This is a problem from T. Ransford's Potential Theory in the Complex Plane.
Let $u:\Delta(0,r)\to\mathbb{R}$ be a function such that $u(x+iy)$ is convex in $x$ for each fixed $y$ and convex in $y$ for each fixed $x$. Prove that $u$ is subharmonic on $\Delta(0,r)$. Give an example to show that the converse is false. $\Delta(0,r)$ denotes the ball of radius $r$ centered at the origin in the complex plane.
This problem has been asked before: Prove that $u$ is upper semicontinuous on $\Delta(0,\rho)$.
I was able to show upper semi continuity but not the submean inequality. How do I go about showing the submean inequality?
Fix $z \in \Delta(0,r)$, and let $\Delta(z,s) \subset\subset \Delta(0,r)$. We need to show that $\frac{1}{2\pi}\int^{2\pi}_{0} u(z+se^{i\theta})d\theta \geq u(z)$.
First note that for $t\in[-s,s]$, $$u(z+t) \leq \frac{1}{2}[u(z+t+i\sqrt{r^2-t^2})+u(z+t-i\sqrt{r^2-t^2})]$$ by convexity of $u$ along $y$-direction.
Let $u_1(t)=u(z+t+i\sqrt{r^2-t^2}), u_2(t)=u(z+t-i\sqrt{r^2-t^2})$ for $t\in[-s,s]$. i.e. they are u parametrized by $x$-coordinates.
Now we reparametrize circular mean integral as follows :
$$ \frac{1}{2\pi}\int^{2\pi}_{0} u(z+se^{i\theta})d\theta = \frac{1}{2\pi}[\int^{\pi}_{0} + \int^{2\pi}_{\pi}] $$
where $$\int^{\pi}_{0} u d\theta = \int^{-s}_{s} u(z+se^{i{cos}^{-1}(t/s)})\frac{-dt}{\sqrt{1-t^2/s^2}}=\int^{s}_{-s} u_1(t)\frac{dt}{\sqrt{1-t^2/s^2}} $$ and similarly $$\int^{2\pi}_{\pi} u d\theta = \int^{s}_{-s} u(z+se^{i{cos}^{-1}(t/s)})\frac{dt}{\sqrt{1-t^2/s^2}}=\int^{s}_{-s} u_2(t)\frac{dt}{\sqrt{1-t^2/s^2}} $$ Note that we are using two different arccos functions here.
Then $$ \frac{1}{2\pi}\int^{2\pi}_{0} u(z+se^{i\theta})d\theta = \frac{1}{2\pi}[\int^{\pi}_{0} + \int^{2\pi}_{\pi}]=\\ \frac{1}{2\pi}\int^{s}_{-s} (u_1(t)+u_2(t))\frac{dt}{\sqrt{1-t^2/s^2}} \geq \frac{1}{2\pi}\int^{s}_{-s} 2u(z+t) \frac{dt}{\sqrt{1-t^2/s^2}} = \\ \frac{1}{\pi}\int^{s}_{-s} u(z+t) \frac{dt}{\sqrt{1-t^2/s^2}} $$
Since $$ \frac{1}{\pi}\int^{s}_{-s}\frac{dt}{\sqrt{1-t^2/s^2}} = 1 $$ and $u$ is convex along $x$-direction, we can apply Jensen's inequality, getting
$$ \frac{1}{\pi}\int^{s}_{-s} u(z+t) \frac{dt}{\sqrt{1-t^2/s^2}} \geq u(\frac{1}{\pi}\int^{s}_{-s}(z+t)\frac{dt}{\sqrt{1-t^2/s^2}}) = u(z) $$