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Next: Homomorphism theorems Up: Decomposition of analytic measures Previous: Orders on locally compact

Analyticity

We continue with the notation of the previous section. Using the order structure on $ \Gamma$ we define some classes of analytic functions on $ G$:

(14) $\displaystyle H^1(G)$ $\displaystyle =$ $\displaystyle \left\{
f\in L^1(G): \widehat{f}=0\ {\rm on}\ (-P)\setminus\{0\}
\right\};$
(15) $\displaystyle H^1_0(G)$ $\displaystyle =$ $\displaystyle \left\{
f\in L^1(G): \widehat{f}=0\ {\rm on}\ -P
\right\};$

and

(16) $\displaystyle H^\infty(G)=\left\{ f\in L^\infty(G): \int_G f(x)g(x)dx=0 \ {\rm for\ all}\ g\in H^1_0(G) \right\}.$

We clearly have

$\displaystyle H^1(G)=\left\{
f\in L^1(G): \widehat{f}=0\ {\rm on}\ \overline{(-P)\setminus\{0\}}
\right\}.$

We can now give the definition of analytic measures in $ {\cal M}_T(\Sigma)$.

Definition 3.1   Let $ T$ be a sup path attaining representation of $ G$ by isomorphisms of $ M(\Sigma)$. A measure $ \mu\in{\cal M}_T(\Sigma)$ is called weakly analytic if the mapping $ t\mapsto T_t\mu(A)$ is in $ H^\infty(G)$ for every $ A\in \Sigma$.

Definition 3.2   Recall the $ T$-spectrum of a weakly measurable $ \mu\in{\cal M}_T(\Sigma)$,

(17) $\displaystyle {\rm spec}_T (\mu)= \bigcap_{f\in {\cal I}(\mu)} \left\{ \chi\in\Gamma:\ \ \widehat{f}(\chi)=0 \right\}.$

A measure $ \mu$ in $ {\cal M}_T(\Sigma)$ is called $ T$-analytic if $ {\rm spec}_T (\mu)\subset \overline{P}$.

That the two definitions of analyticity are equivalent will be shown later in this section.

Since $ {\cal I}(\mu)$ is translation-invariant, it follows readily that for all $ t\in G$,

$\displaystyle {\cal I}(T_t\mu)={\cal I}(\mu),$

and hence

(18) $\displaystyle {\rm spec}_T (T_t(\mu))={\rm spec}_T (\mu).$

We now recall several basic results from spectral theory of bounded functions that will be needed in the sequel. Our reference is [21, Section 40]. If $ \phi$ is in $ L^\infty(G)$, write $ \left[ \phi\right]$ for the smallest weak-* closed translation-invariant subspace of $ L^\infty(G)$ containing $ \phi$, and let $ {\cal I}([\phi])={\cal I}(\phi)$ denote the closed translation-invariant ideal in $ L^1(G)$:

$\displaystyle {\cal I}(\phi)=\{f\in L^1(G): f*\phi=0\}.$

It is clear that $ {\cal I}(\phi)=\{f\in L^1(G): f*g=0, \forall g\in \left[\phi\right]\}$. The spectrum of $ \phi$, denoted by $ \sigma \left[\phi\right]$, is the set of all continuous characters of $ G$ that belong to $ \left[ \phi\right]$. This closed subset of $ \Gamma$ is also given by

(19) $\displaystyle \sigma \left[\phi\right]=Z({\cal I}(\phi)).$

(See [21, Theorem (40.5)].)

Recall that a closed subset $ E$ of $ \Gamma$ is a set of spectral synthesis for $ L^1(G)$, or an $ S$-set, if and only if $ {\cal I}([E])$ is the only ideal in $ L^1(G)$ whose zero set is $ E$.

There are various equivalent definitions of $ S$-sets. Here is one that we will use at several occasions.
A set $ E\subset \Gamma$ is an $ S$-set if and only if every essentially bounded function $ g$ in $ L^\infty(G)$ with $ \sigma[g]\subset E$ is the weak-* limit of linear combinations of characters from $ E$.
(See [21, (40.23) (a)].) This has the following immediate consequence.

