Transference in Spaces of Analytic measures

In the remainder of this paper, we transfer a result from Littlewood-Paley theory in $H^1(\mathbb{R})$ to spaces of measures on which $\mathbb{R}$ is acting. We then show how this transferred result implies with ease several of the main results of Bochner [3], de Leeuw and Glicksberg [6], and Forelli [9].

We start by setting our notation. Let

$$H^\infty(\mathbb{R})=\left\{f\in L^\infty(\mathbb{R}):\ \int_\mathbb{R}f(t)g(t)\, dt=0\ \mbox{for all}\ g\in H^1(\mathbb{R})\right\}\, .$$

Let $M(\Sigma)=M(\O ,\Sigma)$ denote a space of measures and let $T=(T_t)_{t\in\mathbb{R}}$ denote a sup path attaining representation of $\mathbb{R}$ by isomorphisms of $M(\Sigma)$. According to Forelli [9], a measure $\mu\in M(\Sigma)$ is called $T$-analytic if ${\rm spec}_T(\mu)\subset [0,\infty)$. We now introduce another equivalent definition.

Definition 4.1   Suppose that $T$ is a sup path attaining representation of $\mathbb{R}$ 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 $H^\infty(\mathbb{R})$ for every $A\in \Sigma$.

It is easy to see that if $\mu$ is $T$-analytic then it is weakly analytic. The converse is also true. The proof is based on the fact that $[0,\infty)$ is a set of spectral synthesis (see [1, Proposition 1.7]).

For $n\in\mathbb{Z}$, let $m_n\in L^1(\mathbb{R})$ be the function whose Fourier transform is piecewise linear and satisfies

$$\widehat{m_n}(s)=\left\{ \begin{array}{ll} 0 & \mbox{if $s\not\in [2^{n-1},2^{n+1}]$;}\\ 1 & \mbox{if $s=2^n$.} \end{array} \right.$$ (22)

Let $h\in L^1(\mathbb{R})$ be the function whose Fourier transform is piecewise linear and satisfies

$$\widehat{h}(s)=\left\{ \begin{array}{ll} 0 & \mbox{if $s\not\in [-1,1]$,}\\ 1 & \mbox{if $\vert s\vert\leq \frac{1}{2}$.} \end{array} \right.$$ (23)

It is easy to check that

$$\widehat{h}(s)+\sum^\infty_{n=0} \widehat{m_n}(s) =\left\{ \begin... ...mbox{if $s\geq -\frac{1}{2}$,}\\ 0 & \mbox{if $s\leq -1 $,} \end{array} \right.$$ (24)

and that the left side of (24) is continuous and piecewise linear. The following theorem is a consequence of standard facts from Littlewood-Paley theory. We postpone its proof to the end of this section.

Theorem 4.2 (i)   Let $h$ and $m_n$ be as above, $f$ be any function in $H^1(\mathbb{R})$, and $N$ be any nonnegative integer. Then there is a positive constant $a$, independent of $f$ and $N$, such that

$$\Vert h*f+ \sum^N_{n=0} \epsilon_n m_n*f\Vert _1 \leq a \Vert f\Vert _1$$ (25)

for any choice of $\epsilon_n=-1$ or $1$. (ii) For $f\in H^1(\mathbb{R})$, we have

$$h*f+ \lim_{N\rightarrow\infty} \sum^N_{n=0} m_n*f=f$$

unconditionally in $L^1(\mathbb{R})$.
(iii) For $f\in H^\infty (\mathbb{R})$, we have

$$h*f+ \lim_{N\rightarrow\infty} \sum^N_{n=0} m_n*f=f$$ (26)

almost everywhere on $\mathbb{R}$.

Our main theorem is the following.

Theorem 4.3   Let $T$ be a representation of $\mathbb{R}$ in $M(\Sigma)$ that is sup path attaining, and let $h$ and $m_n, n=0,1 ,2 , \ldots$ be as in Theorem 4.2. Suppose that $\mu\in M(\Sigma)$ is weakly analytic. Then

$$\Vert h*_T\mu + \sum_{n=0}^N\epsilon_n m_n*_T \mu\Vert\leq a c^3 C\Vert\mu\Vert$$ (27)

for any choice of $\epsilon_n\in \{-1,1\}$, where $a$ is as in (25), $c$ as in (1) and $C$ as in (7). Moreover,

$$\mu=h*_T\mu+\sum_{n=0}^\infty m_n*_T \mu,$$ (28)

where the series converges unconditionally in $M(\Sigma)$.

