A question on subharmonic functions on the unit disc I have the following question:
Let $u$ be a smooth subharmonic function on the unit disc $\mathbb{D}:=\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} $.
Assume that $u=0$ on the boundary of $\mathbb{D}$ and
$$
\int_{\mathbb{D}}\Delta u=1.
$$
Here $\Delta u$ is the Riesz measure of $u$.
Is there any chance to show that
$$
\Delta u\leq\frac{M}{1-\left|z\right|^{2}}dV
$$
as measures, for some positive constant $M$ (possibly an absolute constant, i.e. independent of $u$)?
Here $dV$ is the standard Lebesgue measure on $\mathbb{D}$.
Thanks for any suggestions.
 A: The first comment already shows that you cannot have $M$ independent of $u$.
But in fact you can construct a function smooth in the closed unit disk for
which $\Delta u$ has no uniform bound as $|z|\to 1$ whatsoever.
Just take
$$u_1(z)=\sum\delta_j\left(\log|z-a_j|-\log|1-a_jz|\right)=-\sum\delta_jG(z,a_j),$$
where $a_j\in(0,1), a_j\to 1$ is a fixed sequence, and $G$ is the Green function. Then smoothen near $a_j$
by replacing this $u$ near $a_j$ by a convolution with an infinitely smooth
radial $\phi_j(z)$ with very small support. That these convolutions will match
away from the points $a_j$ follows from the average property. By taking $\delta_j$ and supports of $\phi_j$ very small you ensure that the function is smooth in the closed disk, while $\Delta u$ tends to infinity arbitrarily fast.
Another way to do the same is to construct $u$ in the form $u(z)=f(\log|z|)$, where $f$ is a convex function. It is pretty clear that a convex function on $(-\infty,0]$ required properties exists,
and $\Delta u(z)=f''(\log|z|)$.