Proposition 3.3   Suppose that $ B$ is an $ S$-set, $ g\in L^\infty(G)$, and $ {\rm spec}(g)\subset B$. (i) If $ f$ is in $ L^1(G)$ and $ \widehat{f}=0$ on $ B$, then $ f*g(x)=0$ for all $ x$ in $ G$. In particular,

$\displaystyle \int_G f(x) g(-x)\, d\, x =0.$

(ii) If $ \mu$ is a measure in $ M(G)$ with $ \widehat{\mu}=0$ on $ B$, then $ \mu*g(x)=0$ for almost all $ x$ in $ G$.

Proof.    Part (i) is a simple consequence of [21, Theorems (40.8) and (40.10)]. We give a proof for the sake of completeness. Write $ g$ as the weak-* limit of trigonometric polynomials, $ \sum_{\chi\in E} a_\chi \chi(x)$, with characters in $ E$. Then

  $\displaystyle \int_G f(x)g(y-x)\, d\,x$ $\displaystyle =$ $\displaystyle \lim \int_G
\sum_{\chi\in E} a_\chi \chi(y)f(x)\chi(-x)\, d\,x$
    $\displaystyle =$ $\displaystyle \lim
\sum_{\chi\in E} a_\chi \chi(y) \widehat{f}(\chi)=0$

since $ \widehat{f}$ vanishes on $ E$.

To prove (ii), assume that $ \mu*g$ is not 0 a.e.. Then, there is $ f$ in $ L^1(G)$ such that $ f*(\mu*g)$ is not 0 a.e.. But this contradicts (i), since $ f*(\mu*g)=(f*\mu)*g$, $ f*\mu$ is in $ L^1(G)$, and $ \widehat{f*\mu}=0$ on $ B$.

The following is a converse of sorts of Proposition 3.3 and follows easily from definitions.

Proposition 3.4   Let $ B$ be a nonvoid closed subset of $ \Gamma$. Suppose that $ f$ is in $ L^\infty(G)$ and

(20) $\displaystyle \int_G f(x)g(x)dx=0$

for all $ g$ in $ L^1(G)$ such that $ \widehat{g}= 0$ on $ -B$. Then $ \sigma[f]\subset B$.

Proof.    Let $ \chi_0$ be any element in $ \Gamma\setminus B$. We will show that $ \chi_0$ is not in the spectrum of $ f$ by constructing a function $ h$ in $ L^1(G)$ with $ \widehat{h}(\chi_0)\neq 0$ and $ h*f= 0$. Let $ U$ be an open neighborhood of $ \chi_0$ not intersecting $ B$, and let $ h$ be in $ L^1(G)$ such that $ \widehat{h}$ is equal to 1 at $ \chi_0$ and to 0 outside $ U$. Direct computations show that the Fourier transform of the function $ g:\ \ t\mapsto h(x-t)$, when evaluated at $ \chi\in\Gamma$, gives $ \overline{\chi(x)}\widehat{h}(-\chi)$, and hence it vanishes on $ -B$. It follows from (20) that $ h*f= 0$, which completes the proof.

A certain class of $ S$-sets, known as the Calderón sets, or $ C$-sets, is particularly useful to us. These are defined as follows. A subset $ E$ of $ \Gamma$ is called a $ C$-set if every $ f$ in $ L^1(G)$ with Fourier transform vanishing on $ E$ can be approximated in the $ L^1$-norm by functions of the form $ h*f$ where $ h\in L^1(G)$ and $ \widehat{h}$ vanishes on an open set containing $ E$.
$ C$-sets enjoy the following properties (see [21, (39.39)] or [22, Section 7.5]).

