,\mathrm{d} y. \end{align*} $$ (doing a change of variable $y ↔ -y$ to get the last line. I hope this helps!