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

Let $X$ be a Banach space non reflexive and $T$ from $l_2(X)$ to $l_2(X)$ a bounded operator defined by: $$T(x_n )=\frac{x_n }{n}.$$ We know that : $$T^{**-1}(l_2(X))=\{x_n^{**} \in l_2(X^{**}) : \...