The following result should be true, at least I think I saw it somewhere before but I can't find it now. Please help me to find a reference, or point out if you don't think it is true.
Given $\Omega\subset\mathbb R^N$ is open bounded, smooth boundary. Assume that $(u_n)\subset W^{1,p}(\Omega)$ and we assume $u_n\to u$ in $L^1$ for $u\in W^{1,p}(\Omega)$. Next, let us define $u(t):=u(t,x_2,x_3,\ldots, x_n)$, i.e., the slicing of $u$. We know, for instance, Evans & Gariepy, that $u(t)\in W^{1,p}(I)$ where $I$ is the compact interval in $\Omega$.
My question: do we have $u_n(t)\to u(t)$ in $L^1$ as well? I think yes, but I can not found an reference showing that.
I also think this result hold for $BV$ functions.
Remark: $u(t)\in W^{1,p}(I)$ holds only for a.e. $(x_2,\dots,x_n)$.
No, the slices need not converge in any sense. Take the standard example that $L^1$ convergence does not imply convergence a.e. Make these functions $f_n$ Lipschitz continuous by replacing rectangles by trapezoids in their graphs. Define $u_n(x,y)=f_n(x)$ for $x,y\in [0,1]$. Now $u_n\to 0$ in $L^1([0,1]^2)$ but for any $x$, the functions $y\mapsto u_n(x,y)$ do not converge: they are constants bouncing around.