When does the limit of $exp(-\alpha T)$ exist as $\alpha\to \infty$?

Question

I have the following situation: $T$ is a bounded linear operator on a Banach space $X$.

Question: when does the limit $$\lim_{\alpha\to \infty} \exp(-\alpha T)$$ exist as a bounded linear operator on $X$?

By "limit" I mean here either norm topology or strong operator topology, but any thoughts on other topologies will be greatly appreciates as well.

Answer 1

A partial answer: When the spectrum of $T$ is included in the real half-plane $H_+=\{ \mbox{Re } z >0 \}$ then the operator $e^{-\alpha T}$ goes to zero in norm. This may be seen from

$$ e^{-\alpha T} = \oint_C e^{-\alpha \lambda} (\lambda- T)^{-1} \frac{d\lambda}{2\pi i} $$ where the contour $C$ (enclosing the spectrum) may be chosen strictly inside $H_+$.

If, however, the spectrum touches the imaginary axis it becomes complicated. For example given the multiplication operator $$ Tf(x) = x f(x), \ x\in [0,1] $$ acting upon $L^1([0,1])$, $e^{-\alpha T}$ goes to zero but not when acting upon $C([0,1])$ (no limit). In both cases the spectrum is $[0,1]$. In the case of a matrix, eigenvalues should either have strictly positive real part or be zero without nilpotent part for the zero eigenvalue.