Assume we have $f \in W^{1, \infty}(\mathbb{R}^n \times \mathbb{R})$. Can we then deduce that the mapping $x \mapsto f(x, y)$ for almost every $y \in \mathbb{R}$ is weakly differentiable?
Intuitively, this seems "weaker" and thus true, but I can not at all see it when i look at the integral equation that belongs to the definition of weak differentiability. Maybe we can choose test functions whose support is a subset of $\mathbb{R}^n \times \lbrace y \rbrace$...
Let me discuss the case $n=1$ only. Let me denote by $f_x$ the weak derivative of $f$ with respect to $x$. Let $\phi,\psi$ be two smooth test functions. Then $$ -\iint f_x(x,y) \phi(x)\psi(y)dx\ dy=\iint f(x,y) \phi_x(x)\psi(y)dx\ dy, $$ using Fubini on both sides gives $$ \int \left( \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx\right) \psi(y) \ dy = 0 $$ for all smooth $\psi$, so by the fundamental theorem of calculus of variations $$ \int -f_x(x,y) \phi(x) + f(x,y) \phi_x(x)\ dx = 0 $$ for almost all $y$, which is the desired weak differentiability of $x \mapsto f(x,y)$ .