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Set $G_\eta:=\{(x,y)\in \mathbb{R}^2|-\eta<x<\eta, 0<y<1\}$. If $u_\eta\geq 0$ is a sequence of subharmonic functions defined on $G_\eta$ such that $$ \int_{G_\eta}|u_\eta|^2dx\wedge dy\leq C\eta, $$ where $C$ is some constant, then can we show that $u_\eta$ is uniformly bounded on the line $0\times [\frac{1}{3},\frac{2}{3}]\subset \mathbb{R}^2$?

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This is incorrect as stated: it just does not scale correctly. Let $v(z)$ be the harmonic function on the unit square $|x|<1,|y|<1,$ which is zero on the vertical sides and $1$ on the horizontal sides. Then we define the subharmonic function $u$ in your rectangle by setting $u(z)=v(z/\eta)$ when $|x|<\eta,\; |y|<\eta$ and zero otherwise. Then $u(0)=v(0)$ is a positive constant, while $\| u\|_2=\| v\|_2\eta$, so $$\int_{G_\eta}u^2\approx c\eta^2.$$

This example suggests the correct inequality: $$u(x)\leq C\| u\|_2/\eta,$$ with an absolute constant $C$. This is a simple consequence of the average property of subharmonic functions. Let $x$ be a point on your interval $(1/3,3/2)$. Let $B$ be the disk of radius $\eta$ centered at $x$. Then the average property says that $$u(x)\leq \int_B u\; dm/(\pi\eta^2),$$ where $dm$ is the Lebesgue area element and $\pi\eta^2$ is the area of the disk. Applying the Schwarz (Cauchy-Bunyakovski-...) inequality, we obtain $$\left(\int_B u\; dm\right)^2\leq\pi\eta^2\int_Bu^2dm\leq\pi\eta^2\| u\|_2^2,$$ so $$\int_Bu\; dm\leq\sqrt{\pi}\eta\| u\|_2,$$ and combining with the first inequality we obtain the result with constant $\sqrt{\pi}$.

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  • $\begingroup$ This doesn't work since the $L^2$ norm of $u$ is bounded by $\sqrt{\eta}$ not $\eta$. $\endgroup$ – foliations Mar 30 '16 at 22:50
  • $\begingroup$ @foliations: I edited. It is of course incorrect as stated. $\endgroup$ – Alexandre Eremenko Mar 30 '16 at 23:09
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No. A counterexample is (essentially) given by $$ u(z) = -\log |z+i\delta|, \quad\quad z=x+iy, y\ge 0 . $$ This function is harmonic and positive near $z=0$, and $|u(x+iy)|\le |u(x)|\in L^2(-1/2,1/2)$, so $$ \int_{-1/2}^{1/2} dx\int_0^{2\eta} dy\, u^2 \lesssim \eta , $$ as required. However, $|u(i\eta)|$ is not bounded as $\delta,\eta\to 0+$.

By rotating and (slightly) rescaling this, we obtain a counterexample in your setting.

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  • $\begingroup$ An easy way to see that his inequality cannot be correct is just by scaling: if you multiply all his functions on constants, sup norms multiply on the same constants, but the integral of the square on the square of the constant. $\endgroup$ – Alexandre Eremenko Mar 31 '16 at 2:30
  • $\begingroup$ @AlexandreEremenko: Yes, but I think what I do is natural, too: solve the Dirchlet problem on a thin rectangle with a boundary value that is large only on a small portion of the boundary and check if the area average of $u^2$ can stay bounded. $\endgroup$ – Christian Remling Mar 31 '16 at 2:35
  • $\begingroup$ Sure. BTW the description in your comment better describes the example in my answer:-) $\endgroup$ – Alexandre Eremenko Mar 31 '16 at 2:39
  • $\begingroup$ @AlexandreEremenko: Well, I tried it with a Poisson kernel first (which fits my description better), but that's too singular after squaring, so switched to a logarithm then... :) $\endgroup$ – Christian Remling Mar 31 '16 at 15:18
  • $\begingroup$ thanks a lot for all of your answers. I know the point in this question. $\endgroup$ – Higgs-Boson Mar 31 '16 at 19:20

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