Harmonic functions and Maximum Modulus Principle. (Real harmonic function implies holomorphic function?)

Question

Theorem. (Maximum Modulus Principle) $\;$If $f$ is a non-constant holomorphic function in a region $\Omega$, then $f$ cannot attains a maximum in $\Omega$.

Problem. Prove the maximum principle for harmonic functions, that is,

(a) If $u$ is a non-constant real-valued harmonic function in a region $\Omega$, then $u$ cannot attain a maximum in $\Omega$.

(b) Suppose that $\Omega$ is a region with compact closure. If $u$ is harmonic in $\Omega$ and continuous in $\overline{\Omega}$, then $$\sup_{z \in \Omega}|u(z)| \leq \sup_{\partial \Omega}|u(z)|$$

My attempt.

(a) Define $g(z) = u_{x}(z) - iu_{y}(z)$. So, $$\frac{\partial}{\partial x}u_{x} = u_{xx}$$ and $$\frac{\partial}{\partial y}u_{y} = -u_{yy}.$$ Thus, $$\frac{\partial}{\partial x}u_{x} - \frac{\partial}{\partial y}u_{y} = 0 \Longrightarrow \frac{\partial}{\partial x}u_{x} = \frac{\partial}{\partial y}u_{y}.$$ Analogously, $$\frac{\partial}{\partial y}u_{x} = -\frac{\partial}{\partial x}u_{y}.$$ Therefore, $g$ is holomorphic and so, there is $f$ such that $f' = g$ and, up to a constant, $u$ is a real part of $f$, i.e, $u$ is holomorphic. Since $u$ is a non-constant holomorphic function in $\Omega$, $u$ cannot attain a maximum in $\Omega$.

(b) Since $u$ is continuous in $\overline{\Omega}$, attains a maximum in $\overline{\Omega}$. But, by the item (a), the maximum cannot attains in $\Omega$, then $$\sup_{z \in \Omega}|u(z)| \leq \sup_{\partial \Omega}|u(z)|.$$

The problem not seems simple to solve. I suspect I've done something wrong. Can someone help me?

Answer 1

I will suggest a different approach. Suppose $u$ attains its maximum at some point $z_0 \in \Omega$. Then $u(z_0)$ is also the maximum of $u$ in some disc $D(z_0,r)$ contained in $\Omega$. There is an analytic function $f$ whose real part is $u$. Now apply Maximum Modulus Principle to $e^{f}$. Since $|e^{f}|=e^{u}$ it follows that $|e^{f}|$ attains maximum at $z_0$. This implies that $e^{f}$ and hence $f$ is a constant and so is $u$. Let $M =u(z_0)$. Using what we just proves you can verify that $\{z\in \Omega : u(z)=M\}$ is open and closed. Since $\Omega$ is connected it follows that $u$ has the constant value $M$ throughout $\Omega$.

Second part follows immediately. The continuous function $|u|$ attains its maximum at some point $z_1$ in $\overset {-} {\Omega}$ and this must be on the boundary. (It is the maximum of $u$ or $-u$). Hence $u(z)\leq \sup_{\partial \Omega}|u|$ for all $z \in \Omega$. We can also apply this to $-u$ so we get $|u(z)|\leq \sup_{\partial \Omega}|u|$ for all $z$.