Weak derivative of a Sobolev function in unbounded domains

Question

I have following setup:

$f\in W^{1,1}(\mathbb{R}^3)$ with $f\geq 0$ and $\int_{\mathbb{R}^3}f=1$, how can I show that the weak derivative of $\tilde{f}(x):=\int_{\mathbb{R}^2}f(x,y,z)\,\text{d}y\text{d}z$ is $\int_{\mathbb{R}^2}\partial_x f(x,y,z)\,\text{d}y\text{d}z$?

The problem arguing with test functions are the different domain dimensions of $f(x,y,z)$ and $\tilde{f}(x)$ ...

I am looking forward for any suggestions and hints :)

Answer 1

This is a guideline not a full answer.

Some tools we would like to use include: applying the DCT so that you could do differentiation under the integral sign.

We need to verify the following:

• $f(x,y,z)$ is integrable in ${\rm d} y {\rm d}z$ for any $x$, this is very straightforward by Fubini. Then, by DCT, the differentiation under integral sign can be applied.
• Then for any $\phi\in C^{\infty}_c(\mathbb{R})$, we need to verify $$ \int_{\mathbb{R}}\tilde{f} \phi_x \,{\rm d}x = - \int_{\mathbb{R}}\left( \int_{\mathbb{R}^2}\partial_x f(x,y,z)\,\mathrm{d}y\mathrm{d}z\right) \phi {\rm d}x, $$ applying the differentiation under integral sign on the right and the integration by parts will give you the proof.