Decomposition of Analytic Measures

Define measures $\mu_{\alpha_0}$ and $d_\alpha$ by their Fourier transforms: $\widehat{\mu_{\alpha_0}}=1_{C_{\alpha_0}}$, and $\widehat{d_\alpha}=1_{C_\alpha\setminus D_\alpha}$. Then we have the following decomposition theorem.

Theorem 5.1   Let $G$ be a locally compact abelian group with an ordered dual group $\Gamma$. Suppose that $T$ is a sup path attaining representation of $G$ in $M(\Sigma)$. Then for any weakly analytic measure $\mu\in M(\Sigma)$ we have that the set of $\alpha$ for which $d_\alpha *_T\mu \ne 0$ is countable, and that

$$\mu = \mu_{\alpha_0}*_T\mu+ \sum_\alpha d_\alpha *_T\mu ,$$ (32)

where the right side converges unconditionally in norm in $M(\Sigma)$. Furthermore, there is a positive constant $c$, depending only upon $T$, such that for any signs $\epsilon_\alpha = \pm 1$ we have

$$\left\Vert \sum_\alpha \epsilon_\alpha d_\alpha*_T\mu \right\Vert \le c \Vert\mu\Vert .$$ (33)

One should compare this theorem to the well-known results from Littlewood-Paley theory on $L^p(G)$, where $1<p<\infty$ (see Edwards and Gaudry [11]). For $L^p(G)$ with $1<p<\infty$, it is well-known that the subgroups $(C_\alpha)$ form a Littlewood-Paley decomposition of the group $\Gamma$, which means that the martingale difference series

$$f= \mu_{\alpha_0}*f+ \sum_\alpha d_\alpha *f$$

converges unconditionally in $L^p(G)$ to $f$. Thus, Theorem 5.1 above may be considered as an extension of Littlewood-Paley Theory to spaces of analytic measures.

The next result, crucial to our proof of Theorem 5.1, is already known in the case that $G = \Bbb T^n$ with the lexicographic order on the dual. This is due to Garling [15], and is a modification of the celebrated inequalities of Burkholder. Our result can be obtained directly from the result in [15] by combining the techniques of [3] with the homomorphism theorem 4.5. However, we shall take a different approach, in effect reproducing Garling's proof in this more general setting.

Theorem 5.2   Suppose that $G$ is a locally compact group with ordered dual $\Gamma$. Then for $f\in H^1(G)$, for any set $\{\alpha_j\}_{j=1}^n$ of indices less than $\alpha_0$, and for any numbers $\epsilon_j \in \{0,\pm 1\}$ ( $1 \le j \le n$), there is an absolute constant $a>0$ such that

$$\left\Vert \sum_{j=1}^n \epsilon_j d_{\alpha_j} *f \right\Vert _1 \leq a \Vert f\Vert _1.$$ (34)

Furthermore,

$$f = \mu_{\alpha_0}*f + \sum_\alpha d_\alpha*f,$$ (35)

where the right hand side converges unconditionally in the norm topology on $H^1(G)$.

Proof. The second part of Theorem 5.2 follows easily from the first part and Fourier inversion.

Now let us show that if we have the result for compact $G$, then we have it for locally compact $G$. Let $\pi_{\alpha_0} :\ \Gamma\rightarrow \Gamma/C_{\alpha_0}$ denote the quotient homomorphism of $\Gamma$ onto the discrete group $\Gamma/C_{\alpha_0}$ (recall that $C_{\alpha_0}$ is open), and define a measurable order on $\Gamma/C_{\alpha_0}$ to be $\pi_{\alpha_0}(P)$. By Remarks 2.2 (c), the decomposition of the group $\Gamma/C_{\alpha_0}$ that we get by applying Theorem 2.1 to that group, is precisely the image by $\pi_{\alpha_0}$ of the decomposition of the group $\Gamma$. Let $G_0$ denote the compact dual group of $\Gamma/C_{\alpha_0}$. Thus if Theorem 5.2 holds for $H^1(G_0)$, then applying Theorem 4.5, we see that Theorem 5.2 holds for $G$.

Henceforth, let us suppose that $G$ is compact. We will suppose that the Haar measure on $G$ is normalized, so that $G$ with Haar measure is a probability space.

