Thursday, 13 November 2014

The idempotent theorem

Let $G$ be a locally compact abelian group and let $M(G)$ be the Banach algebra of regular complex Borel measures on $G$. Given $\mu\in M(G)$ its Fourier transform
$$\hat{\mu}(\gamma) = \int \overline{\gamma}\,d\mu,$$
is a continuous function defined on the Pontryagin dual $\hat{G}$ of $G$. If the measure $\mu$ is "nice" in some way then we expect some amount of regularity from the function $\hat{\mu}$. For instance if $\mu$ is actually an element of the subspace $L^1(G)\subset M(G)$ of measures absolutely continuous with respect to the Haar measure of $G$ then the Riemann-Lebesgue lemma asserts $\hat{\mu}\in C_0(\hat{G})$.

The idempotent theorem of Cohen, Helson, and Rudin describes the structure of measures $\mu$ whose Fourier transform $\hat{\mu}$ takes a discrete set of values, or equivalently, since $\|\hat{\mu}\|_\infty\leq\|\mu\|$, a finite set of values. To describe the theorem, note that we can define $P(\mu)$ for any polynomial $P$ by taking appropriate linear combinations of convolution powers of $\mu$, and moreover we have the relation $\widehat{P(\mu)} = P(\hat{\mu})$, where on the right hand side we apply $P$ pointwise. Thus if $\hat{\mu}$ takes only the values $a_1,\dots,a_n$ then by setting
$$P_i(x) = \prod_{j\neq i} (x-a_j)/(a_i-a_j)$$
we obtain a decomposition $\mu = a_1\mu_1 + \cdots + a_n\mu_n$ of $\mu$ into a linear combination of measures $\mu_i=P_i(\mu)$ whose Fourier transforms $\hat{\mu_i} = P_i(\hat{\mu})$ take only values $0$ and $1$. Such measures are called idempotent, because they are equivalently defined by $\mu\ast\mu=\mu$. By the argument just given it suffices to characterise idempotent measures: this explains the name of the theorem.

The most obvious example of an idempotent measure is the Haar measure $m_H$ of a compact subgroup $H\leq G$. Moreover we can multiply any idempotent measure $\mu$ by a character $\gamma\in\hat{G}$ to obtain a measure $\gamma\mu$ defined by
$$\int f \,d(\gamma\mu) = \int f\gamma\,d\mu.$$
This measure $\gamma\mu$ will again be idempotent, as
$$\int f\,d(\gamma \mu\ast\gamma \mu) = \int\int f(x+y)\gamma(x)\gamma(y)\,d\mu(x)d\mu(y) = \int\int f(x+y) \gamma(x+y)\,d\mu(x)d\mu(y) = \int f\gamma\,d\mu.$$
If we add or subtract two idempotent measures then though we may not have again an idempotent measure we certainly have a measure whose Fourier transform takes integer values. On reflection, it feels more natural in the setting of harmonic analysis to require that $\hat{\mu}$ takes values in a certain discrete subgroup than to require that it take values in $\{0,1\}$, so we relax our restriction so. The idempotent theorem states that we have already accounted for all those $\mu$ such that $\hat{\mu}$ is integer-valued.

Theorem (the idempotent theorem): For every $\mu\in M(G)$ such that $\hat{\mu}$ is integer-valued there is a finite collection of compact subgroups $G_1,\dots,G_k\leq G$, characters $\gamma_1,\dots,\gamma_k\in\hat{G}$, and integers $n_1,\dots,n_k\in\mathbf{Z}$ such that
$$\mu = n_1\gamma_1 m_{G_1} + \cdots + n_k\gamma_k m_{G_k}.$$
As a consequence we deduce a structure theorem for $\mu$ with $\hat{\mu}$ taking finitely many values, as we originally wanted: for every such $\mu$ there is a finite collection of compact subgroups $G_1,\dots,G_k\leq G$, characters $\gamma_1,\dots,\gamma_k\in\hat{G}$, and complex numbers $a_1,\dots,a_k\in\mathbf{C}$ such that
$$\mu = a_1\gamma_1 m_{G_1} + \cdots + a_k \gamma_k m_{G_k}.$$

The theorem was first proved in the case of $G=\mathbf{R}/\mathbf{Z}$ by Helson in 1953: in this case the theorem states simply that if $\hat{\mu}$ is integer-valued then it differs from some periodic function in finitely many places. In 1959 Rudin gave the theorem its present form and proved it for $(\mathbf{R}/\mathbf{Z})^d$. Finally in 1960 Cohen proved the general case, in the same paper in which he made the first substantial progress on the Littlewood problem. The proof was subsequently simplified a good deal, particularly by Amemiya and Ito in 1964. We reproduce their proof here.

