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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Finite-dimensional Riesz representation: every functional is uniquely vv,w

Statement

Let V be a finite-dimensional real or complex inner product space, with the inner product linear in its first argument. For every linear functional f:VF, there is a unique wV such that

f(v)=v,wfor every vV.

The map w(vv,w) is a conjugate-linear bijection from V to its algebraic dual. This includes V=0.

Facts & Assumptions

Given: A finite-dimensional inner product space V and a linear functional f.

[L2]

An orthonormal basis (ei) gives v=iv,eiei (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

[L4]

The algebraic dual consists of all linear functionals from V to its scalar field (Linear functionals and the algebraic dual V=L(V,F)).

Proof

technique · direct
1.1

Choose an orthonormal basis (e0,,en1) by [L1] and define w=i<nf(ei)ei.

L1choose
2.1

For vV, [L2] and linearity of f give f(v)=iv,eif(ei). Conjugate-linearity in the second argument makes the right side equal to v,w.

step 1.1L2L4algebra
3.1

If w is another representative, then v,ww=0 for every v, so [L3] gives w=w.

step 2.1L3
4.1

The assignment w,w is conjugate-linear because the inner product is conjugate-linear in its second argument. Existence makes it surjective and uniqueness makes it injective. When V=0, the chosen basis and both sums are empty and the same argument applies.

step 1.1step 2.1step 3.1L4

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 54 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources