According to (0x6698f5c7) , we have
\begin{equation*} \biggl(\frac{\lvert S\rvert}{12}\biggr)^{2 \frac{\log M}{\log 2}}\sup_{x\in I} \lvert f(x)\rvert \le \sup_{x\in S} \lvert f(x)\rvert, \end{equation*}since \(M>1\) (why?) the exponent on the left hand side is positive and we may apply (0x6698f127) . This yields
\begin{equation*} \sup_{x\in I}|f(x)|\le 12^{2\frac{\log M}{\log 2}}\biggl(\frac{2}{|S|}\biggr)^{2 \frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S)} \le \biggl(\frac{24}{|S|}\biggr)^{2 \frac{\log M}{\log 2} + \frac{1}{p}} \lVert f\rVert_{L^p(S)}. \end{equation*}Using Hölder’s inequality we estimate the left-hand side from below with
\begin{equation*} \lVert f\rVert_{L^p(I)}\le \sup_{x\in I}|f(x)|. \end{equation*}Note, since \(|I|=1\) no constant appears.