An upper semicontinous function which is not subharmonic. Let $\Delta\subset\Bbb C$ be the open unitary disk. Let $\varphi:\Delta\to\Bbb R$ defined as follows: $\varphi(z)=1$ if $\Re z\ge0$, $\varphi(z)=0$ otherwise. So $\varphi$ is upper semicontinous.
In order to prove $\varphi$ is NOT subharmonic, I've to find a compact subset $K\Subset\Delta$ and a real valued function $h\in\mathcal{C}^0(K)\cap\mathcal H(\stackrel{\circ}{K})$ such that


*

*$\varphi\leq h$ on $\partial K$

*$\varphi\nleq h$ on $\stackrel{\circ}{K}$


Here $\mathcal H(\stackrel{\circ}{K})$ is the set of harmonic functions on the interior of $K$.
I tried for hours, but I did't find an answer.
The only thing which gave me a chance for one second was $h(z):=\log\left(\frac{|z|}{a}\right)+1$ on $K:=\bar\Delta_{0,a}$ which is the closure of the disk centered in $0$ of radius $a\in]0,1[$. This function would be right, the problem is it's not defined in $0$. So I tried cutting out a small circle around zero, but we get other obvious problems in satisfying the given conditions.

Take a disk $\lvert z\rvert \leqslant r$ for some $0 < r < 1$, and define the boundary values by
$$h_0(re^{i\vartheta}) = l(\vartheta),$$
where
$$l(t) = \begin{cases} \frac{2}{\pi}(t+\pi) &, -\pi \leqslant t \leqslant -\frac{\pi}{2}\\ \quad 1 &, -\frac{\pi}{2} \leqslant t \leqslant \frac{\pi}{2}\\ \frac{2}{\pi}(\pi-t) &, \frac{\pi}{2} \leqslant t \leqslant \pi.\end{cases}$$
Then let $h$ be the solution to the Dirichlet problem with boundary values $h_0$. Since $h_0$ is not constant, $h$ is not constant, and by the maximum principle, $h(z) < 1$ for $\lvert z\rvert < r$. But then $\varphi(r/2) = 1 > h(r/2)$.
