On logarithms of measures

Abstract

Let A be a Banach algebra and let $x\in A$ have the property that its spectrum does not separate 0 from infinity. It is well known that x has a logarithm, i.e., there exists $y\in A$ with $x=e^y$. We will use this statement to identify measures defined on a locally compact group to have logarithms. Also, we will show that the converse of the above statement is in general not true. Our results will be related to infinitely divisible probability measures.

1 Introduction

Throughout this paper A will be a complex Banach algebra with identity 1 and invertible group $A^{-1}$. We denote the spectrum of $x\in A$ by $\sigma (x)=\{\lambda \in {\mathbb {C}}\,:\, \lambda - x\notin A^{-1} \}$ and the spectral radius of x by r(x) with $r(x)=\sup \{|\,\lambda \, |\,:\,\lambda \in \sigma (x)\,\}$. The spectral radius of a Banach algebra element x can be calculated by the well known formula $r(x)=\lim _{n\rightarrow \infty }\parallel x^n\parallel ^{1/n}$. If it is important to emphasize the relevant algebra A, then the spectrum and spectral radius of $x\in A$ will be denoted by $\sigma (x,A)$ and r(x, A) respectively. If an element $x\in A$ has the property that $\sigma (x)=\{0\}$, or equivalently that $r(x)=0$, then it is called quasinilpotent. We call a subspace I of A an ideal if $IA\subset I$ and $AI\subset I$. By exp$\,A$ we designate the set exp$\,A=\{e^x:\,x\in A\}$. We will say $x\in A$ has a logarithm if $x\in \,$exp$\,A$. For convenience we will denote the set $\{\lambda \in {{\mathbb {C}}} \,:\,|\lambda |=1\,\}$ by ${\mathbb {T}}$. Suppose A is a Banach algebra and $x\in A$ is invertible. If 0 belongs to the unbounded component of ${{\mathbb {C}}}\backslash \sigma (x)$ (the complement of the spectrum), then we will say that the spectrum of x does not separate 0 from infinity. The following result in spectral theory of Banach algebras is well known:

Theorem 1.1

[3, Theorem 3.3.6] Let A be a Banach algebra. Suppose $x\in A$ has a spectrum which does not separate 0 from infinity. Then there exists $y\in A$ such that $x=e^y$.

We mention some corollaries of Theorem 1.1 that can assist one to identify logarithms in a Banach algebra.

Corollary 1.2

Let A be a Banach algebra and $x\in A$.

• 1. $x\in$ exp$\,A+$ exp$\,A$.

• 2. If there exists $n\in {\mathbb {N}}$ with $x^n=1$, then x has a logarithm.

• 3. If the spectrum of x has at most 0 as a limit point, then $\lambda -x$ has a logarithm for all $\lambda \notin \sigma (x)$.

• 4. If $\sigma (x)\subset {\mathbb {R}}$, then $\lambda -x$ has a logarithm for all $\lambda \notin \sigma (x)$.

If X is a Banach space then the collection ${{\mathcal {L}}}(X)$ of bounded operators on X is a Banach space with the operator norm. If one defines multiplication in ${{\mathcal {L}}}(X)$ as composition of operators it becomes a Banach algebra with identity the identity operator. Let E be a complex Banach lattice. An operator $T:E\rightarrow E$ is called regular if it can be written as a linear combination over ${\mathbb {C}}$ of positive operators. The space of regular operators on E will be denoted by ${{\mathcal {L}}}^r(E)$ and it is a subspace of ${{\mathcal {L}}}(E)$. When ${{\mathcal {L}}}^r(E)$ is provided with the r-norm

$$\begin{aligned} \parallel T\parallel _r=\inf \{\parallel S\parallel :\,S\in \mathcal{L}(E),\,S\ge 0,\,|Tz|\le S|z|\,\text{ for } \text{ all }\,z\in E\} \end{aligned}$$

