The theorem of F. and M. Riesz

Jordan Bell
July 1, 2014

1 Totally ordered groups

Suppose that G is a locally compact abelian group and that PG is a semigroup (satisfies P+PP) that is closed and satisfies P(-P)={0} and P(-P)=G. We define a total order on G by xy when y-xP. We verify that this is indeed a total order. (We remark that nowhere in this do we show the significance of P being closed; but in this note we shall be speaking about discrete abelian groups where any set is closed.)

If xy and yz, then y-xP and z-yP and hence z-x=(z-y)+(y-x)P+PP, showing that xz, so is transitive. If xy and yx then y-xP and x-yP, the latter of which is equivalent to y-x=-(x-y)-P, hence y-xP(-P), and then P(-P)={0} implies that y-x=0, i.e. x=y, so is antisymmetric. If x,yP then y-x is either 0, in which case x=y, or it is contained in one and only one of P and -P, and then respectively x<y or y<x, showing that is total.

Moreover, the total order induced by the semigroup P is compatible with the group operation in G: if xy and zG, then (y+z)-(x+z)=y-xP, showing that x+zy+z.

We say that G with the total order induced by P is a totally ordered group. We shall use the following lemma in the next section.11 1 Walter Rudin, Fourier Analysis on Groups, p. 194, Theorem 8.1.2.

Lemma 1.

Suppose that Γ is a discrete abelian group. Γ can be totally ordered if and only if γΓ having finite order implies that γ=0.

2 Functions of analytic type

If G is a compact abelian group, then G is connected if and only if γG^ having finite order implies that γ=0.22 2 Walter Rudin, Fourier Analysis on Groups, p. 47, Theorem 2.5.6. Combined with Lemma 1, we get that a compact abelian group is connected if and only if its dual group can be ordered.

Suppose in the rest of this section that G is a connected compact abelian group, and let be a total order on G^ induced by some semigroup. We say that a function fL1(G) is of analytic type if γ<0 implies that f^(γ)=0, and we say that a measure μM(G) is of analytic type if γ<0 implies that μ^(γ)=0. (We denote by M(G) the set of regular complex Borel measures on G.) For 1p, we denote by Hp(G) those elements of Lp(G) that are of analytic type. We emphasize that the notion of a function or measure being of analytic type depends on the total order on G^.

We remind ourselves that when is a σ-algebra on a set X and μ is a measure on , if A and μ(E)=μ(AE) for all E then we say that μ is concentrated on A. Measures λ,μ on are said to be mutually singular if they are concentrated on disjoint sets.

Let m be the Haar measure on G such that m(G)=1, and suppose that σ is a positive element of M(G). The Lebesgue decomposition tells us that there is a unique pair of finite Borel measures σs and σa on G such that (i) σ=σs+σa, (ii) σa is absolutely continuous with respect to m, and (iii) σs and m are mutually singular. Then the Radon-Nikodym theorem tells us that there is a unique nonnegative wL1(m) such that dσa=wdm. Thus,

dσ=dσs+wdm.

We define Ω to be the set of all trigonometric polynomials Q on G such that Q^(γ)=0 for γ0. We also define K={1+Q:QΩ}. KL2(σ), and we denote by K¯ its closure in the Hilbert space L2(σ).

Lemma 2.

K¯ is a convex set.

Proof.

Let f,gK be distinct and let 0t1. There are Pn,QnΩ such that 1+Pnf and 1+Qng, and

(1-t)f+tg=limn((1-t)(1+Pn)+t(1+Qn))=limn(1+(1-t)Pn+tQn).

For each n, (1-t)Pn+tQnΩ, so we have written (1-t)f+tg as a limit of elements of K, showing that (1-t)f+tgK¯ and hence that K¯ is convex. ∎

As K¯ is a closed convex set in the Hilbert space L2(σ), there is a unique ϕK¯ such that d(0,K¯)=0-ϕ (namely, that attains the infimum of the distance of elements of K¯ to the origin), which we can write as

ϕ=infQΩ1+Q.

ϕ is the unique element of K¯ such that

ϕ,ψ-ϕ=0,ψK¯.

The following lemma establishes properties of ϕ.33 3 Walter Rudin, Fourier Analysis on Groups, p. 199, Lemma 8.2.2.

Lemma 3.
  1. 1.

    ϕ=0 almost everywhere with respect to σs.

  2. 2.

    ϕwL2(m) and |ϕ|2w=ϕ2 almost everywhere with respect to m.

  3. 3.

    If ϕ>0 and h=1ϕ, then hH2(m) and h^(0)=1.

Proof.

