The theorem of F. and M. Riesz

1 Totally ordered groups

Suppose that $G$ is a locally compact abelian group and that $P\subset G$ is a semigroup (satisfies $P+P\subset P$) that is closed and satisfies $P\cap (-P)=\{0\}$ and $P\cup (-P)=G$. We define a total order on $G$ by $x\le y$ when $y-x\in P$. 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 $x\le y$ and $y\le z$, then $y-x\in P$ and $z-y\in P$ and hence $z-x=(z-y)+(y-x)\in P+P\subset P$, showing that $x\le z$, so $\le$ is transitive. If $x\le y$ and $y\le x$ then $y-x\in P$ and $x-y\in P$, the latter of which is equivalent to $y-x=-(x-y)\in -P$, hence $y-x\in P\cap (-P)$, and then $P\cap (-P)=\{0\}$ implies that $y-x=0$, i.e. $x=y$, so $\le$ is antisymmetric. If $x,y\in P$ 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 or , showing that $\le$ is total.

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

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.

Lemma 1.

Suppose that $\mathrm{\Gamma }$ is a discrete abelian group. $\mathrm{\Gamma }$ can be totally ordered if and only if $\gamma \mathrm{\in }\mathrm{\Gamma }$ having finite order implies that $\gamma \mathrm{=}\mathrm{0}$.

2 Functions of analytic type

If $G$ is a compact abelian group, then $G$ is connected if and only if $\gamma \in \widehat{G}$ having finite order implies that $\gamma =0$. 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 $\le$ be a total order on $\widehat{G}$ induced by some semigroup. We say that a function $f\in L^1(G)$ is of analytic type if implies that $\widehat{f}(\gamma )=0$, and we say that a measure $\mu \in M(G)$ is of analytic type if implies that $\widehat{\mu }(\gamma )=0$. (We denote by $M(G)$ the set of regular complex Borel measures on $G$.) For $1\le p\le \mathrm{\infty }$, we denote by $H^p(G)$ those elements of $L^p(G)$ that are of analytic type. We emphasize that the notion of a function or measure being of analytic type depends on the total order $\le$ on $\widehat{G}$.

We remind ourselves that when $ℳ$ is a $\sigma$-algebra on a set $X$ and $\mu$ is a measure on $ℳ$, if $A\in ℳ$ and $\mu (E)=\mu (A\cap E)$ for all $E\in ℳ$ then we say that $\mu$ is concentrated on $A$. Measures $\lambda ,\mu$ on $ℳ$ 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 $\sigma$ is a positive element of $M(G)$. The Lebesgue decomposition tells us that there is a unique pair of finite Borel measures ${\sigma }_s$ and ${\sigma }_a$ on $G$ such that (i) $\sigma ={\sigma }_s+{\sigma }_a$, (ii) ${\sigma }_a$ is absolutely continuous with respect to $m$, and (iii) ${\sigma }_s$ and $m$ are mutually singular. Then the Radon-Nikodym theorem tells us that there is a unique nonnegative $w\in L^1(m)$ such that $d{\sigma }_a=wdm$. Thus,

$$d\sigma =d{\sigma }_s+wdm.$$

We define $\mathrm{\Omega }$ to be the set of all trigonometric polynomials $Q$ on $G$ such that $\widehat{Q}(\gamma )=0$ for $\gamma \le 0$. We also define $K=\{1+Q:Q\in \mathrm{\Omega }\}$. $K\subset L^2(\sigma )$, and we denote by $\overline{K}$ its closure in the Hilbert space $L^2(\sigma )$.

Lemma 2.

$\overline{K}$ is a convex set.

Proof.

Let $f,g\in K$ be distinct and let $0\le t\le 1$. There are $P_n,Q_n\in \mathrm{\Omega }$ such that $1+P_n\to f$ and $1+Q_n\to g$, and

$$(1-t)f+tg=\underset{n\to \mathrm{\infty }}{lim}((1-t)(1+P_n)+t(1+Q_n))=\underset{n\to \mathrm{\infty }}{lim}(1+(1-t)P_n+t{Q}_n).$$

For each $n$, $(1-t)P_n+t{Q}_n\in \mathrm{\Omega }$, so we have written $(1-t)f+tg$ as a limit of elements of $K$, showing that $(1-t)f+tg\in \overline{K}$ and hence that $\overline{K}$ is convex. ∎

As $\overline{K}$ is a closed convex set in the Hilbert space $L^2(\sigma )$, there is a unique $\varphi \in \overline{K}$ such that $d(0,\overline{K})=\parallel 0-\varphi \parallel$ (namely, that attains the infimum of the distance of elements of $\overline{K}$ to the origin), which we can write as

$$\parallel \varphi \parallel =\underset{Q\in \mathrm{\Omega }}{inf}\parallel 1+Q\parallel .$$

$\varphi$ is the unique element of $\overline{K}$ such that

$$⟨\varphi ,\psi -\varphi ⟩=0,\psi \in \overline{K}.$$

The following lemma establishes properties of $\varphi$.

