Problem 7

Exercise 5.7 (Maximum modulus principle and its sibling)   Let $\Omega\subseteq\mathbb{C}$ be a connected domain and let $f(z)$ be holomorphic on $\Omega$. Show that neither $\Re[f(z)]$ nor $\vert f(z)\vert$ attain a maximum on $\Omega$ unless $f$ is constant.

Proof. This is the maximum modulus principle, which the question is asking us to prove. Suppose $\vert f(z_0)\vert\geq \vert f(z)\vert$ over $\Omega$. Find a power series

$$f(z) = a_0 + a_1(z-z_0) + \cdots$$ (5.38)

in a region about $z_0$. If $f$ is constant, we are done, so suppose $a_1\neq 0$. In this case, $f$ is a locally an open mapping, so select $r>0$ such that $f(B_r(z_0))$ is open. Note that $a_0$ lies in this set, so select $\delta>0$ such that

$$B_\delta(a_0) \subseteq f(B_r(z_0))$$ (5.39)

Let $a_0=a+bi$. If $a>0$, find $a_0+\delta/2=f(z_0+w)$ where $\vert w\vert<r$. Then

$$\vert f(z_0+w)\vert = \vert(a+\delta/2) + bi\vert = \sqrt{(a+\delta/2)^2 + b^2} > \sqrt{a^2+b^2} = \vert a_0\vert = \vert f(z_0)\vert$$ (5.40)

violates that $\vert f(z_0)\vert$ is the maximum modulus. Similarly if $a<0$, subtract $\delta/2$ to find the same contradiction.

To prove the maximum real part principle, suppose $\Re[f(z_0)]\geq \Re[f(z)]$ and pass $f$ to the exponential function. We have:

$$e^{f(z)} = e^{\Re f(z) + \Im f(z)}$$ (5.41)

which implies

$$\vert e^{f(z)}\vert = e^{\Re f(z)}$$ (5.42)

The previous result shows that if this function has a maximum, then the function is constant, which indicates $\Re f(z)$ is constant by the monotonicity of the exponential function, completing the proof. $\qedsymbol$