Let \(R>0\). Suppose \(f\) is holomorphic on \(\ts{z\st \abs{z} < 3R}\). Let \[ M_{R}:=\sup _{|z| \leq R}|f(z)|, \quad N_{R}:=\sup _{|z| \leq R}\left|f^{\prime}(z)\right| \]
Estimate \(M_{R}\) in terms of \(N_{R}\) from above.
Estimate \(N_{R}\) in terms of \(M_{2 R}\) from above.
Solution
First note that by the maximum modulus principal, it suffices to consider sups on the boundary, i.e. \[ M_R = \sup_{\abs{z} = R}\abs{f(z)}, \qquad N_R = \sup_{\abs{z} = R} \abs{f'(z)} .\]
The first estimate: stuck!
The second estimate: suppose \(z_0 \in \DD_R(0)\), then any \(D_R(z_0)\) is contained in \(D_{2R}(0)\), So for any such \(z_0\), apply Cauchy’s integral formula centered at \(z_0\): \[ f^{(1)}(z_0) &= {1\over 2\pi i }\oint_{\bd\DD_{R}(z_0)} {f(\xi)\over (\xi-z_0)^2 }\dxi \\ \implies \abs{ f^{(1)}(z_0)} &\leq {1\over 2\pi} \oint_{\bd\DD_{R}(z_0)} \abs{f(\xi)\over (\xi-z_0)^2 }\dxi \\ &= {1\over 2\pi} \oint_{\bd\DD_R(z_0)} { \abs{f(\xi)} \over \abs{\xi-z_0}^2 } \dxi \\ &= {1\over 2\pi} \oint_{\bd\DD_R(z_0)} { \abs{f(\xi)} \over R^2 } \dxi \\ &\leq {1\over 2\pi} R^{-2} \oint_{\bd\DD_R(z_0)} { \sup_{z\in \bd\DD_{R}(z_0) } \abs{f(z)} } \dxi \\ &= {1\over 2\pi} R^{-2} \sup_{\bd\DD_R(z_0) } \abs{f(z)} \cdot 2\pi R \\ &= \inverseof{R} \sup_{\bd\DD_R(z_0) } \abs{f(z)} \\ &\leq \inverseof{R} M_{2R} ,\] where we’ve used in the last step that \(\DD_R(z_0) \subseteq \DD_{2R}(0)\), and sups can only get larger when taken over larger sets. Since this was an arbitrary \(z_0\in \DD_R(0)\), this holds for all \(z\) with \(\abs{z} \leq R\). Since taking sups preserves inequalities, we have \[ \abs{f'(z_0)} \leq \inverseof{R} M_{2R}\, \forall \abs{z} \leq R \implies N_R\da \sup_{\abs{z} \leq R}\abs{f'(z)} \leq \inverseof{R} M_{2R} .\]