Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31
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The parallelogram law in L2

Statement

For u,v∈L2(μ) one has

∥u+v∥22+∥u−v∥22=2∥u∥22+2∥v∥22.

Facts & Assumptions

Given: Classes u,v∈L2(μ) with measurable representatives f,g.

[L2]

Products of two L2 functions lie in L1 (Generalized Holder inequality puts products into Lr).

[L3]

The Lebesgue integral is linear on L1 (The Lebesgue integral is linear on L1(μ)).

Proof

technique · Expand $|f + g|^2 + |f - g|^2$ pointwise to $2|f|^2 + 2|g|^2$ and integrate
1.1L2L3givenalgebra

Because f,g∈L2(μ), step [L2] puts fg in L1(μ), so every term [L2, L3, given, algebra] in the algebraic expansions below is integrable. Pointwise, ∣f+g∣2+∣f−g∣2=(f+g)2+(f−g)2=2f2+2g2. Integrating and using [L3] gives ∫∣f+g∣2 dμ+∫∣f−g∣2 dμ=2∫∣f∣2 dμ+2∫∣g∣2 dμ.

2.1step 1.1L1∎

Rewriting the four integrals as ∥u+v∥22, ∥u−v∥22, [step 1.1, L1] ∥u∥22, and ∥v∥22 is legitimate by [L1]. That yields the parallelogram identity.

Depends on

Used by

Dependency tree · two levels

17 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