First note that if $\hat{\mu}$ is integer-valued then $\mu$ is supported on a compact subgroup. Indeed by inner regularity there is a compact set $K$ such that $|\mu|(K^c)<0.1$, the set $U$ of all $\gamma\in\hat{G}$ such that $|1-\gamma|<0.1/\|\mu\|$ on $K$ is then open, and if $\gamma\in U$ then
$$\|\gamma\mu-\mu\| = \int_G |\gamma-1|\,d|\mu| \leq \int_K + \int_{K^c} < 0.1 + 0.1 < 1.$$
But if $\gamma\mu\neq\mu$ then
$$\|\gamma\mu-\mu\|\geq \|\widehat{\gamma\mu}-\hat{\mu}\|_\infty \geq 1,$$
so $\gamma\mu=\mu$ for all $\gamma\in U$. Thus $\Gamma=\{\gamma\in\hat{G}: \gamma\mu=\mu\}$ is an open subgroup of $\hat{G}$, so by Pontryagin duality its preannihilator $\Gamma^\perp = \{g\in G: \gamma(g)=1 \text{ for all }\gamma\in\Gamma\}$ is a compact subgroup of $G$. Clearly $\mu$ is supported on $\Gamma^\perp$. Thus from now on we assume $G$ is compact.

Fix a measure $\mu\in M(G)$ and let $A=\{\gamma\mu: \gamma\in\hat{G}\}$.
Lemma 1: If $\nu$ is a weak* limit point of $A$ then $\|\nu\|<\|\mu\|$.
Proof: Fix $\epsilon>0$ and suppose we could find $f\in C(G)$ such that $\|f\|_\infty\leq 1$ and $\int f\,d\nu > (1-\epsilon)\|\mu\|$. Let $\gamma\mu$ be close enough to $\nu$ that $\Re\int f\gamma\,d\mu > (1-\epsilon)\|\mu\|$. Write $\mu = \theta|\mu|$ and $f\gamma\theta = g + ih$. Then if $Z$ is the complex number $Z = \int (g+i|h|)\,d|\mu|$, then $|Z|\leq\|\mu\|$ and
$$\Re Z = \int g \,d|\mu| = \Re\int f\gamma\,d\mu > (1-\epsilon)\|\mu\|,$$
so we must have
$$\Im Z = \int |h|\,d|\mu| \leq (1-(1-\epsilon)^2)^{1/2}\|\mu\| \leq 2\epsilon^{1/2}\|\mu\|.$$
Thus also
$$\int |1 - f\gamma\theta| \,d|\mu| \leq \int |1 - g|\,d|\mu| + \int |h|\,d|\mu| \leq 3\epsilon^{1/2}\|\mu\|.$$
But if this holds for both $\gamma_1\mu$ and $\gamma_2\mu$, say with $\gamma_1\mu\neq\gamma_2\mu$, then we have
$$ 1\leq \|\gamma_1\mu-\gamma_2\mu\| \leq \int |\gamma_1 - f\gamma_1\gamma_2\theta|\,d|\mu| + \int |\gamma_2 - f\gamma_1\gamma_2\theta|\,d|\mu| \leq 6\epsilon^{1/2}\|\mu\|,$$
so $\epsilon \geq 1/(36\|\mu\|^2)$, so
$$\|\nu\| \leq \|\mu\| - \frac{1}{36\|\mu\|}.\square$$
Lemma 2: If $\nu$ is a weak* limit point of $A$ then $\nu$ is singular with respect to the Haar measure $m_G$ of $G$.
Proof: By the Radon-Nikodym theorem we have a decomposition $\mu = f m_G + \mu_s$ for some $f\in L^1(G)$ and some $\mu_s$ singular with respect to $m_G$. By the Riemann-Lebesgue lemma then $\nu$ is a limit point of $\{\gamma\mu_s:\gamma\in\hat{G}\}$. Thus for any open set $U$ and $f\in C(G)$ such that $\|f\|_\infty\leq 1$ and $f=0$ outside of $U$ we have
$$\left|\int f\,d\nu\right| \leq \sup_\gamma \left|\int f\gamma \,d\mu_s\right| \leq |\mu_s|(U),$$
so $|\nu|(U)\leq |\mu_s|(U)$. This inequality extends to Borel sets in the usual way, so $\nu$ is singular.$\square$

The theorem follows relatively painlessly from the two lemmas. Fix $\mu\in M(G)$ with $\hat{\mu}$ integer-valued and let $A = \{\gamma\mu: \int\gamma\,d\mu\neq 0\}$. Then $\overline{A}$ is weak* compact, so because $\|\cdot\|$ is lower semicontinuous in the weak* topology there is some $\nu\in\overline{A}$ of minimal norm. Since $\int d\nu$ is an integer different from $0$ we must have $\nu\neq 0$. Thus by Lemma 1 the set $\{\gamma\nu: \int\gamma\,d\nu\neq 0\}$ is finite. But this implies that
$$\nu = (n_1 \gamma_1 + \cdots + n_k \gamma_k) m_H\qquad(\ast)$$
for some $n_1,\dots,n_k\in\mathbf{Z}$, $\gamma_1,\dots,\gamma_k\in\hat{G}$, and $H=\{\gamma:\gamma\nu=\nu\}^\perp$ the support group of $\nu$. In particular $\nu$ is absolutely continuous with respect to $m_H$, so because $\nu|_H$ is in the weak* closure of $\{\gamma\mu|_H:\gamma\in\hat{G}\}$ we deduce from Lemma 2 that $\nu|_H = \gamma\mu|_H$ for some $\gamma$.  Thus $\mu|_H$ is a nonzero measure of the form $(\ast)$ and we have an obvious mutually singular decomposition
$$\mu = \mu|_H + (\mu-\mu|_H).$$
Since $\|\mu-\mu|_H\| = \|\mu\| - \|\mu|_H\|\leq\|\mu\|-1$ the theorem follows by induction.

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