We write c=ϕ. Let 1+QnK such that 1+Qnϕ. If gL2(σ) and ϕ+gK¯, then ϕ,(ϕ+g)-ϕ=0, i.e. ϕ,g=0. Let γ>0. On the one hand, γΩ so ϕ+γ=limn1+(Qn+γ)K¯, hence ϕ,γ=0 and so γ,ϕ=0. On the other hand, define g=ϕγ, which satisfies

ϕ+g=ϕ(1+γ)=limn(1+Qn)(1+γ)=limn1+γ+Qn+Qnγ,

and because γ>0, each term of γ+Qn+Qnγ belongs to Ω, showing that ϕ+gK¯, from which we get ϕ,g=0 and so g,ϕ=0. We have proved that

Gx,γϕ(x)¯𝑑σ(x)=0,γ>0, (1)

and

Gx,γ|ϕ(x)|2𝑑σ(x)=0,γ>0. (2)

Taking the complex conjugate of (2) gives

Gx,γ|ϕ(x)|2𝑑σ(x)=0,γ<0.

Defining dλ=|ϕ|2dσ we have λM(G). The above and (2) give

λ^(γ)=0,γ0.

As well,

λ^(0)=G|ϕ|2𝑑σ=c2.

Because λM(G) and λ^L1(G^), there is some fL1(G) such that dλ=fdm, defined by

f(x)=G^λ^(γ)x,γ𝑑mG^(γ),γG^,

where mG^ is the Haar measure on G^ that assigns measure 1 to each singleton.44 4 Walter Rudin, Fourier Analysis on Groups, p. 30. That is, dλ=fdm where f(x)=c2mG^({0})=c2, hence dλ=c2dm. Combined with dλ=|ϕ|2dσ we get

|ϕ|2dσ=c2dm.

Therefore |ϕ|2dσ is absolutely continuous with respect to m, and because |ϕ|2dσ=|ϕ|2dσs+|ϕ|2wdm, it follows that |ϕ|2dσs=0, that is, that ϕ(x)=0 for σs-almost all xG, proving the first claim. Furthermore, |ϕ|2dσ=|ϕ|2wdm and using |ϕ|2dσ=c2dm we get |ϕ(x)|2w(x)=c2 for m-almost all xG. Because wL1(m) and |ϕw|2=c2w, we get ϕwL2(m), proving the second claim.

So far we have not supposed that c>0. If indeed c>0, then |h|2=|ϕ|-2=c-2w, giving hL2(m). For γG^,

Gh(x)x,γ𝑑m(x) = G|ϕ(x)|-2ϕ(x)¯x,γ𝑑m(x)
= c-2Gx,γϕ(x)¯w(x)𝑑m(x)
= c-2Gx,γϕ(x)¯𝑑σ(x).

This and (1) yield

Gh(x)x,γ𝑑m(x)=0,γ>0,

in other words,

h^(γ)=0,γ<0,

namely, h is of analytic type, i.e. hH2(m). Moreover, for each n we check that Qn+ϕK¯ and hence that 1+Qn,ϕ=1,ϕ, giving

c2h^(0)=Gϕ¯𝑑σ=G(1+Qn)ϕ¯𝑑σ.

This is true for all n, so we obtain

c2h^(0)=G|ϕ|2𝑑σ=ϕ2=c2,

i.e. h^(0)=1, proving the third claim. ∎

The above lemma is used to prove the following theorem.55 5 Walter Rudin, Fourier Analysis on Groups, p. 200, Theorem 8.2.3. The proof of this theorem in Rudin is not long, but I don’t understand the first step in his proof so I have not attempted to write it out.

Theorem 4.

Suppose that G is a connected compact abelian group and that μM(G) is of analytic type. If the Lebesgue decomposition of μ is

dμ=dμs+fdm,

where μs and m are mutually singular and fL1(m), then μsM(G) is of analytic type and f is of analytic type, and μ^s(0)=0.

3 The theorem of F. and M. Riesz

We are now equipped to prove the theorem of F. and M. Riesz.66 6 Walter Rudin, Fourier Analysis on Groups, p. 201, §8.2.4.

Theorem 5 (F. and M. Riesz).

If μM(T) and μ^(n)=0 for every negative integer n, then μ is absolutely continuous with respect to Haar measure.

Proof.

Write dμ=dμs+fdm, where μs and m are mutually singular and fL1(m). Theorem 4 tells us that μs is of analytic type, i.e. μ^s(n)=0 for n<0, and that μ^s(0)=0. Therefore, if μs0 then there is a minimal positive integer n0 for which μ^s(n0)0. Defining λ^(n)=μ^s(n0+n), we get that λM(𝕋) and that λ and m are mutually singular. But λ^(n)=μ^s(n0+n)=0 for n<0, so λ is of analytic type, and therefore Theorem 4 says that μ^s(n0)=λ^(0)=0 (because λ and m are mutually singular), a contradiction. Hence μ^s(n)=0 for all n, which implies that μs=0. But this means that μ is absolutely continuous with respect to m, completing the proof. ∎