Alternate proof to maximum modulus principle

Question

Let $D$ be an open disc with boundary $C$, $w \in D$.

I'm trying to show that there exists a constant $M$ such that $$ |f(w)| \leq M\sup_{z \in C}|f(z)| $$ for any $f$ continuous and analytic. I'm also asked to prove that $M=1$ by looking at $f(z)^n$.

What I got is:

Let $R$ be the radius of $C$.

$$ f(w) = |\frac{1}{2\pi i}\int_{C} \frac{f(z)}{z-w}dz| \leq \frac{1}{2\pi} \max_{z \in C}\frac{|f(z)|}{|z-w|}2\pi R $$

Then I don't know what to do with $\max_{z \in C}\frac{|f(z)|}{|z-w|}$ because both the denominator and the numerator depend on $z$.

Answer 1

This is the "power" trick.

Let $A=max_{z \in C}|f(z)|$ which exists by compactness and continuity. By Cauchy we can assume $A > 0$ as otherwise $f=0$ so nothing to prove. Fix $w \in D$ and let $B(w)>0$ the distance from $w$ to $C$ which again exists and is positive by compactness, hence $|w-z| \ge B(w), z \in C$ - so far analyticity is used "only" to show that $A=0$ implies $f=0$ but that actually is deceptive since it is a key part of the proof.

Applying Cauchy and the triangle inequality for integrals, we get:

$|f(w)| \le \frac{AR}{B(w)} \le M(w) max_{z \in C}|f(z)|, M(w) >0$.

But now the power trick means that we apply the same tactic to $f^n$ which has the same properties as $f$ (in this case analytic, continuos). Since $R, B(w)$ (hence $M(w)$) do not change as they depend only on the geometry of the problem, while obviously $A$ changes to $A^n$ we get:

$|f(w)|^n \le \frac{A^nR}{B(w)}=M(w)A^n$

Taking $n$th root we get:

$|f(w)| \le M(w)^{\frac{1}{n}}A$ and now letting $n \to \infty$ finishes the proof as $M(w)^{\frac{1}{n}} \to 1$

This power trick (sometimes called tensoring trick) applies in many situations and cool proofs of results like Kronecker approximation theorem can be obtained with it.

Edit later - just for clarity, I want to note that in this proof we first show that there is a required $M(w)$ for a fixed $w$ which apriori could be unbounded in $w$ and then we show that $M(w)=1$ for all $w$ so we get both results (the existence of $M$ and the fact that is $1$ simultaneously)

Answer 2

We know that powers of analytic functions are analytic and hence that $\displaystyle f(w)^n=\frac{1}{2\pi i}\int_C \frac{f(z)^n}{z-w} \mathrm{d}z$.

In other words, $\displaystyle \frac{1}{2\pi i}\int_C \frac{1}{z-w}\frac{f(z)^n}{f(w)^n} \mathrm{d}z=1$.

Suppose that the maximum modulus principle is false and $\displaystyle\frac{|f(z)|}{|f(w)|}<1$ for all $z\in C$. Then $\displaystyle \lim_{n\rightarrow \infty}\frac{f(z)^n}{f(w)^n}=0$ for all $z \in C$. This contradicts the fact that $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{2\pi i}\int_C \frac{1}{z-w}\frac{f(z)^n}{f(w)^n} \mathrm{d}z=1$