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.


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 $H(\Omega)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.