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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)
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Holder's inequality for integrals, including the endpoint cases

Statement

Let (X,A,μ) be a measure space, let p,q∈[1,∞] be conjugate exponents, and let f,g be measurable real-valued functions.

  1. If 1≤p<∞ and q<∞ with f∈Lp(μ) and g∈Lq(μ), then ∫∣fg∣ dμ≤∥f∥p∥g∥q.
  2. If p=1 and q=∞ with f∈L1(μ) and g∈L∞(μ), then ∫∣fg∣ dμ≤∥f∥1∥g∥∞.
  3. If p=∞ and q=1 with f∈L∞(μ) and g∈L1(μ), then ∫∣fg∣ dμ≤∥f∥∞∥g∥1.

In every case the right-hand side is finite, so fg is integrable.

Facts & Assumptions

Given: A measure space (X,A,μ), conjugate exponents p,q∈[1,∞], and measurable real-valued functions f,g in the spaces named in the relevant clause of the Statement.

[L1]
[L2]

For 0<r<∞, membership in Lr(μ) means ∫∣h∣r dμ<∞, while L∞(μ) means finite essential supremum (The function space Lp(μ) for 0<p<∞, The space L∞(μ) of essentially bounded measurable functions).

[L3]

A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere).

[L4]

If h∈L∞(μ), then ∣h∣≤∥h∥∞ almost everywhere (The essential supremum is attained as the least essential bound).

[L5]

Young's inequality says uv≤up/p+vq/q for u,v≥0 when 1<p,q<∞ are conjugate (Young's inequality for conjugate real exponents).

[L6]

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

[L7]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

[L8]

The integral of a nonnegative function over a null set is zero (A nonnegative integral over a null set vanishes).

Proof

technique · For $1 < p < infinity$, normalize the nonzero norms and apply the published Young inequality pointwise before integrating. Treat the endpoint pairs $(1,\infty)$ and $(\infty,1)$ separately from the essential-bound definition
1.1L2L3L8given

Assume first 1<p,q<∞, and put A:=∥f∥p and B:=∥g∥q. If A=0 or B=0, then the corresponding power integral is 0, so [L3] makes the corresponding function vanish outside a measurable null set. By [L8], ∫∣fg∣ dμ=0. Thus only the case A,B>0 remains.

1.2L2L4L6L7L8given

For the endpoint pair (p,q)=(1,∞), let M:=∥g∥∞. By [L4] there is a measurable null set N outside which ∣fg∣≤M∣f∣. On N, [L8] gives zero integral for ∣fg∣χN and M∣f∣χN. Integrating the inequality on X∖N with [L6] and [L7] therefore gives ∫∣fg∣ dμ≤M∫∣f∣ dμ=∥g∥∞∥f∥1.

2.1step 1.1L1L2L5L6L7algebra

In the remaining strict-exponent case, Young's inequality applied pointwise to u=∣f∣/A and v=∣g∣/B gives [step 1.1, L1, L2, L5, L6, L7, algebra] ∣f∣∣g∣AB≤∣f∣ppAp+∣g∣qqBq. Integrating and using additivity, monotonicity, homogeneity, and the definitions of A and B yields ∫∣fg∣ dμ≤BpAp−1∫∣f∣p dμ+AqBq−1∫∣g∣q dμ=ABp+ABq=AB.

2.2step 1.2given

The case (p,q)=(∞,1) is identical after exchanging f and g. [step 1.2, given] ∫∣fg∣ dμ≤∥f∥∞∥g∥1.

3.1step 2.1step 1.2step 2.2L2∎

Step 2.1 proves the strict-exponent case, and steps 1.2 and 2.2 prove the two endpoint cases. In every case the right-hand side is finite by [L2], so fg is integrable.

Depends on

Used by

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