Proof. To prove (27), combine Theorems 1.6 and 4.2. Inequality (27) states that the partial sums of the series $h*_T \mu+\sum_{n=0}^\infty m_n*_T \mu$ are unconditionally bounded. To prove that they converge unconditionally, we recall the Bessaga-Peczynski Theorem from [2]. This theorem tells us that for any Banach space, every unconditionally bounded series is unconditionally convergent if and only if the Banach space does not contain an isomorphic copy of $c_0$. Now since $M(\Sigma)$ is weakly complete and $c_0$ is not (see [7, Chap IV.9, Theorem 3, and IV.13.9]), we conclude that $M(\Sigma)$ does not contain $c_0$. Applying the Bessaga-Peczynski Theorem, we infer that there is a measure $\eta\in M(\Sigma)$ such that

$$\eta=h*_T\mu+\sum_{n=0}^\infty m_n*_T \mu$$ (29)

unconditionally in $M(\Sigma)$. Moreover, $\eta$ is weakly measurable, because of (29). It remains to show that $\eta=\mu$. By Proposition 2.4, it is enough to show that for every $A\in \Sigma$, we have

$$T_t\mu(A)=T_t\eta(A)$$ (30)

for almost every $t \in \mathbb{R}$. Since $\mu$ is weakly analytic, the function $t\mapsto T_t\mu(A)$ is in $H^\infty(\mathbb{R})$. By Theorem 4.2 (iii), we have

$$T_t\mu(A)=h*T_t\mu(A)+\sum_{n=0}^\infty m_n* T_t \mu(A)$$ (31)

for almost every $t \in \mathbb{R}$. On the other hand, from the unconditional convergence of the series in (29), (1), and (3), it follows that

$$T_t\eta(A)=h*T_t\mu(A)+\sum_{n=0}^\infty m_n* T_t \mu(A)$$ (32)

for all $t \in \mathbb{R}$. Comparing (31) and (32), we see that (30) holds, completing the proof of the theorem.

Remarks 4.4   It is interesting to note that Theorem 4.3 implies the classical F. and M. Riesz theorem for measures defined on the real line. To see this, consider the representation of $\mathbb{R}$ acting by translation on the Banach space of complex regular Borel measures on $\mathbb{R}$. It is easy to see that a regular Borel measure is analytic if and only if its Fourier-Stieltjes transform is supported in $[0,\infty)$. In this case, each term in (28) belongs to $L^1(\mathbb{R})$, being the convolution of an $L^1(\mathbb{R})$ function with a regular Borel measure. Thus, the unconditional convergence of the series in (28) implies that the measure $\mu$ is absolutely continuous.

This argument provides a new proof of the F. and M. Riesz Theorem, based on Littlewood-Paley theory and the result of Bessaga-Peczynski [2]. Also, it can be used to prove the following version of Bochner's generalization of the F. and M. Riesz Theorem.

Theorem 4.5   Suppose that $G$ is a locally compact abelian group with dual group $\Gamma$, and $\psi:\ \Gamma \rightarrow \mathbb{R}$ is a continuous homomorphism. Suppose that $\nu \in M(G)$ is such that, for every real number $s$, $\psi^{-1}((-\infty,s])\cap {\rm supp}(\widehat{\nu})$ is compact. Then $\nu$ is absolutely continuous with respect to Haar measure on $G$. That is, $\nu\in L^1(G)$.

Proof. Let $\phi:\ \mathbb{R}\rightarrow G$ denote the continuous adjoint homomorphism of $\psi$. Define a representation $T=(T_t)_{t\in\mathbb{R}}$ of $\mathbb{R}$ on the regular Borel measures $M(G)$ by

$$T_t(\mu)(A)=\mu(A+\phi(t)),$$

for all $\mu\in M(G)$ and all Borel subsets $A\subset G$. By Example 2.2, $T$ is sup path attaining, and every measure $\mu\in M(G)$ is weakly measurable. Moreover, $\mu\in M(G)$ is weakly analytic (equivalently, $T$-analytic) if and only if ${\rm supp}(\widehat{\mu})\subset \psi^{-1}([0,\infty))$ (see [6]).

To prove the theorem, we can, without loss of generality, suppose that ${\rm supp}(\widehat{\nu})\subset \psi^{-1}([0,\infty))$. Otherwise, we consider the measure $\overline{\chi} \nu$, where $\psi(\chi) +\psi({\rm supp}(\widehat{\nu}))\subset [0,\infty)$.

