Sequence of subharmonic functions on shrinking domains 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 $0\times [\frac{1}{3},\frac{2}{3}]\subset \mathbb{R}^2$?
 A: 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 $\sqrt{\pi}$. 
A: 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.