it becomes a Banach algebra which contains the unit of $\mathcal{L}(E)$, see for instance the remarks preceding Lemma 1.1 in [1]. The spectrum of a regular operator T in $\mathcal{L}^r(E)$ is termed the o-spectrum of T and it is denoted by $\sigma _o (T)$. This concept was introduced by Schaefer [9] and it was extensively investigated in [1, 2]. An operator $T\in {{\mathcal {L}}}^r(E)$ is called r-compact if it can be approximated in the r-norm by operators of finite rank, [1]. These operators will be denoted by ${{\mathcal {K}}}^r(E)$. The center ${{\mathcal {Z}}}(E)$ of E consists of all operators $T\in {{\mathcal {L}}}(E)$ for which there exists $c>0$ such that $|Tx|\le c|x|$ for all $x\in E$.

Let X be a locally compact group with identity e. By a measure on X we understand a complex regular Borel measure. By $C_0(X)$ we denote the space of all continuous complex valued functions vanishing at infinity. It is well known that the space ${{\mathcal {M}}}(X)$ of all bounded measures on X may be identified with the dual space of $C_0(X)$, [5, C.18]. Note that ${{\mathcal {M}}}(X)$ is a vector space over the complex numbers $\mathbb C$. Define for $\mu \in {{\mathcal {M}}}(X)$ the variation of $\mu$ by

$$\begin{aligned} |\mu |(X)=\sup \left\{ \sum _{j=1}^m|\mu (X_j)|:\,(X_j)_{j=1}^m\, \text{ is } \text{ a } \text{ measurable } \text{ partition } \text{ of } \text{ X }\right\} . \end{aligned}$$

If we let $\parallel \mu \parallel =|\mu |(X)$, then $\parallel \mu \parallel$ is a norm on ${{\mathcal {M}}}(X)$ and ${{\mathcal {M}}}(X)$ is a Banach space with this norm. For $\mu ,\nu \in {{\mathcal {M}}}(X)$ the ordering $\mu \le \nu$, i.e., $\mu (E)\le \nu (E)$ for all Borel sets $E\subset X$ together with the variation norm induce the structure of a Banach lattice on ${{\mathcal {M}}}(X)$. The positive cone in $\mathcal{M}(X)$ will be denoted by ${{\mathcal {M}}}(X)_+$.

By the above identification a measure $\mu \in {{\mathcal {M}}}(X)$ corresponds to the linear functional

$$\begin{aligned} f\mapsto \int _Xfd\mu \quad (f\in C_0(X)). \end{aligned}$$

This allows one to define a convolution product $\mu *\nu$ for $\mu ,\,\nu \in {{\mathcal {M}}}(X)$ by

$$\begin{aligned} \int _Xfd(\mu *\nu )=\int _X\int _X\,f(st)d\mu (s)d\nu (t)\quad (f\in C_0(X)). \end{aligned}$$

If $x\in X$ and if $E\subset X$ is a Borel set, let $\delta _x$ be the measure defined by

$$\begin{aligned} \delta _x(E)=\left\{ \begin{array}{ll} 1 &{} \text{ when } x\in E\text{, }\\ 0 &{} \text{ when } x\notin E\text{. } \end{array} \right. \end{aligned}$$

With the variation norm and convolution product, ${{\mathcal {M}}}(X)$ becomes a Banach algebra with identity $\delta _e$. If X is a abelian group, we will denote the identity by 0 and in this case it is well known that ${{\mathcal {M}}}(X)$ is a commutative Banach algebra.

2 A counter example

In this section we provide a counter example to show that in general the converse of Theorem 1.1 is not true.

Let X be a locally compact abelian group and let $1\le p\le \infty$. Consider the complex Banach space $L^p(X)$, where the measure is Haar measure. The translation operator on $L^p(X)$ associated with $x\in X$ is the operator $T_{x,p}$ on $L^p(X)$ defined by

$$\begin{aligned} (T_{x,p}f)(t)=f(t-x) \end{aligned}$$

where $f\in L^p(X)$ and $t\in X$ a.e..

