I am working on the following problem:
Let $v$ be continuous on $|z|<1$ and subharmonic on $0<|z|<1$. Show that $v$ is subharmonic on all of $|z|<1.$
Here, we say $\varphi$ is subharmonic on $U$ if any closed disc $D(z,r) \subset U$ of center $z$ and radius $r$ one has $$\varphi(z) \leq \frac{1}{2\pi} \int_0^{2\pi} \varphi(z+ re^{i\theta}) \, d\theta.$$
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My attempt:
It is enough to show that for some $r$, we have $$v(0) \leq \frac{1}{2\pi} \int_0^{2\pi} v(re^{it}) \, dt.$$ There is a hint: consider the $$v_{\epsilon}(z)=v(z)+\epsilon \log |z|$$ I have no idea how to use this hint since $v_{\epsilon}(z)$ behaves even worse near $0$. It will go to infinity when $z\to 0$. So what does this hint mean?
Any hints and answers are welcomed!
No, $v_\epsilon(0)=-\infty$, not $\infty$. And that's better, not worse: It's obvious that $v_\epsilon$ is subharmonic since it's subharmonic in $0<|z|<1$ and equals $-\infty$ at $0$.
The tricky part is that $$\lim_{\epsilon\to0}v_\epsilon(0)\ne v(0).$$
Major thing about subhamonic functions: The inequality in the definition is required to hold only for $0<r<\delta(z)$ (this is important, for example this is why $v_\epsilon$ is obviously subharmonic: for $z\ne0$ take $\delta(z)=|z|$), but if $u$ is subharmonic then the inequality holds for all $r$.
So assume $0<r<1/2$ and $0<|z|<r$. Then $$v_\epsilon(z)\le\frac1{2\pi}\int_0^{2\pi}v_\epsilon(z+re^{it})\,dt.$$ Let $\epsilon\to0$ in that inequality, and then let $z\to0$.