# Weak derivative of a Sobolev function in unbounded domains

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 :)

• $$f(x,y,z)$$ 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.