If $1<p<\infty$ and if $x\in X$ has infinite order, then by [6, Theorem 20.18] the spectrum of the operator $T_{x,p}$ in the Banach algebra ${{\mathcal {L}}}(L^p(X))$ is the set ${\mathbb {T}}$. Hence the spectrum of $T_{x,p}$ in ${{\mathcal {L}}}(L^p(X))$ separates 0 from infinity. However, in view of [6, Theorem 20.19] there exists an operator $A_{x,p}\in {{\mathcal {L}}}(L^p(X))$ with $T_{x,p}=e^{iA_{x,p}}$, i.e., $T_{x,p}$ has a logarithm.

3 The spectrum of $\delta _x$

Let X be a locally compact abelian group and $x\in X$. We show how to calculate the spectrum of the measure $\delta _x$. The Banach algebra ${{\mathcal {M}}}(X)$ is often represented in the Banach algebra ${{\mathcal {L}}}(L^p(X))$ via the mapping $\mu \mapsto T_{\mu , p}$ where $T_{\mu ,p}f=\mu *f, \,f\in L^p(X) \quad (1\le p\le \infty )$ and $\mu *f(t)=\int _X f(t-s)d\mu (s)$   $t\in X$. The bounded operators $T_{\mu ,p}$ are called convolution operators. This representation has the inconvenience that the spectrum of $\mu$ in the Banach algebra ${{\mathcal {M}}}(X)$ does not in general coincide with the spectrum of the operator $T_{\mu , p}$ in the Banach algebra ${{\mathcal {L}}}(L^p(X))$.

By noting that $T_{\mu ,p}$ is positive whenever $\mu$ is positive, we know that every convolution operator is regular since every measure in ${{\mathcal {M}}}(X)$ is a linear combination of positive measures. Thus, the mapping $\mu \mapsto T_{\mu ,p}$ is actually a representation of ${{\mathcal {M}}}(X)$ into the Banach algebra ${{\mathcal {L}}}^r(L^p(X))$. W. Arendt proved in [1, Proposition 3.3] that if X is amenable, then the mapping $\mu \mapsto T_{\mu ,p}$ is an isometric algebra and lattice isomorphism onto a full subalgebra of ${{\mathcal {L}}}^r(L^p(X))$ for $1\le p<\infty$. This implies that the spectrum of $\mu$ in $\mathcal{M}(X)$ coincides with the spectrum of $T_{\mu ,p}$ in $\mathcal{L}^r(L^p(X))$, i.e. $\sigma (\mu )=\sigma _0 (T_{\mu ,p})$.

Theorem 3.1

Let X be a locally compact abelian group and let $x\in X$ have infinite order and suppose $\mu =\delta _x$.

• 1. If $p=1,\,\infty$, then

  $$\begin{aligned} \sigma (\mu )=\sigma _0 (T_{\mu ,p} )=\sigma (T_{\mu ,p})={\mathbb {T}}. \end{aligned}$$

• 2. If $1< p <\infty$, then

  $$\begin{aligned} \sigma (\mu )=\sigma _0 (T_{\mu ,p} )\supset \sigma (T_{\mu ,p})={\mathbb {T}}. \end{aligned}$$

Proof

It is easy to see that $T_{\mu ,p}=T_{x,p}$ for $1\le p\le \infty$.

(i):

This follows from [1, Theorem 3.4 1] and [6, Theorem 20.18] since X is abelian.

(ii):

This follows from [1, Theorem 3.4 1] and [6, Theorem 20.18] since X is amenable (it is abelian).$\square$

4 Compact groups

Let X be a compact group. In this section we give an illustration of Corollary 1.2. To accomplish this we invoke convolution operators on $L^p(X)$ with $1\le p\le \infty$. The spaces $L^p(X)$ are understood relative to the left Haar measure m on X. By invoking the Radon Nikodym theorem, $L^1(X)$ can be identified with an ideal in ${{\mathcal {M}}}(X)$. In fact, $L^1(X)$ is a closed Banach algebra ideal in ${{\mathcal {M}}}(X)$. Hence, ${{\mathcal {M}}}(X)/L^1(X)$ is a Banach algebra under the quotient norm. The natural homomorphism $q:{{\mathcal {M}}}(X)\rightarrow {{\mathcal {M}}}(X)/L^1(X)$ is defined by $q(\mu )=\mu +L^1(X)$. Let

$$\begin{aligned} {{\mathcal {M}}}_0(X)=\{\mu \in {{\mathcal {M}}}(X):\,\sigma (\mu +L^1(X),\, \mathcal{M}(X)/L^1(X)) =\{0\}\,\}. \end{aligned}$$

