2
$\begingroup$

Say, we're given smooth functions $f_n$, $n=1,2,3,...$ defined on a smooth bounded domain $\Omega\subset\mathbb{R}^d$ satisfying

  1. $\Delta f_n\ge 0$ (subharmonic)
  2. $f_n\ge 0$
  3. $\int_\Omega f_n=I>0$ for all $n\in\mathbb{N}$
  4. ${f_n}_{|\partial\Omega}=n$

Then, say $B\subset\subset \Omega$. Can we conclude that $\int_B f_n\to 0$?

When I visualize these functions, I suspect this might be true, but I can't come up with a proof nor a counterexample. Any help would be appreciated.

$\endgroup$
5
  • 4
    $\begingroup$ No, this is wrong: $n=1$, $\Omega=(-1,1)$, $f_n=1$ on $[-1+1/n^2, 1-1/n^2]$ and then move up linearly (you can make this smooth, of course, if desired). This doesn't satisfy (3) exactly, but you can multiply by suitable constants. $\endgroup$ Oct 10 '20 at 14:39
  • $\begingroup$ Sorry, I deleted my previous comment after I noticed your edit - looks like we were out of sync. OK but now isn't the issue that $g_n<1$ in the interior? And possibly $g_n\to 0$ in the interior because we're normalizing by factors which increase.. $\endgroup$
    – Fozz
    Oct 10 '20 at 14:53
  • $\begingroup$ If I normalize by dividing by $(1/2)\int f_n$, then these factors converge to $1$, so $g_n\to 1$ locally uniformly on $\Omega$. $\endgroup$ Oct 10 '20 at 15:00
  • $\begingroup$ Right, right, I see. OK looks good $\endgroup$
    – Fozz
    Oct 10 '20 at 15:02
  • $\begingroup$ One thing to keep in mind is these boundary blow up solutions; $ \Delta u= u^p$ in $\Omega$ with $u=\infty$ on $ \partial \Omega$. This might give you some intuition also about what can happen $\endgroup$
    – Math604
    Oct 12 '20 at 23:17
2
$\begingroup$

Let $\Omega$ be the unit ball, $B$ some smaller concentric ball, and $u_n(x)=1$ for $|x|\leq 1-1/n$ and $u_n(x)=n(n-1)|x|+2n-n^2$ for $1-1/n\leq|x|\leq 1$. Then your conditions 1,2,4 are satisfied exactly, and 3 is satisfied approximately (integrals tend to a positive constant), so a slight modification will give you constant integrals, if really needed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.