Since each one of the subgroups $C_\alpha$, and $D_\alpha$ ( $\alpha< \alpha_0$) is open, it follows that their annihilators in $G$, $G_\alpha=A(G,C_\alpha)$, and $A(G,D_\alpha)$, are compact. Let $\mu_\alpha$ and $\nu_\alpha$ denote the normalized Haar measures on $A(G,C_\alpha)$ and $A(G,D_\alpha)$, respectively. We have $\widehat{\mu}_\alpha=1_{C_\alpha}$ (for all $\alpha$), and $\widehat{\nu}_\alpha=1_{D_\alpha}$ (for all $\alpha\neq\alpha_0$), so that $d_\alpha=\mu_\alpha -\nu_\alpha$.

For each $\alpha$, let ${\cal B}_\alpha$ denote the $\sigma$-algebra of subsets of $G$ of the form $A+G_\alpha$, where $A$ is a Borel subset of $G$. We have ${\cal B}_{\alpha_1}\subset {\cal B}_{\alpha_2}$, whenever $\alpha_1>\alpha_2$. It is a simple matter to see that for $f\in L^1(G)$, the conditional expectation of $f$ with respect to ${\cal B}_\alpha$ is equal to $\mu_\alpha*f$ (see [11, Chapter 5, Section 2]).

We may suppose without loss of generality that $\alpha_1>\alpha_2>\ldots>\alpha_n$. Thus the $\sigma$-algebras ${\cal B}_{\alpha_k}$ form a filtration, and the sequence $(d_{\alpha_1}*f, d_{\alpha_2}*f,\ldots,d_{\alpha_n}*f)$ is a martingale difference sequence with respect to this filtration.

In that case, we have the following result of Burkholder [7, Inequality (1.7)], and [8]. If $0<p<\infty$, then there is a positive constant $c$, depending only upon $p$, such that

$$\left\Vert \sup_{1 \le k \le n} \left( \sum_{j=1}^k \epsilon_j d_... ... \sup_{1 \le k \le n} \left( \sum_{j=1}^k d_{\alpha_j}*f\right) \right\Vert _p.$$ (36)

Lemma 5.3   For any index $\alpha$, $0<p<\infty$, and $f\in H^1(G)\cap L^p(G)$, we have almost everywhere on $G$

$$\left\vert \mu_\alpha *f \right\vert^p \leq \mu_\alpha*\left\vert f\right\vert^p,$$ (37)

where $\mu_\alpha$ is the normalized Haar measure on the compact subgroup $G_\alpha=A(G,C_\alpha)$.

Proof. The dual group of $G_\alpha$ is $\Gamma/C_\alpha$ and can be ordered by the set $\pi_\alpha (P)$, where $\pi_\alpha$ is the natural homomorphism of $\Gamma$ onto $\Gamma/C_\alpha$.

Next, by convolving with an approximate identity for $L^1(G)$ consisting of trigonometric polynomials, we may assume that $f$ is a trigonometric polynomial. Then we see that for each $x \in G$ that the function $y\mapsto f(x+y)$, $y\in G_\alpha$, is in $H^1(G_\alpha)$. To verify this, it is sufficient to consider the case when $f$ is a character in $P$. Then

$$f(x+y)= f(x) \pi_\alpha (f)(y),$$

and by definition $\pi_\alpha(f)$ is in $H^1(G_\alpha)$.

Now we have the following generalization of Jensen's Inequality, due to Helson and Lowdenslager [16, Theorem 2]. An independent proof based on the ideas of this section is given in [3]. For all $g\in H^1(G)$

$$\left\vert\int_G g(x) d x\right\vert \leq \exp \int_G\log \vert g(x)\vert d x.$$ (38)

Apply (38) to $y\mapsto f(x+y)$, $y\in G_\alpha$ to obtain

$$\left\vert\int_{G_\alpha} f(x+y) d \mu_\alpha(y)\right\vert \leq \exp \int_{G_\alpha}\log \vert f(x+y)\vert d \mu_\alpha(y).$$

Extending the integrals to the whole of $G$ (since $\mu_\alpha$ is supported on $G_\alpha$), raising both sides to the $p$th power, and then applying the usual Jensen's inequality for the logarithmic function on finite measure spaces, we obtain

$$\left\vert\int_G f(x+y) d \mu_\alpha(y)\right\vert^p \leq \exp \int_G \log \vert f(x+y)\vert^p d \mu_\alpha(y)$$

$$\leq \int_G \vert f(x+y)\vert^pd \mu_\alpha(y).$$

Changing $y$ to $-y$, we obtain the desired inequality.

