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Completeness of the complex Haar L1 and L2 spaces and density of Cc

Statement

Assume AC. Let μ be a Radon measure on an LCH space X. Then the complex spaces L1(X,μ;C) and L2(X,μ;C) (Complex Haar L^p spaces and compactly supported functions) are complete, and Cc(X;C) is dense in both of them.

Facts & Assumptions

Given: An LCH space X with a Radon measure μ, the complex spaces Lp(X,μ;C) for p∈{1,2}, the real spaces Lp(μ), and AC.

[F1]

For p∈{1,2} the complex space Lp(X,μ;C) consists of the almost-everywhere equivalence classes of measurable complex functions f with ∥f∥p=(∫X∣f∣p dμ)1/p<∞, on classes it is a complex normed space, and Cc(X;C) denotes the continuous complex-valued functions of compact support (Complex Haar L^p spaces and compactly supported functions).

[F2]

Assume countable choice (in particular, the Axiom of Choice suffices). Let (X,A,μ) be a measure space and let 1≤p≤∞. Then Lp(μ) is complete (Riesz-Fischer completeness of Lp for 1≤p≤∞).

[F3]

Assume Dependent Choice. If μ is a Radon measure on an LCH space X and 1≤p<∞, then Cc(X) is dense in Lp(μ) (C_c(X) is dense in L^p(mu) for a Radon measure).

[F4]

If 0≤f≤g are measurable then ∫f dμ≤∫g dμ (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F5]

Let (N,0,σ) be a Peano system, in particular N. For any set A, any a∈A and any f:A→A there is a unique g:N→A with g(0)=a and g(σ(n))=f(g(n)) for all n (The recursion theorem).

[F6]

DC is the statement that for every nonempty set X, every relation R entire on X and every a∈X there is x:N→X with x0=a and xnRxn+1 for every n (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain).

[F7]

ACω is the statement that for every family (Xn)n∈N of nonempty sets there is f with domain N and f(n)∈Xn for every n (The Axiom of Countable Choice (ACω)).

[F8]

Cc(X;F) denotes the continuous F-valued functions with compact support, for F=R or C, and Cc(X) denotes the real space Cc(X;R) (Compact support, Cc(X), and C0(X)).

[A1]

AC is assumed, in the choice-function form that every family of nonempty sets has a choice function (The Axiom of Choice).

Proof

technique · direct
1.1

Discharge the Dependent Choice hypothesis of [F3] from AC. Let R be entire on a nonempty set X and let a∈X. The family {R[x]:x∈X} of nonempty sets has, by [A1], a choice function c with c(R[x])∈R[x] for every x∈X; put s(x):=c(R[x]), so that xRs(x) for every x∈X. Applying [F5] to the set X, the point a and the function s gives g:N→X with g(0)=a and g(n+1)=s(g(n)); then g(n)Rg(n+1) for every n. This is exactly the conclusion required by [F6], so DC holds.

A1F3F5F6
1.2

Discharge the countable choice hypothesis of [F2] from AC. Let (Xn)n∈N be a family of nonempty sets. Its range {Xn:n∈N} is a family of nonempty sets, so by [A1] there is a choice function c on that family with c(Xn)∈Xn for every n; the function f:N→⋃nXn, f(n):=c(Xn), satisfies f(n)∈Xn for every n, which is the conclusion required by [F7]. So ACω holds.

A1F2F7
1.3

For every z∈C one has ∣Re⁡z∣≤∣z∣, ∣Im⁡z∣≤∣z∣ and ∣z∣≤∣Re⁡z∣+∣Im⁡z∣; consequently for all z,w∈C also ∣Re⁡z−Re⁡w∣=∣Re⁡(z−w)∣≤∣z−w∣ and ∣Im⁡z−Im⁡w∣≤∣z−w∣, and for real a,b one has (∣a∣+∣b∣)2≤2∣a∣2+2∣b∣2.

algebra
2.1

Let (fn) be a Cauchy sequence in L1(X,μ;C). Its real and imaginary parts are Cauchy in the real space L1(μ): by the first two inequalities of step 1.3 and monotonicity [F4], ∫∣Re⁡fn−Re⁡fm∣ dμ≤∫∣fn−fm∣ dμ and likewise for the imaginary parts, the two left-hand sides being the norms of the real classes Re⁡fn−Re⁡fm and Im⁡fn−Im⁡fm. The same argument with ∣⋅∣2 in place of ∣⋅∣ and the last inequality of step 1.3 shows that a Cauchy sequence in L2(X,μ;C) has real and imaginary parts that are Cauchy in the real space L2(μ).

F1F4step 1.3
2.2

Complex density. Let f∈Lp(X,μ;C) with p∈{1,2} and let ϵ>0. The real and imaginary parts of f are real classes in Lp(μ) by [F1], and p is finite, so by [F3] under the Dependent Choice of step 1.1 there are real u,v∈Cc(X) with ∥Re⁡f−u∥p<ϵ/2 and ∥Im⁡f−v∥p<ϵ/2. Then u+iv∈Cc(X;C) by [F8], and for p=1 steps 1.3 and [F4] give ∥f−(u+iv)∥1≤∥Re⁡f−u∥1+∥Im⁡f−v∥1<ϵ, while for p=2 the last inequality of step 1.3 and [F4] give ∥f−(u+iv)∥22=∫∣(Re⁡f−u)+i(Im⁡f−v)∣2 dμ≤2∥Re⁡f−u∥22+2∥Im⁡f−v∥22<ϵ2. Thus Cc(X;C) is dense in Lp(X,μ;C) for both p.

F1F3F4F8step 1.1step 1.3
3.1

Let (fn) be a Cauchy sequence in L1(X,μ;C), or in L2(X,μ;C); fix p∈{1,2} accordingly. By step 2.1 the real sequences (Re⁡fn) and (Im⁡fn) are Cauchy in the real space Lp(μ), so by [F2] under the countable choice of step 1.2 there are g,h∈Lp(μ) with ∥Re⁡fn−g∥p→0 and ∥Im⁡fn−h∥p→0.

F2step 1.2step 2.1
4.1

Recombination, for p∈{1,2} as in step 3.1. Define f:=g+ih, the class of the measurable complex function g+ih, which lies in Lp(X,μ;C) because ∣f∣≤∣g∣+∣h∣ and ∣g∣+∣h∣∈Lp(μ) for the finite p by steps 1.3 and [F4] in the case p=1 and by the last inequality of step 1.3 and [F4] in the case p=2. Then ∥fn−f∥p→0 by the same two computations applied to fn−f=(Re⁡fn−g)+i(Im⁡fn−h) together with step 3.1, so (fn) converges in Lp(X,μ;C). Hence both complex spaces are complete.

F1F4step 1.3step 3.1
5.1

Combining the density of step 2.2 with the completeness of step 4.1, the complex spaces L1(X,μ;C) and L2(X,μ;C) are complete and Cc(X;C) is dense in both. ∎

step 2.2step 4.1

Remarks

  • Where choice is spent. Steps 1.1 and 1.2 are the only uses of [A1]: they supply the Dependent Choice hypothesis of the density theorem [F3] and the countable choice hypothesis of Riesz–Fischer completeness [F2]. The Cauchy-sequence reduction and the recombination are choice-free.
  • Why the complex spaces are treated locally. The published completeness theorem [F2] and density theorem [F3] are statements about the real spaces of The space Lp(μ) as the quotient by null functions; the complex statements asserted here are obtained by the componentwise reduction above rather than assumed.

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