Convolution powers of singular-symmetric measures
1979; Mathematical Society of Japan; Volume: 31; Issue: 1 Linguagem: Inglês
10.2969/jmsj/03110001
ISSN1881-1167
Autores Tópico(s)Advanced Topics in Algebra
ResumoL^{1}(G)$ be the group algebra on $G$ .In [7], Taylor showed that: There are a compact topological abelian semigroup $S$ and an isometric isomorphism $\theta$ of $M(G)$ into $M(S)$ such that;(a) $\theta(M(G))$ is a $weak-*dense$ subalgebra of $M(S)$ ; (b) $\hat{S}$ , the set of all continuous semicharacters on $S$ , separates the points of $S$ ;(c) for $f\in S,$ $\mu\rightarrow\int_{s}fd\theta\mu$ $(\mu\in M(G))$ is a non-zero complex homomor- phism of $M(G)$ ;(d) for a non-zero complex homomorphism $F$ of $M(G)$ , there is an $f\in\hat{S}$ such that $ F(\mu)=\int_{s}fd\theta\mu$ for $\mu\in M(G)$ .We can consider that $\hat{S}$ is the maximal ideal space of $M(G),\hat{G}\subset\hat{S}$ , and the Gelfand transform of $\mu\in M(G)$ is given by $\hat{\mu}(f)=\int_{s}fd\theta\mu(f\in\hat{S})$ .A closed subspace (ideal, subalgebra) $N\subset M(G)$ is called an L-subspace (L-ideal, L-sub- algebra) if $L^{1}(\mu)\subset N$ for every $\mu\in N$, where $ L^{1}(\mu)=\{\lambda\in M(G);\lambda$ is absolutely continuous with respect to $\mu(\lambda\ll\mu)$ }.We denote by Rad $L^{1}(G)$ the radical of $L^{1}(G)$ in $M(G)$ , that is, Rad $L^{1}(G)=$ { $\mu\in M(G);\hat{\mu}(f)=0$, for all $f\in\hat{S}\backslash \hat{G}$ }.We put $\mathfrak{L}(G)=\sum_{\tau}$ Rad $L^{1}(G_{\tau})$ , where $\tau$ runs through over L. C. A. group topologies on $G$ which are stronger than the original one.Then $\mathfrak{L}(G)\subset M(G)$ and $\mathfrak{L}(G)$ is an L-subalgebra ([2]).For $\mu\in M(G)$ , we put $\mu^{*}(E)=\overline{\mu(-E)}$ for every Borel subset $E$ of $G$ .We denote by $\mathfrak{M}$ the set of all symmetric measures of $M(G)$ , that is, $\mathfrak{M}=$ { $\mu\in M(G);\hat{\mu}^{*}(f)=\overline{\hat{\mu}(f})$ for every $f\in\hat{S}$ }.Then it is easy to show that $\mathfrak{L}(G)\subset \mathfrak{M}$ .A measure $\mu\in \mathfrak{M}$ is called singular-symmetric if $\mu$ is singular with 2 K. IZUCHI $\mathfrak{L}(G)(\mu\perp \mathfrak{L}(G))$ .In [4], the author shows that if $\overline{R}$ is the Bohr compactifica- tion of the real line $R$ , then there exists a singular-symmetric measure $\mu$ on $\overline{R}$ .Moreover it is easy to show that $\mu$ (constructed in [4]) has the property $\mu*\mu\in \mathfrak{L}(\overline{R})$ .By the same method as in [4], we can construct $\mu$ on an infinite compact abelian group $G$ whose dual group has an infinite independent subset, such that $\mu$ is singular-symmetric with $\mu*\mu\in \mathfrak{L}(G)$ .In this paper, we show THEOREM.Let $G$ be an infinite compact abelian group.If $G$ has an infnite independent subset, then there exis $ts$ a singular-symmetric measure $\mu$ on $G$ such that $\mu^{n}$ is singular-symmetric for every posiiive integer $n$ , where $\mu^{n}=\mu^{n-1}*\mu$ \ l a n g l e $n\geqq 2$ ) and $\mu^{1}=\mu$ .2. Proof of theorem.Let $G$ be an infinite compact abelian group such that $\hat{G}$ has an infinite independent subset $E$ which we may suppose to generate $\hat{G}$ without loss of generality.Then there is a family of infinite subsets of $E$ , { $E_{n,i}$ ; $n=1,2,$ $\cdots$ , $i=1,2,$ $\cdots,$ $2^{n}$ }, which satisfies the following properties:1) For $n\geqq 1,$ $\cup\{E_{n,i} ; 1\leqq i\leqq 2^{n}\}=E$ ;2) for $1\leqq i<j\leqq 2^{n},$ $E_{n.i}\subsetneqqE_{n.j}$ and $E_{n,j}\backslash E_{n}i$ is an infinite set;3) $E_{n+1,k}\subset E_{n.i}$ for $k<2i$ and $E_{n+1,2i}=E_{n,:}(1\leqq i\leqq 2^{n})$ .Let $H_{n,i}$ be the subgroup of $\hat{G}$ generated by $E_{n,:}$ , then $\{H_{n.i}\}_{n.:}$has the following properties by 1), 2) and 3):4) For $n\geqq 1$ and $1\leqq i<j\leqq 2^{n},$ $H_{n.i}\subsetneqqH_{n.j},$ $H_{n.j}/H_{n.i}$ is an infinite group, and $H_{n.2n}=\hat{G}$;5) $H_{n}.:\supsetneqH_{n+1\cdot k}$ and $H_{n.i}/H_{n+1.k}$ is an infinite group for $k<2i$ , and $H_{n.i}=H_{n+1.2i}$ for $1\leqq i\leqq 2^{n}$ .By the above facts 4) and 5), we have: 6) For $n\leqq s$ and $1\leqq i\leqq 2^{s-n}j,$ $H_{n.j}\supsetH_{s.i}$ and $H_{n,j}/H_{s,:}$ is an infinite group if $i\neq 2^{s-n}j$ , and $H_{n.j}=H_{s.2}s-n_{j}$ .Let $G_{n.i}$be the annihilator of $H_{n,i}$ in $G(G_{n.i}=H_{n,i}^{\perp}\subsetG)$ , then $G_{n}$ .: is a compact subgroup of $G$ and $\{G_{n,i}\}_{n.i}$satisfies the following by 4), 5) and 6): 7) For $n\geqq 1$ and $1\leqq i<j\leqq 2^{n},$ $G_{n.l}\supseteqqG_{n,j},$ $G_{n.i}/G_{n.j}$ is an infinite compact group, and $G_{n,2n}=\{0\}$ , where $0$ is the unit element of $G$ ; 8) $G_{n.i}\subsetneqqG_{n+1k}$ and $G_{n+1.k}/G_{n.i}$ is an infinite compact group for $k<2i$ , and $G_{n.i}=G_{n+1,8:}$for $1\leqq i\leqq 2^{n}$ ; 9) for $n\leqq s$ and $1\leqq i\leqq 2^{s-n}j,$ $G_{n,j}\subset G_{s,i}$ and $G_{s.i}/G_{n.i}$ is an infinite compact group if $i\neq 2^{s-n}j$ , and $G_{n.j}=G_{s.2}s-n_{j}$ .
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