Suppose \(z_0\in \mathbb{R}\), \(f(z)\) is holomorphic on \(D(z_0,5r)\), \(I\subseteq \mathbb{R}\) is a bounded, \(z_0\in I\), \(S\subseteq I\) is a measurable subset with nonzero measure and \(p\ge 1\). If \(\lvert f(z_0)\rvert\ge 1\) and \(M=\max_{|z-z_0|\le 4r} |f(z)|\), then
\begin{equation*} \lVert f\rVert_{L^p(I)} \le \biggl(\frac{24\lvert I\rvert}{|S|}\biggr)^{2 \frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S)}. \end{equation*}We use (0x6698f0dc) . Set \(r=\lvert I\rvert\) and \(\widetilde{f}(z)=f(rz)\). Then \(\widetilde{f}\) is an analytical function on the disk \(D(\frac{z_0}{r}, 5)\) with \(\widetilde{f}(\frac{z_0}{r})\ge 1\). By setting \(\widetilde{I}=\frac{I}{r}:=\{\frac{x}{r}\mid x\in I\}\) and \(\widetilde{S}=\frac{S}{r}\), we allowed to use (0x6698f0dc) and we obtain
\begin{equation*} \lVert \widetilde{f}\rVert_{L^p(\widetilde{I})} \le \biggl(\frac{24}{|\widetilde{S}|}\biggr)^{2 \frac{\log \widetilde{M}}{\log 2} + \frac{1}{p}} \lVert \widetilde{f}\rVert_{L^p(\widetilde{S})}, \end{equation*}with \(\widetilde{M}=\max_{|z-\frac{z_0}{r}|\le 4} |\widetilde{f}(z)|\). Using \(M=\widetilde{M}\), \(|\widetilde{S}|=\frac{|S|}{r}\) and \(\lVert \widetilde{f}\rVert_{L^p(\widetilde{I})}=r^{-\frac{1}{p}}\lVert f\rVert_{L^p(I)}\) (same is true for \(\lVert \widetilde{f}\rVert_{L^p(\widetilde{S})}\)) the stated inequality follows.