Let \(f\) be holomorphic in a neighborhood of the closed disk \(\overline{D}_r(z_0)\). Show that for any \(s < r\), there exists a constant \(c > 0\) (depending on \(r\) and \(s\)) such that: \[\|f\|_{(\infty, s)} \le c \|f\|_{(1, r)}\] where \(\|f\|_{(\infty, s)} = \sup_{z \in D_s(z_0)} |f(z)|\) and \(\|f\|_{(1, r)} = \iint_{D_r(z_0)} |f(z)| \, dx \, dy\).
Solution
Let \[ \delta=r-s>0. \] For \(w\in D_s(z_0)\) one has \[ D_\delta(w)\subset D_r(z_0). \]
1. The area mean-value formula for holomorphic functions gives\[ f(w)=\frac1{\pi\delta^2}\iint_{D_\delta(w)}f(z)\,dA(z). \]Hence\[ |f(w)| \le\frac1{\pi\delta^2}\iint_{D_\delta(w)}|f(z)|\,dA(z) \le\frac1{\pi(r-s)^2}\|f\|_{(1,r)}. \]
2. Taking the supremum over \(w\in D_s(z_0)\) gives\[ \|f\|_{(\infty,s)} \le\frac1{\pi(r-s)^2}\|f\|_{(1,r)}. \]Thus one may take\[ c=\frac1{\pi(r-s)^2}. \]