# 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 $P\subset G$ 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 $x\leq y$ 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 $x\leq y$ and $y\leq z$, 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 $x\leq z$, so $\leq$ is transitive. If $x\leq y$ and $y\leq x$ 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 $\leq$ 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 $x or $y, showing that $\leq$ is total.

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

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.11 1 Walter Rudin, Fourier Analysis on Groups, p. 194, Theorem 8.1.2.

###### Lemma 1.

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

## 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 $\gamma=0$.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 $\leq$ 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 $\gamma<0$ implies that $\hat{f}(\gamma)=0$, and we say that a measure $\mu\in M(G)$ is of analytic type if $\gamma<0$ implies that $\hat{\mu}(\gamma)=0$. (We denote by $M(G)$ the set of regular complex Borel measures on $G$.) For $1\leq p\leq\infty$, 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 $\leq$ on $\widehat{G}$.

We remind ourselves that when $\mathscr{M}$ is a $\sigma$-algebra on a set $X$ and $\mu$ is a measure on $\mathscr{M}$, if $A\in\mathscr{M}$ and $\mu(E)=\mu(A\cap E)$ for all $E\in\mathscr{M}$ then we say that $\mu$ is concentrated on $A$. Measures $\lambda,\mu$ on $\mathscr{M}$ 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 $\Omega$ to be the set of all trigonometric polynomials $Q$ on $G$ such that $\hat{Q}(\gamma)=0$ for $\gamma\leq 0$. We also define $K=\{1+Q:Q\in\Omega\}$. $K\subset L^{2}(\sigma)$, 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 $0\leq t\leq 1$. There are $P_{n},Q_{n}\in\Omega$ such that $1+P_{n}\to f$ and $1+Q_{n}\to g$, and

 $(1-t)f+tg=\lim_{n\to\infty}((1-t)(1+P_{n})+t(1+Q_{n}))=\lim_{n\to\infty}(1+(1-% t)P_{n}+tQ_{n}).$

For each $n$, $(1-t)P_{n}+tQ_{n}\in\Omega$, 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 $\phi\in\overline{K}$ such that $d(0,\overline{K})=\|0-\phi\|$ (namely, that attains the infimum of the distance of elements of $\overline{K}$ to the origin), which we can write as

 $\|\phi\|=\inf_{Q\in\Omega}\|1+Q\|.$

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

 $\langle\phi,\psi-\phi\rangle=0,\qquad\psi\in\overline{K}.$

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

###### Lemma 3.
1. 1.

$\phi=0$ almost everywhere with respect to $\sigma_{s}$.

2. 2.

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

3. 3.

If $\|\phi\|>0$ and $h=\frac{1}{\phi}$, then $h\in H^{2}(m)$ and $\hat{h}(0)=1$.

###### Proof.

We write $c=\|\phi\|$. Let $1+Q_{n}\in K$ such that $1+Q_{n}\to\phi$. If $g\in L^{2}(\sigma)$ and $\phi+g\in\overline{K}$, then $\langle\phi,(\phi+g)-\phi\rangle=0$, i.e. $\langle\phi,g\rangle=0$. Let $\gamma>0$. On the one hand, $\gamma\in\Omega$ so $\phi+\gamma=\lim_{n\to\infty}1+(Q_{n}+\gamma)\in\overline{K}$, hence $\langle\phi,\gamma\rangle=0$ and so $\langle\gamma,\phi\rangle=0$. On the other hand, define $g=\phi\gamma$, which satisfies

 $\phi+g=\phi(1+\gamma)=\lim_{n\to\infty}(1+Q_{n})(1+\gamma)=\lim_{n\to\infty}1+% \gamma+Q_{n}+Q_{n}\gamma,$

and because $\gamma>0$, each term of $\gamma+Q_{n}+Q_{n}\gamma$ belongs to $\Omega$, showing that $\phi+g\in\overline{K}$, from which we get $\langle\phi,g\rangle=0$ and so $\langle g,\phi\rangle=0$. We have proved that

