# Positive subharmonic functions with constant integral blowing up at boundary

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.

• 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. – Christian Remling Oct 10 at 14:39
• 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.. – Fozz Oct 10 at 14:53
• 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 $\Omega$. – Christian Remling Oct 10 at 15:00
• Right, right, I see. OK looks good – Fozz Oct 10 at 15:02
• 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 – Math604 Oct 12 at 23:17

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.