Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31
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Proportional functions realise the equality case of Holder

Example

Fix 1<p<∞, let q be its conjugate exponent, and work on [0,1] with Lebesgue measure. For positive constants a,b, define

f:=aχ[0,1],g:=bχ[0,1].

Then

∫01∣fg∣ dλ=∥f∥p∥g∥q.

Facts & Assumptions

Given: Constants a,b>0, an exponent 1<p<∞, and its conjugate q.

[L1]

Equality in Holder holds when ∣f∣p and ∣g∣q are proportional almost everywhere (Equality in Holder's inequality for 1<p<∞).

Verification

technique · Choose nonnegative functions with $|f|^p$ and $|g|^q$ proportional almost everywhere and invoke the equality theorem
1.1given

The functions satisfy [given] ∣f∣p=apχ[0,1],∣g∣q=bqχ[0,1], so ∣f∣p=(ap/bq)∣g∣q almost everywhere.

2.1L1step 1.1∎

Applying [L1] gives equality in Holder: [L1, step 1.1] ∫01∣fg∣ dλ=∥f∥p∥g∥q.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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