Consider a good interval \(I\) of length \(a\), i.e.

\begin{equation*} \lVert f^{(k)}\rVert_{L^p(I)}\le (Cb)^k \lVert f\rVert_{L^p(I)}, \end{equation*}

for some numerical constant \(C>0\) and all \(k\in \mathbb{N}\). We want to show

\begin{equation*} |f^{(k)}(x)|\le \Bigl(\frac{2}{a}\Bigr)^{1/p}(CBb)^k \lVert f\rVert_{L^p(I)} \end{equation*}

for a suitable constant \(B>1\). If no good point exists, then for every \(x\in I\) we have

\begin{equation*} \frac{2}{a}\int_{I} |f|^p \le \sum_{k=0}^{\infty} \frac{1}{B^{kp}(Cb)^{kp}} |f^{(k)}(x)|^p. \end{equation*}

Integrating both sides over \(I\) leads to

\begin{align*} 2\int_{I} |f|^p &\le \sum_{k=0}^{\infty} \frac{1}{B^{kp}(Cb)^{kp}} \int_{I} |f^{(k)}(x)|^p \d x\\ &\le \sum_{k=0}^{\infty} \frac{1}{B^{kp}}\int_{I} |f|^p \\ &=\frac{B^p}{B^{p}-1}\int_{I} |f|^p, \end{align*}

where we used the good interval property and \(B>1\). Set \(B=3\) and we get

\begin{equation*} 2\int_{I} \lvert f\rvert^p\le \frac{3}{2}\int_{I} \lvert f\rvert^p. \end{equation*}

This contradiction proves our claim.