Hi again, this is my solution to part b. (Oops, I didn't realize P4 had 3 parts, so I did this problem before proving the maximum principle for subharmonic functions).

Let's prove that $u \geq v$. $u$ is harmonic and $v$ is subharmonic, so

\begin{equation}

\Delta{u} = 0 \\

\Delta{v} \geq 0

\end{equation}

Define $g=v-u$. Then:

\begin{equation}

\Delta g \geq 0\\

g|_\Sigma = 0

\end{equation}

We only need to prove that $g\leq0$ to obtain our desired result.

Let's assume, on the other hand that $g>0$. Then since $g$ is zero on the boundary, $g$ must have a maximum that is on the interior of $\Omega$.

Let $y$ denote the point on $\Omega$ where $g$ is maximum. Then we draw a ball around it $B(y,r)$ such that it remains inside the domain $\Omega$.

Now we assume the result of 2b, that $g(y)$ does not exceed the average of $g$ on this ball. But $g(y)$ is the global maximum, and also the maximum on this ball. The only way these can hold simultaneously is if $g = g(y)$ on this entire ball. If on $B(r,y)$, $g\neq g(y)$, then $g \leq g(y)$, and $g(y)$ would certainly exceed the average of $g$ on $B(r,y)$. Thus we conclude $g = g(y)$ on $B(r,y) \in \Omega$.

Now within this ball we take a new point $y'$. Then $g(y')$ is also the global maximum on $\Omega$. We repeat this over and over again and discover that on a connected $\Omega$, $g$ is identically equal to $g(y)$ and so $g>0$ on the boundary. This contradicts $g|_\Sigma = 0$. Therefore our original assumption cannot be correct, and so $g\leq0$. This proves that $u\geq v$.

Doing the exact same thing with $f=u-v$ proves that $v\geq u$