Does restricting variables of a Sobolev function produces a Sobolev function?

Question

To simplify I will ask in $\mathbb{R}^2$, but I think it will work in any dimension, if I have an open and bounded domain $\Omega$ and a function $u \in W^{1,p}(\Omega)$, is it true that for almost every $x$ the function $u_{x}(y) = u(x,y)$ is in $W^{1,p}(\Omega_{x})$? Where

$$ \Omega_{x} = \{y \in \mathbb{R} \colon (x,y) \in \Omega\}. $$

I tried using Lebesgue Differentiation Theorem, but I got stuck on how to construct the test functions that would be useful, any tips or reference would help me a lot.

Answer 1

There is a very useful theorem about Sobolev spaces sometimes referred to as "Absolute Continuity on Lines". In the theorem, $\mathcal{L}^{N}$ is the $N$-dimensional Lebesgue measure.

Let $\Omega \subseteq \mathbb{R}^{N}$ be an open set and let $1 \leq p < \infty$. A function $f \in L^{p}(\Omega)$ belongs to the space $W^{1, p}(\Omega)$ if and only if it has a representative $\bar{f}$ that is absolutely continuous on $\mathcal{L}^{N-1}$ a.e. line segments of $\Omega$ that are parallel to the coordinate axes, and whose first order (classical) partial derivatives belong to $L^{p}(\Omega)$. Moreover, the (classical) partial derivatives of $\bar{f}$ agree $\mathcal{L}^{N}$ a.e. with the weak derivatives of $f$.

I'll stick to two dimensions from here to keep it simple, but this theorem should also help with general dimensions. In two dimensions, this theorem says that for $u \in W^{1, p}(\Omega)$, there is a representative $\bar{u}$ such that the functions $\bar{u}_{x}(y)$ are absolutely continuous with partial derivatives in $L^{p}(\Omega_{x})$ for almost every $x$. Combine that with the fact that $\bar{u}_{x} \in L^{p}(\Omega_{x})$ for almost every $x$ due to $u$ being in $L^{p}(\Omega)$ (using Fubini's theorem), and you have that $\bar{u}_{x} \in W^{1, p}(\Omega_{x})$ for almost every $x \in \mathbb{R}$.