Let $S=\psi^{-1}([0,\infty))$. Then $S$ is a ${\cal T}$-set. Applying Theorem 4.3, we see that

$$\nu=h*_T\nu +\sum_{n=1}^\infty m_n*_T \nu,$$ (33)

unconditionally in $M(G)$. For $f\in L^1(\mathbb{R})$, a straightforward calculation shows that $\widehat{f*_T\nu}(\chi)=\widehat{f}(\psi (\chi))\widehat{\nu}(\chi)$ for all $\chi\in\Gamma$. Since $\psi^{-1}((-\infty,s])\cap {\rm supp}(\widehat{\nu})$ is compact for every $s\in \mathbb{R}$, it follows that ${\rm supp}(\widehat{h*_T\nu})$ and ${\rm supp}(\widehat{m_n*_T\nu})$ are compact. Thus $h*_T\nu$ and $m_n*_T\nu$ are in $M(G)\cap L^2(G)$, and hence they belong to $L^1(G)$. As a consequence, (33) implies that $\nu\in L^1(G)$.

With Theorem 4.3 in hand, we can derive with ease several fundamental properties of analytic measures that were obtained previously by de Leeuw-Glicksberg [6], and Forelli [9]. We note however, that the techniques in [6] and [9] do not apply in our more general settings.

Theorem 4.6   Let $T$ be a representation of $\mathbb{R}$ in $M(\Sigma)$ that is sup path attaining, and let $\mu$ be a weakly analytic measure in $M(\Sigma)$. Then the mapping $t\mapsto T_t \mu$ is continuous from $\mathbb{R}$ into $M(\Sigma)$.

Proof. Using the uniform continuity of translation in $L^1(\mathbb{R})$, it is a simple matter to show that for any function $f\in L^1(\mathbb{R})$, and any weakly measurable $\mu\in M(\Sigma)$, the mapping $t\mapsto f*_T T_t\mu$ is continuous. Now use Theorem 4.3 to complete the proof.

Theorem 4.6 is very specific to representations of $\mathbb{R}$ or $\mathbb{T}$, in the sense that no similar result holds on more general groups. To see this, consider the group $G=\mathbb{T}\times \mathbb{T}$ with a lexicographic order on the dual group $\mathbb{Z}\times \mathbb{Z}$. Let $\mu_0$ denote the normalized Haar measure on the subgroup $\{(x,y):\ y=0\}$, and consider the measure $e^{-i x}\mu_0$. Its spectrum is supported on the coset $\{(m,1):\ m\in \mathbb{Z}\}$ and thus it is analytic with respect to the regular action of $G$ by translation in $M(G)$. Clearly, the measure $e^{-i x}\mu_0$ does not translate continuously, and so a straightforward analog of Theorem 4.6 fails in this setting.

The following application concerns bounded operators ${\cal P}$ from $M(\Sigma)$ into $M(\Sigma)$ that commute with $T$ in the following sense:

$${\cal P}\circ T_t=T_t\circ {\cal P}$$

for all $t \in \mathbb{R}$.

Theorem 4.7   Suppose that $T$ is a representation of $\mathbb{R}$ that is sup path attaining, and that ${\cal P}$ commutes with $T$. Let $\mu\in M(\Sigma)$ be weakly analytic. Then ${\cal P}\mu$ is also weakly analytic.

Proof. First note that by Theorem 4.6, the mapping $t\mapsto T_t\mu(A)$ is continuous, and hence measurable.

Now suppose that $g \in H^1(\mathbb{R})$. Again, by Theorem 4.6, the map $t\mapsto g(t) T_t\mu$ is Bochner integrable. Let

$$\nu = \int_\mathbb{R}g(t) T_t \mu dt .$$

Then by properties of the Bochner integral, and since $\mu$ is weakly analytic, we have that for all $A\in \Sigma$

$$\nu(A) = \int_\mathbb{R}g(t) T_t\mu(A) dt = 0 .$$

Hence $\nu = 0$. Therefore, for all $A\in \Sigma$ we have

$$\int_\mathbb{R}g(t) T_t({\cal P}\mu)(A) dt = \int_\mathbb{R}g(t) {\cal P}(T_t\mu)(A) dt = {\cal P}\nu(A) = 0.$$

Since this is true for all $g \in H^1(\mathbb{R})$, it follows that ${\cal P}\mu$ is weakly analytic.

Definition 4.8   Let $T$ be a sup path attaining representation of $G$ in $M(\Sigma)$. A weakly measurable $\sigma$ in $M(\Sigma)$ is called quasi-invariant if $T_t\sigma$ and $\sigma$ are mutually absolutely continuous for all $t\in G$. Hence if $\sigma$ is quasi-invariant and $A\in \Sigma$, then $\vert\sigma\vert(A)=0$ if and only if $\vert T_t(\sigma)\vert(A)=0$ for all $t\in G$.