This means that a measure $\mu$ in ${{\mathcal {M}}}(X)$ belongs to $\mathcal{M}_0(X)$ if the image of $\mu$ under the natural homomorphism in the quotient algebra ${{\mathcal {M}}}(X)/L^1(X)$ is quasinilpotent.

Theorem 4.1

[8, Theorem 3.5] Let X be a compact group and let $1\le p\le \infty$. If $\mu \in {{\mathcal {M}}}_0(X)$, then

$$\begin{aligned} \sigma (\mu )=\sigma _0 (T_{\mu ,p})=\sigma (T_{\mu ,p}). \end{aligned}$$

Recall that an ideal I in a Banach algebra is called inessential if for every $x\in I$, the spectrum $\sigma (x)$ of x has at most 0 as an accumulation point.

Theorem 4.2

Let X be a compact group and $1\le p\le \infty$. If $\mu \in \mathcal{M}_0 (X)$, then $\lambda -\mu$ has a logarithm for all $\lambda \notin \sigma (\mu )$.

Proof

We claim that the ideal $L^1(X)$ in ${{\mathcal {M}}}(X)$ is an inessential ideal: If $f\in L^1(X)$ then the measure $f\cdot m$ corresponds to the convolution operator $T_{{f\cdot m},p}$ in ${{\mathcal {L}}}(L^p(X))$ with $1\le p\le \infty$. In view of [1, Proposition 4.1] $T_{{f\cdot m},p}\in {{\mathcal {K}}}^r(L^p(X))$. Since the r-norm is finer than the operator norm, $T_{{f\cdot m},p}$ is a compact operator. In view of Theorem 4.1, $\sigma (f\cdot m )$ has at most 0 as a limit point.

Since $\mu \in \mathcal{M}_0(X)$, it follows from [3, Corollary 5.7.5] that $\sigma (\mu )$ has at most 0 as a limit point. By Corollary 1.2.3, $\lambda -\mu$ has a logarithm. $\square$

In our next remarks we illustrate that there are many measures in ${{\mathcal {M}}}(X)$ belonging to the set ${{\mathcal {M}}}_0(X)$.

• If $\nu \in L^1(X)$, then $\nu \in {{\mathcal {M}}}_0(X)$ because $q(\nu )=\nu +L^1(X)=L^1(X)$ is the zero element in the quotient algebra $\mathcal{M}(X)/L^1(X)$, which is quasinilpotent.

• If $\nu \in L^1(X)$ and $\mu \in {{\mathcal {M}}}(X)$ then $\nu *\mu \,,\,\mu *\nu \in {{\mathcal {M}}}_0(X)$ because $L^1(X)$ is an ideal in ${{\mathcal {M}}}(X)$.

• If $\mu \in {{\mathcal {M}}}_0(X)$ and $\nu \in L^1(X)$ then $\mu +\nu \in {{\mathcal {M}}}_0(X)$.

• If $\nu \in {{\mathcal {M}}}_0(X)$ and $\mu \in {{\mathcal {M}}}(X)$ with $0\le \mu \le \nu$, then $\mu \in {{\mathcal {M}}}_0(X)$, see [8, Proposition 3.3].

• If $\mu \in {{\mathcal {M}}}(X)$ is quasinilpotent and $\nu \in L^1(X)$, then $\mu +\nu \in {{\mathcal {M}}}_0(X)$.