Lemma 3.

1. $\varphi =0$ almost everywhere with respect to ${\sigma }_s$.

2. $\varphi w\in L^2(m)$ and ${|\varphi |}^{2}w={\parallel \varphi \parallel }^2$ almost everywhere with respect to $m$.

3. If $\parallel \varphi \parallel >0$ and $h=\frac{1}{\varphi }$, then $h\in H^2(m)$ and $\widehat{h}(0)=1$.

Proof.

We write $c=\parallel \varphi \parallel$. Let $1+Q_n\in K$ such that $1+Q_n\to \varphi$. If $g\in L^2(\sigma )$ and $\varphi +g\in \overline{K}$, then $⟨\varphi ,(\varphi +g)-\varphi ⟩=0$, i.e. $⟨\varphi ,g⟩=0$. Let $\gamma >0$. On the one hand, $\gamma \in \mathrm{\Omega }$ so $\varphi +\gamma ={lim}_{n\to \mathrm{\infty }}1+(Q_n+\gamma )\in \overline{K}$, hence $⟨\varphi ,\gamma ⟩=0$ and so $⟨\gamma ,\varphi ⟩=0$. On the other hand, define $g=\varphi \gamma$, which satisfies

$$\varphi +g=\varphi (1+\gamma )=\underset{n\to \mathrm{\infty }}{lim}(1+Q_n)(1+\gamma )=\underset{n\to \mathrm{\infty }}{lim}1+\gamma +Q_n+Q_n\gamma ,$$

and because $\gamma >0$, each term of $\gamma +Q_n+Q_n\gamma$ belongs to $\mathrm{\Omega }$, showing that $\varphi +g\in \overline{K}$, from which we get $⟨\varphi ,g⟩=0$ and so $⟨g,\varphi ⟩=0$. We have proved that

$${\int }_G⟨x,\gamma ⟩\overline{\varphi (x)}𝑑\sigma (x)=0,\gamma >0,$$ (1)

and

$${\int }_G⟨x,\gamma ⟩{|\varphi (x)|}^2𝑑\sigma (x)=0,\gamma >0.$$ (2)

Taking the complex conjugate of (2) gives

Defining $d\lambda ={|\varphi |}^{2}d\sigma$ we have $\lambda \in M(G)$. The above and (2) give

$$\widehat{\lambda }(\gamma )=0,\gamma \ne 0.$$

As well,

$$\widehat{\lambda }(0)={\int }_G{|\varphi |}^2𝑑\sigma =c^2.$$

Because $\lambda \in M(G)$ and $\widehat{\lambda }\in L^1(\widehat{G})$, there is some $f\in L^1(G)$ such that $d\lambda =fdm$, defined by

$$f(x)={\int }_{\widehat{G}}\widehat{\lambda }(\gamma )⟨x,\gamma ⟩𝑑m_{\widehat{G}}(\gamma ),\gamma \in \widehat{G},$$

where $m_{\widehat{G}}$ is the Haar measure on $\widehat{G}$ that assigns measure $1$ to each singleton. That is, $d\lambda =fdm$ where $f(x)=c^2{m}_{\widehat{G}}(\{0\})=c^2$, hence $d\lambda =c^{2}dm$. Combined with $d\lambda ={|\varphi |}^{2}d\sigma$ we get

$${|\varphi |}^{2}d\sigma =c^{2}dm.$$

Therefore ${|\varphi |}^{2}d\sigma$ is absolutely continuous with respect to $m$, and because ${|\varphi |}^{2}d\sigma ={|\varphi |}^{2}d{\sigma }_s+{|\varphi |}^{2}wdm$, it follows that ${|\varphi |}^{2}d{\sigma }_s=0$, that is, that $\varphi (x)=0$ for ${\sigma }_s$-almost all $x\in G$, proving the first claim. Furthermore, ${|\varphi |}^{2}d\sigma ={|\varphi |}^{2}wdm$ and using ${|\varphi |}^{2}d\sigma =c^{2}dm$ we get ${|\varphi (x)|}^{2}w(x)=c^2$ for $m$-almost all $x\in G$. Because $w\in L^1(m)$ and ${|\varphi w|}^2=c^{2}w$, we get $\varphi w\in L^2(m)$, proving the second claim.