We can use Theorem 4.7 to generalize a result of de Leeuw-Glicksberg [6] and Forelli [9], concerning quasi-invariant measures. In this application, it is necessary to restrict to sup path attaining representations given by isometries of $M(\Sigma)$. We need a lemma.

Lemma 4.9   Suppose that $T$ is a linear isometry of $M(\Sigma)$ onto itself. Let $\mu$, $\nu\in M(\Sigma)$. Then,
(a) $\mu$ and $\sigma$ are mutually singular (in symbols, $\mu\perp\sigma$) if and only if $T\mu\perp T\sigma$;
(b) $\mu<<\sigma$ if and only if $T\mu<<T\sigma.$

Proof. For (a), simply recall that two measures $\mu$ and $\sigma$ are mutually singular if and only if $\Vert\mu+\sigma\Vert=\Vert\mu\Vert+\Vert\sigma\Vert$, and $\Vert\mu-\sigma\Vert=\Vert\mu\Vert+\Vert\sigma\Vert$. For (b), it is clearly enough to prove the implication in one direction. So suppose that $\mu<<\sigma$ and write $\mu=\mu_1+\mu_2$ where $T\mu_1<<T\sigma$ and $T\mu_2\perp T\sigma$. Then $T\mu_1\perp T\mu_2$. Hence $\mu_1\perp \mu_2$, and hence $\mu_2<<\mu<<\sigma$. But $T\mu_2\perp T\sigma$ implies that $\mu_2\perp \sigma$. So $\mu_2=0$. Thus $T\mu=T\mu_1<<T\sigma$.

Theorem 4.10   Suppose that $T$ is a sup path attaining representation of $\mathbb{R}$ by isometries of $M(\Sigma)$. Suppose that $\mu\in M(\Sigma)$ is weakly analytic, and $\sigma$ is quasi-invariant. Write $\mu=\mu_a+\mu_s$ for the Lebesgue decomposition of $\mu$ with respect to $\sigma$. Then both $\mu_a$ and $\mu_s$ are weakly analytic. In particular, the spectra of $\mu_a$ and $\mu_s$ are contained in $[0,\infty)$.

Proof. Let ${\cal P}(\mu)=\mu_s$. Since $\sigma$ is quasi-invariant, the operator ${\cal P}$ commutes with $T$ by Lemma 4.9. Now apply Theorem 4.7.

Let us finish with an example to show that the hypothesis of sup path attaining is required in these results. The next example is a variant of Example 2.5.

Example 4.11   Let $\Sigma_1$ denote the sigma algebra of countable and co-countable subsets of $\mathbb{R}$, let $\Sigma_2$ denote the Borel subsets of $\mathbb{R}$, and let $\Sigma = \Sigma_1 \otimes \Sigma_2$ denote the product sigma algebra on $\mathbb{R}\times \mathbb{R}$. Let $\nu_1:\Sigma_1 \to \{0,1\}$ be the measure that takes countable sets to 0 and co-countable sets to $1$, let $\delta_t:\Sigma_1\to\{0,1\}$ be the measure that takes sets to $1$ if they contain $t$, and to 0 otherwise, and let $\nu_2$ denote the measure on $\Sigma_2$ given by

$$\nu_2(A) = \int_A \exp(-x^2) \, dx .$$

Let $\nu = \nu_1\otimes \nu_2$, let $\theta = \delta_0\otimes\nu_2$, and let $\mu = \nu-\theta$. Finally, let $T_t$ be the representation given by $T_t(x,y) = (x+t,y+t)$.

Then, we see that $\nu$ is quasi-invariant, and that $\theta$ and $\nu$ are mutually singular. Arguing as in Example 2.5, we see that $\mu$ is weakly analytic. However, the singular part of $\mu$ with respect to $\nu$ is $-\theta$, and it may be readily seen that this is not weakly analytic, for example

$$T_t\theta(\mathbb{R}\times[-1,1]) = \int_{-1}^1 \exp(-(x-t)^2) \, dx ,$$

is not in $H^\infty(\mathbb{R})$.

