Prove that $u$ is upper semicontinuous on $\Delta(0,\rho)$. 
Let $u:\Delta(0,\rho)\rightarrow \mathbb{R}$ 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 $\Delta(0,\rho)$.

My Try:
I use the following definition for subharmonic functions.
$u:U\rightarrow \mathbb{R}$ is subharmonic if $u$ is upper semicontinuous on $U$ and satisfy the submean inequality on $U$.
So here I proved that $u$ satisfy the submean inequality. But failed to prove that it is upper semicontinuous on $\Delta(0,\rho)$. 
I need to prove that $\limsup_{(x,y)\rightarrow (x_0,y_0)}\leq u(x_0+iy_0)$
I know that $\lim_{y\rightarrow y_0} u(x_0+iy)=u(x_0+iy_0)$ and $\lim_{x\rightarrow x_0} u(x+iy_0)=u(x_0+iy_0)$. But I could not use these to show the result. Everytime I try I got the other way around. i. e. $\limsup_{(x,y)\rightarrow (x_0,y_0)}\geq u(x_0+iy_0)$. Any suggestion please..
 A: You need to show that $\{u<t\}$ is open. Say $u(z)<t$.
Since convex functions are continuous there exists $\alpha>0$ so that $$u(z\pm\alpha)<t.$$And now for the same reason there exists $\beta>0$ so that $$u(z\pm\alpha\pm i\beta)<t.$$
So $u<t$ at every corner of that rectangle. Separate convexity shows that $u<t$ on the boundary of the rectangle, and then separate convexity shows that $u<t$ in the interior of the rectangle.

In Detail: First we note that 
Lemma If $f$ is convex on $[a,b]$, $f(a)<t$ and $f(b)<t$ then $f<t$ on $(a,b)$. 
I can prove that if you want; I'm not sure what aspect of this we don't get, so I don't know if I need to.
Now. We have $u<t$ at each of the four points $z\pm\alpha\pm i\beta$. In particular $u(z+\alpha+i\beta) < t$ and $u(z-\alpha+i\beta) < t$. So the lemma, with $f(x)=u(z+x+i\beta)$, shows that $$u(z+x+i\beta)<t\quad(-\alpha\le x\le\alpha).$$ Similarly the lemma, with $f(x)=u(z+x-i\beta)$, shows that$$u(z+x-i\beta)<t\quad(-\alpha\le x\le\alpha).$$ 
Now fix $x\in[-\alpha,\alpha]$. We have $$u(z+x+i\beta)<t$$and$$u(z+x-i\beta)<t,$$so another application of the lemma, this time with $f(y)=u(z+x+iy)$,  shows that $$u(z+x+iy)<t\quad(-\beta\le y\le\beta).$$
