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I am struggling with Evans PDE problem 5.10. #12. Here is the question: Let $V\subset \subset U$ be open sets. Show by example that if we have $\|D^hu\|_{L^1(V)}\le C$ for all $0< |h|<\frac{1}{2}\operatorname{dist}(V,\partial U)$, it does not necessarily follow that $u\in W^{1,1}(V)$.

I have no idea why the bigger open set $U$ is relevant. I would very much appreciate your help.

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$U$ is there just to make sure the difference quotient in $V$ is well-defined. –  Shuhao Cao Nov 11 '12 at 23:40
    
I see, but am still in trouble proving this. –  Pooya Nov 12 '12 at 10:54
    
Maybe use the following idea. The assumption gives a conditions for elements why are not too near of the boundary of $U$, but doesn't prevent a bad behaviour near it. –  Davide Giraudo Nov 12 '12 at 21:31

1 Answer 1

Here $D^h u = (u(\cdot+h)-u(\cdot))/|h|$, if I recall correctly. The issue is that while the $L^1$ norm of $D^hu$ controls the total variation of $u$, it does not control how concentrated this variation can be, e.g., it does not prevent the variation from being supported on a set of measure zero. The uniform boundedness of $\|D^h u\|_{L^1}$ yields $u\in BV$, which is a larger space than $W^{1,1}$. (In contrast, the $L^p$ norm of function $f$ for $p>1$ does control the degree of concentration of $f$, via Hölder's inequality $|\int_E f|\le \|f\|_p |E|^{1/p'}$.)

For a concrete example, take $u$ to be any discontinuous function of bounded variation, such as the characteristic function of a ball $B$ contained in $V$. The difference $\chi_B(\cdot+h)-\chi_B(\cdot)$ has $L^1$ norm of magnitude $|h|$. The computation is particularly easy in one dimension.

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