# Prove $f$ attains its maximum on boundary

Let $$f$$ be holomorphic, bounded on $$|z|>1$$ and continuous on $$|z|\geq 1$$. Prove that $$|f|$$ attains maximum on $$|z|=1$$, i.e $$\sup\limits_{|z|\geq 1}|f(z)|=\max\limits_{|z|=1}|f(z)|.$$

My attempt:
Consider $$g(z)=f\left(\dfrac{1}{z}\right)$$. From hypothesis, we know that $$g$$ is holomorphic, bounded on $$0<|z|<1$$ and continuous on $$0<|z|\leq 1$$. By Maximum modulus principle, $$g$$ is constant or attains its maximum on $$|z|=1$$. I can't go further. Could someone help me or have other ways to deal with problem? Thanks in advance!

Great start! You're on the right track by considering the function $$g(z) = f\left(\frac{1}{z}\right)$$. Let's continue from where you left off and complete the proof.
As you mentioned, define $$g(z) = f\left(\frac{1}{z}\right)$$. From the given conditions, $$g$$ is holomorphic and bounded on $$0 < |z| < 1$$ and continuous on $$0 < |z| \leq 1$$. By the Maximum Modulus Principle, either $$g$$ is constant or it attains its maximum modulus on the boundary, i.e., on $$|z| = 1$$. If $$g$$ is constant, then $$f$$ is also constant (since $$f(z) = g\left(\frac{1}{z}\right)$$), and the result trivially holds. If $$g$$ is not constant, then $$\max_{|z|=1} |g(z)| > |g(z)|$$ for all $$|z| < 1$$. Now, let's relate this back to $$f$$. For any $$z$$ with $$|z| > 1$$, we have $$\left|\frac{1}{z}\right| < 1$$. Therefore, $$|f(z)| = \left|g\left(\frac{1}{z}\right)\right| < \max_{|z|=1} |g(z)| = \max_{|z|=1} \left|f\left(\frac{1}{z}\right)\right| = \max_{|z|=1} |f(z)|$$ This means that for all $$z$$ with $$|z| > 1$$, $$|f(z)|$$ is strictly less than $$\max_{|z|=1} |f(z)|$$. Since $$f$$ is continuous on $$|z| \geq 1$$, we can conclude that $$\sup_{|z|\geq 1} |f(z)| = \max_{|z|=1} |f(z)|$$ This completes the proof.
• You are missing an important step: holomorphic extension of $g$ to zero. Commented Mar 14 at 15:27
• @MoisheKohan Is it possible to show $g$ is holomorphic extension to zero, with the fact that $f$ is bounded on $|z|>1$?