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Is $e^{(rA)} = (e^{A})^r$ when $r \in \mathbb{R}$ and $A$ is an element of a Banach algebra?

Clearly if $n$ is an integer, then

$e^{(nA)} = e^{A+A \cdots +A} = e^{A}e^{A}\cdots e^{A} = (e^{A})^n$,

where the second equality follows from the Baker-Hausdorff lemma and the fact that [A,A]=0. On the other hand, I think the equality is not generally true when $r \in \mathbb{C}$. But what about the reals?

Many thanks for your thoughts!

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    $\begingroup$ How do you define the non-integer powers for arbitrary Banach algebra elements? $\endgroup$
    – fedja
    Commented Dec 24, 2010 at 15:35
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    $\begingroup$ Using the binomial series, I guess? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 24, 2010 at 15:36
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    $\begingroup$ The conventional definition is to use the limit of (1 + x/n)^n as n goes to infinity. (There's a slightly different variants that's more likely to converge, but I can't remember what it is.) $\endgroup$
    – arsmath
    Commented Dec 24, 2010 at 16:09
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    $\begingroup$ The Taylor expansion of the logarithm diverges more often than not. There is no reason to expect it to converge for $e^A$. Of course, if everything is defined as a series and you can meaningfully plug series into series justifying all passages to the limits, the identity holds. On the other hand, if the right hand side just fails to exist or has multiple values like non-integer powers of complex numbers (the first example to look at), the question hardly makes sense. $\endgroup$
    – fedja
    Commented Dec 24, 2010 at 16:59
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    $\begingroup$ You are asking why $(f\circ g)(A)= f(g(A))$. The answer is, it is just a matter of functional calculus; check any textbook on the topic. $\endgroup$
    – Pietro Majer
    Commented Dec 24, 2010 at 17:38

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A particular case of a Banach algebra is the one-dimensional case $\mathbb C$, right?
And according to Pietro, if you can prove this in that case, you get the general case by saying "functional calculus". (At least for $A$ where it applies, such as normal $A$.)
But is it true for $\mathbb C$?

$A = 2\pi i$, $r=1/2$, $1 = \exp(2\pi i)$ and $1^{1/2} =^{?} \exp(\pi i) = -1$ .

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  • $\begingroup$ Oh dear -- you're of course correct, the claim is not generally true. Thanks! $\endgroup$
    – soulphysics
    Commented Dec 26, 2010 at 17:21

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