TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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$L^{p}$ norms converge to the essential supremum for essentially bounded $L^{r}$ functions

Statement

Let $0 < r < \infty$ , let $f \in L^{r} (μ) \cap L^{\infty} (μ)$ , and put $M : = ∥ f ∥_{\infty}$ . Then $f \in L^{p} (μ)$ for every finite $p \geq r$ and

$\lim_{p \to \infty} ∥ f ∥_{p} = M .$

Facts & Assumptions

Given: A real exponent $r > 0$ and a function $f \in L^{r} (μ) \cap L^{\infty} (μ)$ .

[L1]

If $M = ∥ f ∥_{\infty} < \infty$ , then $∣ f ∣ \leq M$ almost everywhere and $M$ is the least essential bound (The essential supremum is attained as the least essential bound).

[L2]

Membership in $L^{r} (μ)$ and $L^{\infty} (μ)$ is defined in The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L3]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

Proof

technique · The upper bound is $\|f\|_p^p\le\|f\|_\infty^{p-r}\|f\|_r^r$. For the lower bound, every $\varepsilon$ below the essential supremum leaves a set of positive measure where $|f|$ exceeds $\|f\|_\infty-\varepsilon$, forcing the $p$-norm above that level as $p$ grows

1.1

If $M = 0$ , then [L1] gives $∣ f ∣ = 0$ almost everywhere, so $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ = 0$ for every finite $p \geq r$ . Hence $∥ f ∥_{p} = 0 = M$ for all such $p$ , and the conclusion follows in this case.

L1L3given

1.2

Assume from now on that $M > 0$ . For every finite $p \geq r$ , one has $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ \leq M^{p - r} \int ∣ f ∣^{r} d μ = M^{p - r} ∥ f ∥_{r}^{r},$ because [L1] gives $∣ f ∣^{p} = ∣ f ∣^{r} ∣ f ∣^{p - r} \leq M^{p - r} ∣ f ∣^{r}$ almost everywhere. Thus $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq M^{1 - r / p} ∥ f ∥_{r}^{r / p} .$ In particular, ${lim sup}_{p \to \infty} ∥ f ∥_{p} \leq M .$

L1L2L3givenalgebra

2.1

Fix $ε$ with $0 < ε < M$ . Because $M$ is the least essential bound, the set $E_{ε} : = {∣ f ∣ > M - ε}$ has positive measure. For every finite $p \geq r$ , step 1.2 gives $∥ f ∥_{p} < \infty$ , so $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ \geq \int_{E_{ε}} ∣ f ∣^{p} d μ \geq (M - ε)^{p} μ (E_{ε})$ forces $μ (E_{ε}) < \infty$ . Therefore $∥ f ∥_{p} \geq (M - ε) μ (E_{ε})^{1 / p},$ and letting $p \to \infty$ gives ${lim inf}_{p \to \infty} ∥ f ∥_{p} \geq M - ε .$

L1L3step 1.2given

3.1

Because $0 < ε < M$ was arbitrary in the case $M > 0$ , step 2.1 yields ${lim inf}_{p \to \infty} ∥ f ∥_{p} \geq M$ . Combined with step 1.2, this proves $∥ f ∥_{p} \to M$ , while step 1.1 already handled the case $M = 0$ .

step 1.1step 1.2step 2.1∎

Depends on

Used by

A two-step function shows the Lᵖ norm converging to the essential supremum Example

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral, Theorem 8.1 (standard reference, not scraped)
John K. Hunter, Measure Theory, Section 17 (standard reference, not scraped)

LpLp norms converge to the essential supremum for essentially bounded LrLr functions