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 $0\times [\frac{1}{3},\frac{2}{3}]\subset \mathbb{R}^2$?
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 (CauchyBunyakovski...) 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 $\sqrt{\pi}$.

$\begingroup$ 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
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.

$\begingroup$ 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. $\endgroup$ – HiggsBoson Mar 31 '16 at 19:20