Let \(f\) be holomorphic in a neighborhood of \(\overline{D_r(z_0)}\). Show that for any \(s < r\), there exists a constant \(c > 0\) such that \[ \|f\|_{(\infty, s)} \leq c \|f\|_{(1, r)} ,\] where \(\displaystyle \|f\|_{(\infty, s)} = \sup_{z \in D_s(z_0)}|f(z)|\) and \(\displaystyle \|f\|_{(1, r)} = \iint_{D_r(z_0)} |f(z)|\,dx\,dy\).
Solution
Let \[ \delta=r-s>0. \] Fix \(z\in D_s(z_0)\). Since \(|z-z_0|<s\), the closed disk \(\overline{D_\delta(z)}\) is contained in \(D_r(z_0)\).
1. For every \(0<\rho<\delta\), the mean-value formula gives\[ f(z)=\frac1{2\pi}\int_0^{2\pi}f(z+\rho e^{i\theta})\,d\theta. \]Multiply by \(2\rho\) and integrate from \(0\) to \(\delta\). Using polar coordinates,\[ \delta^2 f(z) =\frac1\pi\iint_{D_\delta(z)}f(w)\,dA(w). \]Thus\[ f(z)=\frac1{\pi\delta^2}\iint_{D_\delta(z)}f(w)\,dA(w). \]
2. Taking absolute values,\[ |f(z)| \le\frac1{\pi\delta^2}\iint_{D_\delta(z)}|f(w)|\,dA(w) \le\frac1{\pi(r-s)^2}\|f\|_{(1,r)}. \]
3. Taking the supremum over \(z\in D_s(z_0)\) yields\[ \|f\|_{(\infty,s)} \le\frac1{\pi(r-s)^2}\|f\|_{(1,r)}. \]Hence one may take\[ c=\frac1{\pi(r-s)^2}. \]