$|f(z)|\leq M$ on $C$, show that $|f(z)|\leq M$ for all points within $C$.

Question

A function $f(z)$ is analytic within a closed contour $C$ (and continuous on $C$). If $f(z)\neq 0$ within $C$ and $|f(z)|\leq M$ on $C$, show that $|f(z)|\leq M$ for all points within $C$.

Hint: Consider $w(z)=\frac{1}{f(z)}$

I am unable to make any progress in this question, it lies under the section of Cauchy Integral Formula. Any help is appreciated. Thanks.

Answer 1

Note I suspect you stated the problem wrong, because what you say you want has nothing to do with $w=1/f$, it's just the Maximum Modulus Principle.

My conjecture is the problem actually says that if $|f|\ge M$ on $C$ then $|f|\ge M$ in the interior. Assuming that's what we want:

This is just the Maximum Modulus Principle applied to $w=1/f$.

Of course MMP gets stated in various ways; some versions contain what we need to prove as a special case, other versions require a little argument. There's a trick for getting MMP from the Cauchy Integral Formula:

For convenience let's say $A(C)$ is the class of functions analytic inside $C$ and continuous up to $C$, and assume that $C$ has winding number $1$ about every point in the interior. Say the interior is $V$, and define $$||w|||_C=\sup_{\xi\in C}|w(\xi)|.$$

CIF If $w\in A(C)$ and $z\in V$ then $w(z)=\frac1{2\pi i}\int_C\frac{w(\xi)\,d\xi}{\xi-z}.$

Cor. (MMP with a constant). If $z\in V$ there exists $c_z$ such that $|w(z)|\le c_z||w||_C$ for every $w\in A(C)$.

And now the trick:

Cor. (MMP). You can take $c_z=1$ in the previous corollary.

Proof: If $w\in A(C)$ and $n$ is a positive integer then $w^n\in A(C)$, so $$|w(z)|^n\le c_z||w^n||_C=c_z||w||_C^n.$$Take $n$-th roots and let $n\to\infty$.

Answer 2

If $f$ is analytic and has no zero inside the contour $C$, then $w=\frac{1}{f}$ also has no zero and analytic inside $C$.

Now, $\zeta=|f(z)|$ has an extreme point at $z$ if $f'(z)f(z)=0$. Let, $D=|f'|^4-|f|^2|\partial_{xx}f|^2$. For, $D>=<0$, $\zeta$ has minima, saddle point, maxima respectively. (This properties hold for any analytic $f(z)$).

Now, $w(z)$ has no zero, hence at it's extreme points, $w'(z)=0$ must be zero. So, $D \leq 0$ at extreme points, implying that there are no minima.....$(1)$

Given $|w(z)| \geq M$ on $C$. Let, there exists a point $z_0$ inside the contour for which $|w(z_0)| <M$ then obviously there must be a small contour $C'$ which cuts $|w(z)|=M$ plane and $|w(z)|=M$ on that contour. It guarantees that there is a minimum point inside $C'$ which is a contradiction to $(1)$. So, $|w(z)| \geq M$. So, $|w(z)|$ can never be $<M$ inside $C$.

[This extreme point and existence of minima maxima, can also be used to prove that inside a level curve, there exists a zero of an analytic function. Also, this proves the fundamental theorem of Algebra.]