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Complex Lp duality from real Lp duality

Statement

Assume the Axiom of Countable Choice ACω. Let (X,A,μ) be any measure space, let 1<p<, and let q be conjugate to p. Every bounded complex-linear functional Λ:Lp(μ;C)C has a unique hLq(μ;C) such that

Λ([f])=Xfhdμ([f]Lp(μ;C)),

where the pairing is bilinear, with no conjugation. Moreover Λ=hq.

Facts & Assumptions

Given: ACω, an arbitrary measure space, conjugate exponents 1<p,q<, and a bounded complex-linear Λ:Lp(μ;C)C.

[F1]

Under Countable Choice, every bounded real-linear functional on real Lp over an arbitrary measure space is uniquely integration against a real Lq density, with equality of norms (For 1<p<, the same representation theorem holds on arbitrary measure spaces).

[F2]

Complex Lp is the a.e. quotient of measurable finite-valued complex functions with finite p-norm; real and imaginary parts, conjugation, products, and positive powers have the stated measurability conventions, and bilinear tests use fs without conjugation (Complex Lp classes and Euclidean test-function conventions).

[F3]

Complex Hölder makes the bilinear pairing bounded, and complex Lp has the quotient norm with fp=fp (Complex Holder, Minkowski, and the quotient norm).

[F4]

Countable Choice selects from every countable family of nonempty sets (The Axiom of Countable Choice (ACω)).

Proof

technique · represent the real and imaginary parts on the real-valued subspace, then use a normalized phase test for the norm and uniqueness
1.1

Regard real Lp(μ) as the real-valued subspace of complex Lp(μ;C). The maps A(u)=ReΛ(u) and B(u)=ImΛ(u) are bounded real-linear functionals there, with A(u),B(u)Λup. Applying [F1] twice gives real a,bLq(μ) such that A(u)=ua and B(u)=ub for every real uLp. Put h=a+ibLq(μ;C); component inequalities in [F3] make its q-norm finite.

F1F2F3given
2.1

For real-valued u, componentwise complex integration gives Λ(u)=A(u)+iB(u)=u(a+ib)=uh. If f=u+iv is an arbitrary complex Lp class, [F3] puts its real and imaginary parts in real Lp, and complex linearity gives Λ(f)=Λ(u)+iΛ(v)=uh+ivh=fh. All identities depend only on a.e. classes by the quotient and integration conventions in [F2]–[F3].

F2F3step 1.1algebra
3.1

Hölder [F3] gives fhfphq, hence Λhq. If h=0 a.e., step 2.1 gives Λ=0 and equality follows. Otherwise define v=0 on {h=0} and v=hq2h where h0. Then v=hq1 and vh=hq pointwise. Since (q1)p=q, [F2]–[F3] give vLp, vp=hqq1, and Λ(v)=hq=hqq. Testing on v/vp proves Λhq, including the closed unit-norm endpoint.

F2F3step 2.1algebra
4.1

If kLq(μ;C) gives the same pairing functional, put d=hk. Then fd=0 for every fLp. If d were nonzero, the phase test of step 3.1 with d in place of h would produce vLp with vd=dqq>0, a contradiction. Thus d=0 in Lq, so the density is unique.

F2F3step 3.1discharge-contradiction
5.1

Steps 2.1–4.1 prove existence, equality of norms, and uniqueness. Countable Choice is used only inside the real arbitrary-measure representation [F1], whose construction makes countably many local choices; applying that theorem to A and B requires only two instances and no stronger choice principle. The empty and zero-measure spaces have only zero Lp classes and are covered by the h=0 branch of step 3.1; the forbidden endpoints p=1, never enter because 1<p,q<.

F1F4step 1.1step 2.1step 3.1step 4.1

Remarks

The absence of a conjugate in the displayed pairing is deliberate. It is why the phase test contains h: multiplication then gives the nonnegative real function hq.

Depends on

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Sources