I was reading Reed & Simon where the following definition of an analytic vector is given.
Definition: Let $A$ be an operator on a Hilbert space $\mathscr{H}$. The set $C^{\infty}(A) := \bigcap_{n=1}^{\infty}D(A^{n})$ is called the $C^{\infty}$-vectors for $A$. A vector $\varphi \in C^{\infty}(A)$ is called an analytic vector for $A$ if: $$\sum_{n=0}^{\infty}\frac{\|A^{n}\varphi\|}{n!}t^{n} < \infty$$ for some $t > 0$.
Note that, in the above definition, the operator $A$ need not to be bounded or self-adjoint.
My question is very simple: suppose $\varphi$ is an analytic vector for $A$, for some $t>0$. Does it imply that relation: $$e^{tA}\varphi = \sum_{n=0}^{\infty}\frac{(tA)^{n}}{n!}\varphi$$ hold? In other words, to evaluate $e^{tA}$ at $\varphi$ one simply Taylor expands $e^{tA}$? This is getting me a bit confused because of the following points:
- It seems that this is the whole point of the definition, however Reed & Simon don't explicitly state this (as far as I remember), which made me think it might not be true then.
- Moreover, if $A$ is self-adjoint operator then $e^{tA}$ is defined by the Spectral Theorem, and this Taylor expansion might be equivalent to the application $e^{tA}\varphi$, where $\varphi$ is analytic and $e^{tA}$ is given by the Spectral Theorem. However, if $A$ is not self-adjoint, then what does $e^{tA}$ even mean in the first place? Should I just define its application at analytic vectors by its Taylor expansion?