Since closed subgroups are $ C$-sets, we conclude that $ \overline{P} \cap \overline{(-P)}$, and $ C_\alpha$, for all $ \alpha$, are $ C$-sets. >From the definition of $ S_{\alpha_0}$, (13), and the fact that $ C_{\alpha_0}$ is open and closed, it follows that the boundary of $ S_{\alpha_0}$ is the closed subgroup $ \psi_{\alpha_0}^{-1}(0)\cap C_{\alpha_0}$. Hence $ S_{\alpha_0}$ is a $ C$-set. For $ \alpha\neq\alpha_0$, the set $ S_\alpha$ is open and closed, and so it has empty boundary, and thus it is a $ C$-set. Likewise $ C_\alpha\setminus D_\alpha$ is a $ C$-set for all $ \alpha\neq\alpha_0$. >From this we conclude that arbitrary unions of $ S_\alpha$ and $ C_\alpha\setminus D_\alpha$ are $ C$-sets, because an arbitrary union of such sets, not including the index $ \alpha_0$, is open and closed, and so it is a $ C$-set.
We summarize our findings as follows.

Proposition 3.5   Suppose that $ P$ is a measurable order on $ \Gamma$. We have:
(i) $ \overline{P}$ and $ \overline{(-P)}$ are $ C$-sets;
(ii) $ S_\alpha$ is a $ C$-set for all $ \alpha$;
(iii) arbitrary unions of $ S_\alpha$ and $ C_\alpha\setminus D_\alpha$ are $ C$-sets.

As an immediate application, we have the following characterizations.

Corollary 3.6   Suppose that $ f$ is in $ L^\infty(G)$, then
(i) $ \sigma[f] \subset S_\alpha$ if and only if $ \int_G f(x) g(x) d x =0$ for all $ g\in L^1(G)$ such that $ \widehat{g}= 0$ on $ -S_\alpha$;
(ii) $ \sigma[f] \subset \Gamma\setminus C_\alpha$ if and only if $ \mu_\alpha *f=0$;
(iii) $ \sigma[f] \subset \overline{P}$ if and only if $ f\in H^\infty(G)$.

Proof. Assertions (i) and (iii) are clear from Propositions 3.5 and 3.4. To prove (ii), use Fubini's Theorem to first establish that for any $ g\in L^1(G)$, and any $ \mu\in M(G)$, we have

$\displaystyle \int_G (\mu*f)(t) g(t)dt=\int_G f(t)(\mu*g)(t)dt.$

Now suppose that $ \sigma[f] \subset \Gamma\setminus C_\alpha$, and let $ g$ be any function in $ L^1(G)$. From Propositions 3.5 and 3.4, we have that $ \int_G f g dt = 0$ for all $ g$ with Fourier transform vanishing on $ \Gamma\setminus C_\alpha$, equivalently, for all $ g=\mu_\alpha*g$. Hence, $ \int_G f (\mu_\alpha*g) dt = \int_G (\mu_\alpha*f) g dt=0$ for all $ g$ in $ L^1(G)$, from which it follows that $ \mu_\alpha *f=0$. The converse is proved similarly, and we omit the details.

Aiming for a characterization of weakly analytic measures in terms of their spectra, we present one more result.

Proposition 3.7   Let $ \mu$ be weakly measurable in $ M(\Sigma)$.
(i) Suppose that $ B$ is a nonvoid closed subset of $ \Gamma$ and $ {\rm spec}_T\mu\subset B$. Then $ \sigma[t\mapsto T_t\mu(A)]\subset B$ for all $ A\in \Sigma$.
(ii) Conversely, suppose that $ B$ is an $ S$-set in $ \Gamma$ and that $ \sigma[t\mapsto T_t\mu(A)]\subset B$ for all $ A\in \Sigma$, then $ {\rm spec}_T\mu\subset B$.