Let us continue with the proof of Theorem 5.2. We may suppose that $f$ is a mean zero trigonometric polynomial, and that the spectrum of $f$ is contained in $\bigcup_{j=1}^n C_{\alpha_j} \setminus D_{\alpha_j}$, that is to say

$$f = \sum_{j=1}^n d_{\alpha_j} * f .$$

By Lemma 5.3, we have that

$$\sup_{1\leq k\leq n} \left\vert \mu_{\alpha_k}*f \right\vert = \left( \sup_{1\leq k\leq n} \left\vert \mu_{\alpha_k}*f \right\vert^{1/2}\right)^2$$

$$\leq \left( \sup_{1\leq k\leq n} \mu_{\alpha_k}*\vert f\vert^{1/2} \right)^2.$$ (39)

Also, we have that $(\mu_{\alpha_j}*\vert f\vert^{1/2})_{j=1}^n$ is a martingale with respect to the filtration $({\cal B}_j)_{j=1}^n$. Hence, by Doob's Maximal Inequality [10, Theorem (3.1), p. 317] we have that

$$\left\Vert \sup_{1\leq k\leq n'} \mu_{\beta_k}*\vert f\vert^{1/2} \right\Vert _2^2 \leq 4 \left\Vert \mu_{\beta_{n'}}*\vert f\vert^{1/2} \right\Vert _2^2$$

$$\leq 4 \left\Vert \vert f\vert^{1/2}\right\Vert _2^2 = 4\Vert f\Vert _1.$$ (40)

The desired inequality follows now upon combining Burkholder's Inequality (36) with (39), and (40).

Proof of Theorem 5.1. Transferring inequality (34) by using Theorem 1.8, we obtain that for any set $\{\alpha_j\}_{j=1}^n$ of indices less than $\alpha_0$, and for any numbers $\epsilon_j \in \{0,\pm 1\}$ ( $1 \le j \le n$), there is a positive constant $c$, depending only upon the representation $T$, such that

$$\left\Vert \sum_{j=1}^n \epsilon_j d_{\alpha_j} *_T \mu \right\Vert \leq c \Vert\mu\Vert.$$ (41)

Now suppose that $\{\alpha_j\}_{j=1}^\infty$ is a countable collection of indices less than $\alpha_0$. Then by Bessaga and Peczynski [5], the series $\sum_{j=1}^\infty d_{\alpha_j} *_T \mu$ is unconditionally convergent. In particular, for any $\delta>0$, for only finitely many $k$ do we have that $\Vert d_{\alpha_k} *_T \mu \Vert > \delta$. Since this is true for all such countable sets, we deduce that the set of $\alpha$ for which $d_\alpha *_T\mu \ne 0$ is countable.

Hence we have that $\sum_\alpha d_{\alpha} *_T \mu$ is unconditionally convergent to some measure, say $\nu$. Clearly $\nu$ is weakly measurable. To prove that $\mu=\nu$, it is enough by Proposition 1.4 to show that for every $A\in \Sigma$, we have $T_t\mu(A)=T_t\nu(A)$ for almost all $t\in G$.

We first note that since for every $f\in L^1(G)$ the series $\mu_{\alpha_0}*f+ \sum_\alpha d_\alpha *f$ converges to $f$ in $L^1(G)$, it follows that, for every $g\in L^\infty(G)$, the series $\mu_{\alpha_0}*g+ \sum_\alpha d_\alpha *g$ converges to $g$ in the weak-* topology of $L^\infty(G)$. Now on the one hand, for $t\in G$ and $A\in \Sigma$, we have $\mu_{\alpha_0}*_TT_t\mu(A)+ \sum_\alpha d_\alpha *_T T_t\mu(A)=T_t\nu(A)$, because of the (unconditional) convergence of the series $\mu_{\alpha_0}*_T\mu+ \sum_\alpha d_\alpha *_T\mu$ to $\nu$. On the other hand, by considering the $L^\infty(G)$ function $t\mapsto T_t(A)$, we have that $\mu_{\alpha_0}*_TT_t\mu(A)+ \sum_\alpha d_\alpha *_T T_t\mu(A)= \mu_{\alpha_0}*T_t\mu(A)+ \sum_\alpha d_\alpha * T_t\mu(A)=T_t\mu(A)$, weak *. Thus $T_t\mu(A)=T_t\nu(A)$ for almost all $t\in G$, and the proof is complete.