# 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.

• 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. Jul 13 at 11:56
• 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". Jul 13 at 12:16
• @Mateusz Ok, how about if I assume $\int_{\mathbb{D}} (-u) \Delta u =1$, instead of $\int_{\mathbb{D}} \Delta u = 1.$ Jul 13 at 13:23
• You can do the same thing that I mentioned above with $u(z)=-\log(1-\log |z|)$. Jul 13 at 14:02
• 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? Jul 13 at 15:13

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 $$\Delta u(z)=f''(\log|z|)$$.