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Consider a sequence $(f_n)$ of functions in the flat torus $T^d$ converging Lebesgue-a.e. to a limit function $f$.

Assume that:

1) $|f_n|(x)\leq 1$ for every $n,x$

2) $\Delta f_n\geq -1$ in the sense of distributions for every $n$

Can we deduce that $$ \lim_{n\to\infty}\int|\nabla f_n|^2 d\mathcal L^d=\int|\nabla f|^2 d\mathcal L^d $$ ?

Notice that $(1)+(2)$ grant that $\int|\nabla f_n|^2 d\mathcal L^d=-\int f_nd\Delta f_n\leq \mathcal L^d(T^d)<\infty$ for every $n$ so that the sequence is bounded in $W^{1,2}$ and thus, given the a.e. convergence, also weakly converging.

The question is then if it is strongly converging in $W^{1,2}$.

In fact, I'm interested in this problem in a much more general setting, the functions being defined on a converging sequence of metric measure spaces with Ricci curvature bounded from below, but after some thinking I realized that I don't know the answer not even in this simplified setting.

Some comments. The assumption on the Laplacians ensures that they are measures with uniformly bounded mass, and the uniform bound on the $W^{1,2}$-norms of the functions grant that $|\Delta f_n(E)|$ can be bounded from above in terms of the capacity of $E$ only.

Therefore to conclude it would be sufficient to show that for every $\epsilon$ there exists $\delta,N$ such that for $n>N$ the set $\{|f_n-f|>\delta\}$ has capacity smaller than $\epsilon$ (here it matters to pick the upper semicontinuous representatives of the functions).

I've searched in the literature and found results which are very close to this one, but not really sufficient for my purposes, see for instance the book `Growth Theory of Subharmonic Functions' by Vladimir Azarin.

Yet, I found no reference for the question I'm asking.

Any help would be appreciated, thanks in advance.

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  • $\begingroup$ Bozhidar Velichkov provided me an explicit example showing that the answer to my question is no. The example is based on the work "Un terme étrange venu d'ailleurs" by Cioranescu-Murat. (I'm not sure what is the MO policy in these cases: should I delete the question? or what?) $\endgroup$ – Nicola Gigli Oct 6 '14 at 22:36
  • $\begingroup$ Sorry for the late reply: I guess the best thing would be to write an answer yourself so that this is kept for the records. $\endgroup$ – Dirk Mar 26 '15 at 13:42
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I'm reporting an example, presented to me by Bozhidar Velichkov, showing that the answer to my question is no.

The example is based on a construction by Cioranescu-Murat given in their paper "Un terme étrange venu d'ailleurs".

They show that there exists a sequence $\Omega_n$ of open subsets with closure contained in the unit ball $B$ with the following property. Let $u_n$ be the solution of $$ \left\{\begin{array}{ll} \Delta u_n =-1 & on\ \ \Omega_n\\ u_n=0 & on\ \ B\setminus\Omega_n \end{array} \right. $$ and notice that:

1) $u_n$ is non-negative

2) the Laplacian of $u_n$ is $\geq -1$ in the sense of distributions in $B$

3) $u_n$ is bounded from above by the solution $v$ of $\Delta v=-1$ on $B$ with 0 boundary condition. In particular they are uniformly bounded in $L^\infty$.

4) $\int_B |\nabla u_n|^2=-\int_{\Omega_n}u_n\Delta u_n=\int_B u_n$

In particular the sequence $(u_n)$ is bounded in $W^{1,2}$ and thus compact in $L^2$. All of this is true regardless of the choice of $\Omega_n$.

What Cioranescu-Murat did was to show that one can choose the $\Omega_n$'s so that $u_n\to u$ in $L^2$ for $u$ such that $\Delta u+1$ is a positive measure $\mu$ (i.e. non-negative and with strictly positive mass).

This provides a counterexample because on one hand we have $$ \int_B|\nabla u_n|^2=\int_Bu_n\to \int_B u $$ and on the other $$ \int_B|\nabla u|^2=-\int u\Delta u=\int u(1-\mu)<\int u $$ (I've been a bit sloppy in the way this last line is written and also in not justifying why $u$ is strictly positive in a set of positive $\mu$-mass, but this should give the idea).

Roughly said, the sets $\Omega_n$ are the whole $B/2$ minus a lot of tiny disjoint balls spread around which decrease in size and increase in number as $n\to\infty$

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  • $\begingroup$ So the discrepancy is related to a (bad behaved) homogeneization problem, to put things in context ? $\endgroup$ – Hachino Apr 3 '15 at 7:07
  • $\begingroup$ yes. Further details can be found, for instance, in section 4.3 of the book Variational Methods in Shape Optimization Problems by Bucur-Buttazzo $\endgroup$ – Nicola Gigli Apr 3 '15 at 7:29

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