Lp norms converge to the supremum norm (${\lim }_{p\to \infty }\Vert f{\Vert }_p=\Vert f{\Vert }_{\infty }$)

On a measure space of finite measure, a bounded measurable function satisfies the displayed limit, where the right side is the essential supremum of its absolute value. The upper estimate is $\Vert f{\Vert }_p\le \mu (X{)}^{1/p}\Vert f{\Vert }_{\infty }$. For every $a<\Vert f{\Vert }_{\infty }$, the set where $|f|>a$ has positive measure, giving the reverse limiting estimate. For a continuous function on an open Euclidean domain, the essential supremum equals the ordinary supremum. Finiteness of the measure is part of this statement.

For a bounded measurable function with positive essential supremum on a finite measure space, put $M_n=\int |f{|}^n$. The Cauchy-Schwarz inequality makes these moments log-convex, so successive ratios are nondecreasing. Their product satisfies $M_n/M_0\le (M_{n+1}/M_n{)}^n$, while each ratio is at most the essential supremum. Positive-measure level sets arbitrarily close to that supremum make $\mu (E{)}^{-1/n}\Vert f{\Vert }_n$ converge to it. Squeezing then proves the ratio limit.