We end this section by proving Theorem 4.2. We have

$$\sum^\infty_{n=-\infty} \widehat{m_n}(s) =\left\{ \begin{array}{ll} 1 & \mbox{if $s>0$;}\\ 0 & \mbox{if $s\leq 0$.} \end{array} \right.$$ (34)

Recall the Fejér kernels $\{k_a\}_{a>0}$, where

$$\widehat{k_a}(s) =\left\{ \begin{array}{ll} 1-\frac{\vert s\vert}{a} & \mbox{if $\vert s\vert<a$;}\\ 0 & \mbox{otherwise.} \end{array} \right.$$ (35)

By Fourier inversion, we see that

$$m_n(x) =\exp(i 2^n x) k_{2^{n-1}}(x)+ \frac{1}{2} \exp (i 3\ 2^{n-1} x) k_{2^{n-1}} (x).$$ (36)

Theorem 4.12 (i)   Let $f$ be any function in $H^1(\mathbb{R})$, and let $M$ and $N$ be arbitrary positive integers. Then there is a positive constant $a$, independent of $f, M,$ and $N$ such that

$$\Vert\sum_{n=-M}^N\epsilon_n m_n*f\Vert _1\leq a \Vert f\Vert _1$$ (37)

for any choice of $\epsilon_n=-1$ or 1. (ii) Moreover, for $f\in H^1(\mathbb{R})$,

$$\lim_{M,N\rightarrow\infty} \sum^N_{n=-M} m_n*f=f$$

unconditionally in $L^1(\mathbb{R})$.

Proof. The proof of (ii) is immediate from (i) and (34), by Fourier inversion. For part (i), use (36), to write

$$\sum_{n=-M}^N\epsilon_n m_n = \sum_{n=-M}^N\epsilon_n\exp(i 2^n x) k_{2^{n-1}}(x)+ \frac{1}{2} \exp (i 3\ 2^{n-1} x) k_{2^{n-1}} (x)$$

$$= \sum_{n=-M}^{-1}\epsilon_n\exp(i 2^n x) k_{2^{n-1}}(x) +\sum_{n=0}^N\epsilon_n\exp(i 2^n x) k_{2^{n-1}}(x)$$

$$+ \frac{1}{2} \sum_{n=-M}^{-1} \epsilo... ... (x) + \frac{1}{2}\sum_{n=0}^N \epsilon_n \exp (i 3\ 2^{n-1} x) k_{2^{n-1}} (x)$$

$$= K_1(x)+K_2 (x)+K_3 (x)+K_4(x).$$

Hence, to prove (37) it is enough to show that there is a positive constant $a$, independent of $f$ such that $\Vert K_j*f\Vert _1\leq a \Vert f\Vert _1$, for $j=1,2,3,4$. Appealing to [15, Theorem 3, p. 114], we will be done once we establish that:

$$\left\vert\widehat{K_j}\right\vert\leq A,$$ (38)

and the Hörmander condition

$$\sup_{y>0} \int_{\vert x\vert>2y} \vert K_j(x-y)-K_j(x)\vert dx\leq B,$$ (39)

where $A$ and $B$ are absolute constants. Inequality (38) holds with $A=1$, since the Fourier transforms of the summands defining the kernels $K_j$ have disjoint supports and are bounded by 1. Condition (39), is well-known. For a proof, see [8, pp. 138-140, and 7.2.2, p. 142].

Proof of Theorem 4.2. Parts (i) and (ii) follow as in Theorem 4.12, so we only prove (iii). For notational convenience, let

$$\kappa_N(x)=\sum_{n=0}^N m_n(x),\quad N=0,\,1,\, 2\, \ldots$$ (40)

Let $V_{2^N}$ denote the de la Vallée Poussin kernels on $\mathbb{R}$ of order $2^N$. Its Fourier transform is continuous, piecewise linear, and satisfies

$$\widehat{V_{2^N}}(s)=\left\{ \begin{array}{ll} 0 & \mbox{if $s\no... ...2^{n+1},2^{n+1}]$;}\\ 1 & \mbox{if $\vert s\vert\leq 2^n$.} \end{array} \right.$$ (41)

It is well-known that $V_{n}$ is a summability kernel for $L^1(\mathbb{R})$ and, in particular, that $V_{2^N}*f$ converges pointwise almost everywhere to $f$ for all $f\in L^p(\mathbb{R})$, for $1\leq p\leq \infty$. Thus, we will be done, if we can show that for $f\in H^\infty(R)$,

$$V_{2^N}*f=h*f+ \sum^N_{n=0} m_n*f.$$ (42)

Write $V_{2^N}=(h+\kappa_{N})+(V_{2^N}-(h+\kappa_{N}))$, where $\kappa_N$ is as in (40). For $N\geq 1$, the Fourier transform of $(V_{2^N}-(h+\kappa_{N}))$, vanishes on $[-\frac{1}{2},\infty)$. Thus, for $f\in H^\infty (\mathbb{R})$, we have $(V_{2^N}-(h+\kappa_{N}))*f=0$, and so (42) follows, and the proof is complete.

Acknowledgements The work of the second author was supported by a grant from the National Science Foundation (U.S.A.).