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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.

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    $\begingroup$ 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. $\endgroup$
    – Christian Remling
    Commented Oct 10, 2020 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
    Commented Oct 10, 2020 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 $\Omega$. $\endgroup$
    – Christian Remling
    Commented Oct 10, 2020 at 15:00
  • $\begingroup$ Right, right, I see. OK looks good $\endgroup$
    – Fozz
    Commented Oct 10, 2020 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
    Commented Oct 12, 2020 at 23:17

1 Answer 1

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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.

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