Proof. We clearly have $ {\cal I}(\mu)\subset {\cal I}([t\mapsto T_t\mu(A)])$. Hence, $ {\rm spec}_T \mu=Z({\cal I}(\mu))\supset
Z({\cal I}([t\mapsto T_t\mu(A)]))=\sigma[t\mapsto T_t\mu(A)],$ and (i) follows.
Now suppose that $ B$ is an $ S$-set and let $ g\in L^1(G)$ be such that $ \widehat{g}= 0$ on $ -B$. Then, for all $ A\in \Sigma$, we have from Proposition 3.4:

$\displaystyle \int_G g (t) T_t\mu (A)d t = 0.$

Equivalently, we have that

$\displaystyle \int_G g(-t) T_{-t}\mu(A) dt =0.$

Since the Fourier transform of the function $ t\mapsto g(-t)$ vanishes on $ B$, we see that $ {\cal I}(\mu) \supset \{f:\ \ \widehat{f}=0\ {\rm on}\ B\}$. Thus $ Z({\cal I}(\mu))\subset
Z(\{f:\ \ \widehat{f}=0\ {\rm on} \ B\})=B$, which completes the proof.

Straightforward applications of Propositions 3.5 and 3.7 yield the desired characterization of weakly analytic measures.

Corollary 3.8   Suppose that $ \mu\in{\cal M}_T(\Sigma)$. Then,
(i) $ \mu$ is weakly $ T-$analytic if and only if $ {\rm spec}_T\mu\subset \overline{P}$ if and only if $ \sigma [t\mapsto T_t\mu (A)]\subset \overline{P}$, for every $ A\in \Sigma$;
(ii) $ {\rm spec}_T\mu\subset S_\alpha $ if and only if $ \sigma[t\mapsto T_t\mu(A)]\subset S_\alpha$ for every $ A\in \Sigma$.
(iii) $ {\rm spec}_T\mu\subset C_\alpha $ if and only if $ \sigma[t\mapsto T_t\mu(A)]\subset C_\alpha$ for every $ A\in \Sigma$.
(iv) $ {\rm spec}_T\mu\subset \Gamma\setminus C_\alpha $ if and only if $ \sigma[t\mapsto T_t\mu(A)]\subset \Gamma\setminus C_\alpha$ for every $ A\in \Sigma$.

The remaining results of this section are simple properties of measures in $ {\cal M}_T(\Sigma)$ that will be needed later. Although the statements are direct analogues of classical facts about measures on groups, these generalization require in some places the sup path attaining property of $ T$.

Proposition 3.9   Suppose that $ \mu\in{\cal M}_T(\Sigma)$ and $ \nu \in M(G)$. Then $ {\rm spec}_T \nu*_T\mu$ is contained in the support of $ \widehat{\nu}$, and $ {\rm spec}_T \nu*_T\mu\subset {\rm spec}_T \mu$.

Proof. Given $ \chi_0$ not in the support of $ \widehat{\nu}$, to conclude that it is also not in the spectrum of $ \nu*_T\mu$ it is enough to find a function $ f$ in $ L^1(G)$ with $ \widehat{f}(\chi_0)=1$ and $ f*_T(\nu*_T\mu)=0$. Simply choose $ f$ with Fourier transform vanishing on the support of $ \widehat{\nu}$ and taking value 1 at $ \chi_0$. By Fourier inversion, we have $ f*\nu=0$, and since $ f*_T(\nu*_T\mu)=(f*\nu)*_T\mu$, the first part of the proposition follows. For the second part, we have $ {\cal I}(\mu)\subset {\cal I}(\nu*_T\mu)$, which implies the desired inclusion.

We next prove a property of $ L^\infty(G)$ functions similar to the characterization of $ L^1$ functions which are constant on cosets of a subgroup [21, Theorem (28.55)].

Proposition 3.10   Suppose that $ f$ is in $ L^\infty(G)$ and that $ \Lambda$ is an open subgroup of $ \Gamma$. Let $ \lambda_0$ denote the normalized Haar measure on the compact group $ A(G,\Lambda)$, the annihilator in $ G$ of $ \Lambda$ (see [20, (23.23)]. Then, $ \sigma[f]\subset \Lambda$ if and only if $ f=f*\lambda_0$ a. e. This is also the case if and only if $ f$ is constant on cosets of $ A(G,\Lambda)$.

