Let $X$ be a Banach space and $T:X\to X$ be a continuous and linear operator. What is the transpose operator of $e^T?$
I would like to prove that $e^{T'}=(e^T)'$. At least that equality make perfect sense because they are defined in the same domain. But if $f\in X'$
$$ (e^T)'f=f(e^T)=f(\sum \frac{T^k}{k!})=\sum\frac{f(T^k)}{k!}$$ $$ (e^{T'})(f)=e^{T'f}=e^{f(T)}=\sum \frac{(f(T))^k}{k!} $$
I'm not sure if they are the same. If not I'm still looking for some possible expression for $(e^T)'$. Please help me!
For $A,B\in\mathcal{B}(X)$ we have $(AB)'=B'A'$. As the consequence $(T^k)'=(T')^k$. Therefore $$ e^{T'} =\sum\limits_{k=0}^\infty\frac{1}{k!}(T')^k =\sum\limits_{k=0}^\infty\frac{1}{k!}(T^k)' =\left(\sum\limits_{k=0}^\infty\frac{1}{k!}T^k\right)' =(e^T)' $$