I came across the following problem in our last midterm exam. I am completely stuck as to how to begin the solution:
If $|f(z)|\leq$ max $|f(z+re^{it})|$ ($0\leq t\leq 2\pi$), then $|f|$ has no strict local maximum within its domain of analyticity.
Currently I am familiar with Cauchy's Theorem and formula. I am not sure why even the first inequality is true? Can anyone lead to some hints/solutions?
I am going to give you a hint based on the assumption that you are trying to prove the following statement (without the use of the maximum modulus principle):
If $f$ is a (complex) analytic function on a domain $D$, then $|f|$ does not have a strict local maximum.
Pick any $z$ in the domain, and choose any $r$ so that the closure of the disk of radius $r$ centered at $z$ is still inside the domain $D$.
Then, the Cauchy integral formula tells you:
$$f(z) = \dfrac{1}{2\pi i}\int_{|w - z| = r} \dfrac{f(w)}{w-z} dw$$
Now, what happens to the integral when you substitute $w = z + r e^{it}$? Can you obtain the inequality after making that transformation?
Assuming that you were able to complete the computation, what this shows is that the inequality
$$|f(z)| \le \max |f(z+r e^{it})|$$ for all , $t\in [0,2\pi]$ and $z$, $r$ that makes the expressions above defined.
Now, suppose that $|f|$ does have a strict local maximum at some point $c$. What happens if you apply the above inequality to $c$?