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Let $\Omega\subseteq\mathbb{C}$ be open and let $B(X)$ be the Banach algebra of all bounded linear operators on the Banach space $X$. Let $f_{\alpha}:\Omega\to\mathbb{C}$ be such that $f_{\alpha}(z)=e^{-\alpha z}$ for $\alpha\in\mathbb{R}$ and define the family $$e^{-\alpha T}=f_{\alpha}(T).$$

I want to prove the group property in the title.

So far, I have said: let $A:=B(X)$, and let $H(\Omega)$ be the algebra of all holomorphic functions in $\Omega$. Define $A_{\Omega}:=\{T\in B(X):\sigma(T)\subset\Omega\}$ be an open subset of $A$. Note that $\exp\in H(\Omega)$.

Now define $\widetilde{H}(A_{\Omega})$ as the set of all A-valued functions $\widetilde{f}$ with domain $A_{\Omega}$ that arise from a $f\in H(\Omega)$ by the formula $$\widetilde{f}(x)=\frac{1}{2\pi i}\int_{\Gamma}f(\lambda)(\lambda e-x)^{-1}d\lambda,$$ where $\Gamma$ is any contour that surrounds $\sigma(T)$ in $\Omega$. Note also that $\widetilde{H}(A_{\Omega})$ is a complex algebra and the mapping $f\to\widetilde{f}$ is an algebra isomorphism of $H(\Omega)$ onto $\widetilde{H}(A_{\Omega})$.

Now, the resolvent for $|\lambda|>\|T\|$ is $$(\lambda e-T)^{-1}=\lambda^{-1}\left(e-\frac{1}{\lambda}T\right)^{-1}=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n}}T^{n}=\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n},$$ and so integrating around a contour that encloses $|\lambda|\le\|T\|$ gives that $$\widetilde{\exp}(T)=\frac{1}{2\pi i}\int_{\Gamma}e^{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}T^{n}d\lambda=\sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\int_{\Gamma}\frac{e^{\lambda}}{\lambda^{n+1}}d\lambda\right)T^{n}=\sum_{n=0}^{\infty}\frac{1}{n!}T^{n}.$$ Since $\exp\in H(\Omega)$, we can write $$e^{-(\alpha+\beta)z}=e^{-\alpha z}e^{-\beta z},\qquad (z\in \Omega)$$ I don't really know where to go from here though. I think a lot of this might have also been unnecessary.

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If you already know that the holomorphic functional calculus provides an algebra homomorphism between $H(\Omega)$ and $\mathcal{B}(X)$ then just choose some open subset $\Omega$ containing $\sigma(T)$ and apply the homomorphism to the identity $\exp((a+b)x) = \exp(ax)\exp(bx)$ which holds in $H(\Omega)$.

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  • $\begingroup$ Then the exact same logic, I presume, could be applied to, for example, the the identity $(\exp(ax))'=a\exp(x)$ which holds in $H(\Omega)$, by applying the homomorphism between $H(\Omega)$ to $B(X)$ to yield $(\exp(Tx))'=T\exp(Tx)$, if I have understood correct? $\endgroup$ – Jason Born Oct 29 '16 at 14:47
  • $\begingroup$ @user3482534: Not really, the map $H(\Omega) \rightarrow \mathcal{B}(X)$ is a homomorphism of algebras so you can only transfer algebraic relations between $H(\Omega)$ and $\mathcal{B}(X)$. The expression $(\exp(Tx))$ is not even well-defined ($\exp(T)$ is an operator, what is $\exp(Tx)$?) so I don't understand what you mean by differentiating it... $\endgroup$ – levap Oct 29 '16 at 21:10
  • $\begingroup$ I apologise, I should have written $\frac{d}{da}\exp(a T)= T\exp(a T)$, for $a\in\mathbb{R}$. But you are right, we cannot use the algebra homomorphism in this case. $\endgroup$ – Jason Born Oct 31 '16 at 17:58

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