 $\int_{G}\langle x,\gamma\rangle\overline{\phi(x)}d\sigma(x)=0,\qquad\gamma>0,$ (1)

and

 $\int_{G}\langle x,\gamma\rangle|\phi(x)|^{2}d\sigma(x)=0,\qquad\gamma>0.$ (2)

Taking the complex conjugate of (2) gives

 $\int_{G}\langle x,\gamma\rangle|\phi(x)|^{2}d\sigma(x)=0,\qquad\gamma<0.$

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

 $\hat{\lambda}(\gamma)=0,\qquad\gamma\neq 0.$

As well,

 $\hat{\lambda}(0)=\int_{G}|\phi|^{2}d\sigma=c^{2}.$

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

 $f(x)=\int_{\widehat{G}}\hat{\lambda}(\gamma)\langle x,\gamma\rangle dm_{% \widehat{G}}(\gamma),\qquad\gamma\in\widehat{G},$

where $m_{\widehat{G}}$ is the Haar measure on $\widehat{G}$ that assigns measure $1$ to each singleton.44 4 Walter Rudin, Fourier Analysis on Groups, p. 30. That is, $d\lambda=fdm$ where $f(x)=c^{2}m_{\widehat{G}}(\{0\})=c^{2}$, hence $d\lambda=c^{2}dm$. Combined with $d\lambda=|\phi|^{2}d\sigma$ we get

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

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

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

 $\displaystyle\int_{G}h(x)\langle x,\gamma\rangle dm(x)$ $\displaystyle=$ $\displaystyle\int_{G}|\phi(x)|^{-2}\overline{\phi(x)}\langle x,\gamma\rangle dm% (x)$ $\displaystyle=$ $\displaystyle c^{-2}\int_{G}\langle x,\gamma\rangle\overline{\phi(x)}w(x)dm(x)$ $\displaystyle=$ $\displaystyle c^{-2}\int_{G}\langle x,\gamma\rangle\overline{\phi(x)}d\sigma(x).$

This and (1) yield

 $\int_{G}h(x)\langle x,\gamma\rangle dm(x)=0,\qquad\gamma>0,$

in other words,

 $\hat{h}(\gamma)=0,\qquad\gamma<0,$

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

 $c^{2}\hat{h}(0)=\int_{G}\overline{\phi}d\sigma=\int_{G}(1+Q_{n})\overline{\phi% }d\sigma.$

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

 $c^{2}\hat{h}(0)=\int_{G}|\phi|^{2}d\sigma=\|\phi\|^{2}=c^{2},$

i.e. $\hat{h}(0)=1$, 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 $\mu\in M(G)$ 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 $f\in L^{1}(m)$, then $\mu_{s}\in M(G)$ is of analytic type and $f$ is of analytic type, and $\hat{\mu}_{s}(0)=0$.

## 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 $\mu\in M(\mathbb{T})$ and $\hat{\mu}(n)=0$ 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. $\hat{\mu}_{s}(n)=0$ for $n<0$, and that $\hat{\mu}_{s}(0)=0$. Therefore, if $\mu_{s}\neq 0$ then there is a minimal positive integer $n_{0}$ for which $\hat{\mu}_{s}(n_{0})\neq 0$. Defining $\hat{\lambda}(n)=\hat{\mu}_{s}(n_{0}+n)$, we get that $\lambda\in M(\mathbb{T})$ and that $\lambda$ and $m$ are mutually singular. But $\hat{\lambda}(n)=\hat{\mu}_{s}(n_{0}+n)=0$ for $n<0$, so $\lambda$ is of analytic type, and therefore Theorem 4 says that $\hat{\mu}_{s}(n_{0})=\hat{\lambda}(0)=0$ (because $\lambda$ and $m$ are mutually singular), a contradiction. Hence $\hat{\mu}_{s}(n)=0$ 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. ∎