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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.

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    $\begingroup$ Morally, the example $u=\log |z|^2$ would violate your conditions. Of course, it is not smooth, but you can approximate it by $u_\varepsilon = \log(|z|^2+\varepsilon^2)-\log(1+ \varepsilon^2)$ (maybe up to a harmless normalizing factor to get mass one) and get a contradiction. $\endgroup$
    – Henri
    Jul 13 at 11:56
  • $\begingroup$ Also, you can set $\Delta u$ to be equal to, say, $(1-|z|^2)^{-2}$ in a very narrow "thorn" that touches the boundary at $1$, and $0$ outside a twice thicker "thorn". $\endgroup$ Jul 13 at 12:16
  • $\begingroup$ @Mateusz Ok, how about if I assume $\int_{\mathbb{D}} (-u) \Delta u =1$, instead of $\int_{\mathbb{D}} \Delta u = 1.$ $\endgroup$
    – AndewUK
    Jul 13 at 13:23
  • $\begingroup$ You can do the same thing that I mentioned above with $u(z)=-\log(1-\log |z|)$. $\endgroup$
    – Henri
    Jul 13 at 14:02
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    $\begingroup$ What do you mean by "smooth"? Smooth in $\{ z:|z|\leq 1\}$ or smooth only in \{ z:|z|<1\}$ and continuous in the closure of the unit disk? $\endgroup$ Jul 13 at 15:13
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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 $\Delta u(z)=f''(\log|z|)$.

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