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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.

spatially
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1 Answers1

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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.

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  • Thank you! Btw, if I have in addition $u_n\to u$ in $W^{1,p}$, can I have the slice goes to $u$ in some sense? – spatially Nov 15 '15 at 16:15
  • It depends. Consider, as in my answer, the functions for which the slices are constant. For this case, you are asking whether $W^{1,p}$ convergence implies pointwise convergence. This is true only if $p>n-1$, by Sobolev-Morrey. –  Nov 15 '15 at 16:17
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    I see. So if I change to ask the subsequence of $u_n$ but not $u_n$ itself, the result will be better right? – spatially Nov 15 '15 at 16:41
  • From $L^1$, you can get a subsequence converging a.e.. And then use Fubini to get that a.e. slice converges a.e. So at least there is that. –  Nov 15 '15 at 16:43