Proof. Suppose that the spectrum of $ f$ is contained in $ \Lambda$. Since $ \Lambda$ is an $ S$-set, it follows that $ f$ is the weak-* limit of trigonometric polynomials with spectra contained in $ \Lambda$. Let $ \{f_\alpha\}$ be a net of such trigonometric polynomials converging to $ f$ weak-*. Note that, for any $ \alpha$, we have $ \lambda_0*f_\alpha=f_\alpha$. For $ g$ in $ L^1(G)$, we have

$\displaystyle \lim_\alpha \int_G f_\alpha \overline{g}dx= \int_G f\overline{g}dx.$

In particular, we have

$\displaystyle \lim_\alpha \int_G f_\alpha
(\lambda_0*\overline{g})dx= \int_G f
(\lambda_0*\overline{g})dx,$

and so

$\displaystyle \lim_\alpha \int_G (f_\alpha *\lambda_0)
\overline{g}dx= \int_G (f*\lambda_0)
\overline{g}dx.$

Since this holds for any $ g$ in $ L^1(G)$, we conclude that $ \lambda_0 *f_\alpha$ converges weak-* to $ \lambda_0*f$. But $ \lambda_0*f_\alpha=f_\alpha$, and $ f_\alpha$ converges weak-* to $ f$, hence $ f*\lambda_0=f$. The remaining assertions of the lemma are easy to prove. We omit the details.

In what follows, we use the symbol $ \mu_\alpha$ to denote the normalized Haar measure on the compact subgroup $ A(G,C_\alpha)$, the annihilator in $ G$ of $ C_\alpha$. This measure is also characterized by its Fourier transform:

$\displaystyle \widehat{\mu_\alpha}=1_{C_\alpha}$

(see [20, (23.19)]).

Corollary 3.11   Suppose that $ \mu\in{\cal M}_T(\Sigma)$. Then,
(i) $ {\rm spec}_T\mu\subset C_\alpha $ if and only if $ \mu=\mu_\alpha *_T \mu$;
(ii) $ {\rm spec}_T\mu\subset \Gamma\setminus C_\alpha $ if and only if $ \mu_\alpha *_T \mu=0$.

Proof. (i) If $ \mu=\mu_\alpha *_T \mu$, then, by Proposition 3.10, $ \sigma[t\mapsto \mu_\alpha*_T T_t\mu(A)]\subset C_\alpha$. Hence by Corollary 3.8, $ {\rm spec}_T\mu\subset C_\alpha $. For the other direction, suppose that $ {\rm spec}_T\mu\subset C_\alpha $. Then by Corollary 3.8 we have that the spectrum of the function $ t\mapsto T_t\mu(A)$ is contained in $ C_\alpha$ for every $ A\in \Sigma$. By Proposition 3.10, we have that

$\displaystyle T_t\mu(A)=\int_{G_\alpha}T_{t-y}\mu (A) d\mu_\alpha=
T_t(\mu_\alpha*\mu)(A)$

for almost all $ t\in G$. Since this holds for all $ A\in \Sigma$, the desired conclusion follows from Proposition 1.4.
Part (ii) follows from Corollary 3.6 (ii), Proposition 3.7(ii), and the fact that $ \Gamma\setminus C_\alpha$ is an $ S$-set.

Corollary 3.12   Suppose that $ \mu\in{\cal M}_T(\Sigma)$ and $ {\rm spec}_T\mu\subset C_\alpha $, and let $ y\in G_\alpha=A(G,C_\alpha)$. Then $ T_y\mu=\mu$.

Proof. For any $ A\in \Sigma$, we have from Corollary 3.11
  $\displaystyle T_y\mu (A)$ $\displaystyle =$ $\displaystyle T_y(\mu_\alpha *\mu)(A)=\mu_\alpha*T_y\mu(A)$
    $\displaystyle =$ $\displaystyle \int_{G_\alpha} T_{y-x} \mu (A) d\mu_\alpha (y)$
    $\displaystyle =$ $\displaystyle \int_{G_\alpha}T_{-x}\mu(A) d\mu_\alpha (y)
=\mu_\alpha *\mu(A)=\mu(A).$


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Next: Homomorphism theorems Up: Decomposition of analytic measures Previous: Orders on locally compact
Stephen Montgomery-Smith 2002-10-30