So far we have not supposed that $c>0$. If indeed $c>0$, then ${|h|}^2={|\varphi |}^{-2}=c^{-2}w$, giving $h\in L^2(m)$. For $\gamma \in \widehat{G}$,

${\displaystyle {\int }_G}h(x)⟨x,\gamma ⟩𝑑m(x)$ $=$ ${\displaystyle {\int }_G}{|\varphi (x)|}^{-2}\overline{\varphi (x)}⟨x,\gamma ⟩𝑑m(x)$

$=$ $c^{-2}{\displaystyle {\int }_G}⟨x,\gamma ⟩\overline{\varphi (x)}w(x)𝑑m(x)$

$=$ $c^{-2}{\displaystyle {\int }_G}⟨x,\gamma ⟩\overline{\varphi (x)}𝑑\sigma (x).$

This and (1) yield

$${\int }_{G}h(x)⟨x,\gamma ⟩𝑑m(x)=0,\gamma >0,$$

in other words,

namely, $h$ is of analytic type, i.e. $h\in H^2(m)$. Moreover, for each $n\in \mathbb{N}$ we check that $Q_n+\varphi \in \overline{K}$ and hence that $⟨1+Q_n,\varphi ⟩=⟨1,\varphi ⟩$, giving

$$c^2\widehat{h}(0)={\int }_G\overline{\varphi }𝑑\sigma ={\int }_G(1+Q_n)\overline{\varphi }𝑑\sigma .$$

This is true for all $n\in \mathbb{N}$, so we obtain

$$c^2\widehat{h}(0)={\int }_G{|\varphi |}^2𝑑\sigma ={\parallel \varphi \parallel }^2=c^2,$$

i.e. $\widehat{h}(0)=1$, proving the third claim. ∎

The above lemma is used to prove the following theorem. 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 $\mu \mathrm{\in }M\mathit{}\mathrm{(}G\mathrm{)}$ is of analytic type. If the Lebesgue decomposition of $\mu$ is

$$d\mu =d{\mu }_s+fdm,$$

where ${\mu }_s$ and $m$ are mutually singular and $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}m\mathrm{)}$, then ${\mu }_s\mathrm{\in }M\mathit{}\mathrm{(}G\mathrm{)}$ is of analytic type and $f$ is of analytic type, and ${\widehat{\mu }}_s\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}$.

3 The theorem of F. and M. Riesz

We are now equipped to prove the theorem of F. and M. Riesz.

Theorem 5 (F. and M. Riesz).

If $\mu \mathrm{\in }M\mathit{}\mathrm{(}\mathrm{T}\mathrm{)}$ and $\widehat{\mu }\mathit{}\mathrm{(}n\mathrm{)}\mathrm{=}\mathrm{0}$ for every negative integer $n$, then $\mu$ is absolutely continuous with respect to Haar measure.

Proof.

Write $d\mu =d{\mu }_s+fdm$, where ${\mu }_s$ and $m$ are mutually singular and $f\in L^1(m)$. Theorem 4 tells us that ${\mu }_s$ is of analytic type, i.e. ${\widehat{\mu }}_s(n)=0$ for , and that ${\widehat{\mu }}_s(0)=0$. Therefore, if ${\mu }_s\ne 0$ then there is a minimal positive integer $n_0$ for which ${\widehat{\mu }}_s(n_0)\ne 0$. Defining $\widehat{\lambda }(n)={\widehat{\mu }}_s(n_0+n)$, we get that $\lambda \in M(𝕋)$ and that $\lambda$ and $m$ are mutually singular. But $\widehat{\lambda }(n)={\widehat{\mu }}_s(n_0+n)=0$ for , so $\lambda$ is of analytic type, and therefore Theorem 4 says that ${\widehat{\mu }}_s(n_0)=\widehat{\lambda }(0)=0$ (because $\lambda$ and $m$ are mutually singular), a contradiction. Hence ${\widehat{\mu }}_s(n)=0$ for all $n\in \mathbb{Z}$, which implies that ${\mu }_s=0$. But this means that $\mu$ is absolutely continuous with respect to $m$, completing the proof. ∎