Let \(f\) be holomorphic in a neighborhood of the closed disk \(\overline{D}_r(z_0)\). Prove Cauchy’s estimate for derivatives: \[|f^{(n)}(z_0)| \le \frac{n!}{r^n} \max_{|z-z_0| = r} |f(z)|\] and the Mean Value Property: \[f(z_0) = \frac{1}{2\pi} \int_0^{2\pi} f(z_0 + r e^{i\theta}) \, d\theta.\]
Solution
Cauchy’s integral formula for derivatives gives \[ f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_{|z-z_0|=r} \frac{f(z)}{(z-z_0)^{n+1}}\,dz. \] If \[ M_r=\max_{|z-z_0|=r}|f(z)|, \] then the \(ML\) estimate yields \[ |f^{(n)}(z_0)| \le \frac{n!}{2\pi} \frac{M_r}{r^{n+1}}(2\pi r) = \frac{n!}{r^n}M_r. \]
For \(n=0\), parameterize the circle by \[ z=z_0+re^{i\theta},\qquad dz=ire^{i\theta}\,d\theta. \] Cauchy’s formula becomes \[ f(z_0) = \frac1{2\pi i} \int_0^{2\pi} \frac{f(z_0+re^{i\theta})}{re^{i\theta}} ire^{i\theta}\,d\theta = \frac1{2\pi} \int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta. \]