Let $X$ be a Banach space. The other day I asked if all bounded operators $A:X\to X$ satisfy the following property:
(P): All bounded nonzero trajectories $t\mapsto e^{tA}x$ satisfy $$\inf_{t\in \mathbb{R}}|e^{tA}x|>0,$$ where $x\in X$ and $e^{tA}$ is the exponential operator defined by $$e^{tA}:=\sum_{k=0}^{+\infty}\frac{t^kA^k}{k!}.$$ If the operator $A$ is scalar ($A=a\in \mathbb{R}$) or in general $A$ is a matrix then the property (P) holds as explained here. In infinite dimensions, there exist operators which do not satisfy this property as explained by fedga. So later I observed that operators $A$ such that $t\mapsto e^{t A}$ is bounded on $\mathbb{R}$ satisfy this property because $$|x|=|e^{-tA}e^{tA}x|\leq\left(\sup_{t\in \mathbb{R}}|e^{-tA}|\right)|e^{tA}x|.$$ For operators such that $t\mapsto e^{tA}$ is not bounded in $t\in \mathbb{R}$, there's examples like: $A=iI$ or maybe $A$ is of finite rank (because it works for matrices).
So I just feel like if $t\mapsto e^{tA}$ is not bounded in $t$ and $A$ satisfies (P), then $A$ must be of finite rank or something.
I don't know if this is true or not. I am also interested in "exponential" of unbounded operators or what it is known as strongly continuous groups of operators.
