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Let $E=\{z\in\mathbb{C}:0<|z|<1\}$ be the unit punctured disc in $\mathbb{C}$, and $u\leq0$ be a subharmonic function in $E$. Suppose that we have a real number $0<r<1$, and the area in $|z|\leq r$ where $u(z)\leq\frac{1}{2}\log r$ is at least $r^3$.

I want to show that we have $u(z)\leq\frac{1}{3}\log|z|$ for any $r\leq|z|\leq1$.

Here is what I tried. Define $M(x)=\sup_{|z|\leq x}u(z)$, then by Hadamard three circle theorem for subharmonic functions (cf. https://math.stackexchange.com/questions/4607637/three-circle-theorem-subharmonic-functions), it is enough to show that $M(r)\leq\frac{1}{3}\log r$. At this point, I tried to induce a contradiction by assuming $M(r)>\frac{1}{3}\log r$, and applied Harnack inequality to check the area where $u(z)\leq\frac{1}{2}\log r$ is smaller than $r^3$, but it didn't work well.

We can assume $r$ is sufficiently small if it feels comfortable.

I would appriciate for any helps.

Edit: the upper bound may not be $\frac{1}{3}\log |z|$, but I think it should be of the form $\frac{1}{n}\log |z|$ for some $n\in\mathbb{N}$.

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    $\begingroup$ Have you tried to average the logarithm of the Blashke factor over the set where you have the bound? $\endgroup$ Commented Jun 10 at 1:02

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With your constants, it is wrong. Take $$u(z)=\frac{1}{4}\log|z|,$$ this functon violates your inequality $u(z)<(1/3)\log|z|$ everywhere. On the other hand, take $r=1/\pi$. Then $u(z)\leq (1/2)\log r$ in the disk $|z|<r^2$ whose area is $\pi r^4=\pi^{-3}=r^3$.

To obtain an (asymptotically) correct result of the type you want, consider a disk $E$ of area $r^3$ touching the circle $|z|=r$ from inside, and the harmonic function in the ring $\{ z:|z|<1\}\backslash E$ which is equal to $\log r$ on $E$, zero on the unit circle. This function can be written explicitly by mapping the ring conformally onto a concentric ring. It will tell you what the correct result is.

For small $r$ Fedja's suggestion will give you asymptotically the same result.

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  • $\begingroup$ Sorry for silly question, but how such harmonic function can be used to solve the problem? Maybe it is a harmonic majorant of $u$? But I am not sure how to check it. Thanks. $\endgroup$ Commented Jun 13 at 7:24
  • $\begingroup$ @Crisp: This function is just an example which would suggest you what the correct estimate is. Once you figure this out, one can think of how to prove it. $\endgroup$ Commented Jun 13 at 12:31
  • $\begingroup$ It depends on what you really need. Do you really care about 1/3 or 1/4? To obtain an exact estimate, you have to solve an extremal problem about harmonic measure of the set of given area, which may be hard. But if you are satisfied with some estimate, it can be proved. $\endgroup$ Commented Jun 14 at 1:10

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