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.
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.