Let \((\Omega, \mathcal{A}, \mu)\) be a measure space and \(f \in L^p(\Omega, \mu)\) with \(p \ge 1\). If \(\mu(\Omega)< \infty\) a point \(x \in \Omega\) exists with
\[ |f(x)|\ge \frac{\lVert f\rVert_{L^p(\Omega)}}{\mu(\Omega)^{1/p}}. \]
Proof
Assume
\begin{equation*} |f(x)|< \frac{\lVert f\rVert_{L^p(\Omega)}}{\mu(\Omega)^{1/p}}. \end{equation*}for all \(x\in \Omega\). Then
\begin{equation*} \lVert f\rVert_{L^p(\Omega)}^p =\int_{\Omega} |f(x)|^p \,d\mu(x) < \int_{\Omega} \frac{\lVert f\rVert_{L^p(\Omega)}^p}{\mu(\Omega)} \, d \mu(x) =\lVert f\rVert_{L^p(\Omega)}^p, \end{equation*}which is a contradiction.