By the bullet above all quasinilpotent measures in ${{\mathcal {M}} }(X)$ belong to ${{\mathcal {M}}}_0(X)$. However, there are no nonzero positive quasinilpotent measures in ${{\mathcal {M}}}(X)$: Suppose $\mu \in {{\mathcal {M}}}(X)_+$ and $\mu$ is quasinilpotent. Since the norm on ${{\mathcal {M}}}(X)_+$ is multiplicative, i.e., $\parallel \mu *\nu \parallel =\parallel \mu \parallel \,\parallel \nu \parallel$, we get

$$\begin{aligned} r(\mu )=\lim _{n\rightarrow \infty }\parallel \mu ^n\parallel ^{1/n}=\parallel \mu \parallel =0 \end{aligned}$$

and so $\mu =0$.

5 Infinitely divisible probability measures

Let X be a locally compact group. We show how one can use our previous results to identify infinitely divisible probability measures: A measure $\mu$ in ${{\mathcal {M}}}(X)$ is called a probability measure if $\mu \ge 0$ and $\mu (X) = 1$. A probability measure on X is said to be infinitely divisible if for each $n\in {\mathbb {N}}$ there exists a probability measure $\mu _n$ with

$$\begin{aligned} \mu =\mu _n*\cdots *\mu _n=(\mu _n)^n. \end{aligned}$$

We call the measures $\mu _n$ factors of $\mu$. Some authors call them $n^{\text {th}}$ roots of the measure $\mu$. Our definition for a probability measure $\mu$ on X to be infinitely divisible is different from the definition that Parthasarathy uses, see [7, Chapter IV, Sect. 4].

Our first example of an infinitely divisible probability measure is the following: If $\nu \in {{\mathcal {M}}}(X)$ is a positive finite measure, then $\mu =e^{-\nu (X)}e^{\nu }$ is a probability measure. Furthermore, $\mu$ is infinitely divisible because for each n,

$$\begin{aligned} \mu ={(e^{-\nu (X)}})^{1/n}e^{\frac{\nu }{n}}*\cdots *{(e^{-\nu (X)}})^{1/n}e^{\frac{\nu }{n}}. \end{aligned}$$

If $\mu$ is an idempotent probability measure, i.e. $\mu *\mu =\mu$, and $0\not =\mu \not =\delta _e$, then $\mu$ is a noninvertible infinitely divisible probability measure. A third example of an infinitely divisible probability measure is a Gausian measure, see [7, Chapter IV, Sect. 6]. Such a measure is defined on a locally compact abelian group.

In our next result we show how a measure with a logarithm is related to some infinitely divisible probability measure.

Theorem 5.1

Let X be a locally compact group and suppose a measure $\mu \in {{\mathcal {M}}}(X)$ has a logarithm $\lambda$ for some $\lambda \in \mathcal{M}(X)$. Then there exists an infinitely divisible probability measure $\nu \not =\delta _e$ with factors $\nu _n\quad ({n\in \mathbb N})$ such that the sequence $(e^{\frac{\lambda }{n}}- \nu _n )$ is a null sequence.

Proof

Note that $\mu =e^\lambda = (e^{\frac{\lambda }{n}})^n$. Since ${{\mathcal {M}}}(X)$ is a Banach lattice algebra, see [10, Example 2], $\mu \le e^{|\lambda |}$. But

$$\begin{aligned} \nu =e^{-|\lambda |(X)}e^{|\lambda |} \end{aligned}$$

is a probability measure and it is infinitely divisible because $\nu =\nu _n*\ldots *\nu _n$ with

$$\begin{aligned} \nu _n =(e^{-|\lambda |(X)})^{1/n}e^{\frac{|\lambda |}{n}}. \end{aligned}$$

for all $n\in {\mathbb {N}}$. Then $(e^{\frac{\lambda }{n}} - \nu _n )$ is a null sequence because $\lim _n e^{\frac{\lambda }{n}}=\lim _n\nu _n=\delta _e$. $\square$

If X is a Banach space and $x\in X$, then the probability measure $\delta _x$ is infinitely divisible because for each $n\in {\mathbb {N}}$

$$\begin{aligned} \delta _x=\delta _\frac{x}{n}*\ldots *\delta _\frac{x}{n}=(\delta _\frac{x}{n})^n. \end{aligned}$$

If X is a locally compact group and $e\not =x\in X$, then it is not clear whether the measure $\delta _x$ is infinitely divisible.