Let $L : X \rightarrow X$ be a bounded linear operator on the Banach space $( X, \lVert \cdot \rVert_X)$. For $t \geq 0$ define the following operators:
$$
T_t := e^{ t L} := \sum_{ k = 0 }^{ \infty } \frac{ 1 }{ k! } t^k L^k.
$$
Is one supposed to understand $T_t$, $t \geq 0$, in the following way: for $n \in \mathbb{N}$ set
$$
T_t^n := \sum_{ k = 0 }^{ n } \frac{ 1 }{ k! } t^k L^k
$$
and for $u \in X$ set
$$
T_t u := \lim_{ n \rightarrow \infty } T_t^n u = \lim_{ n \rightarrow \infty} \sum_{ k = 0 }^n \frac{ 1 }{ k } t^k L^k u,
$$
where the convergence is understood in norm $\lVert \cdot \rVert_X$. But this would mean that for all $u \in X$ there should exist some $v \in X$ such that
$$
\lim_{ n \rightarrow \infty} \lVert T_t^n u – v\rVert_X = 0 \tag{1}
$$
Alternatively, I know that space of all bounded linear operators from $X$ to $X$ is a Banach space equipped with the operator norm. Hence if we manage to show that $T_t^n$ is convergent or a Cauchy sequence, then we could define
$$
T_t := \lim_{ n \rightarrow \infty } T_t^n = \lim_{ n \rightarrow \infty } \sum_{ k = 0 }^n \frac{ 1 }{ k } t^k L^k,
$$
where the convergence is understood in the operator norm. This would of course mean that $T_t$ is itself a bounded linear operator, and since
$$
\lVert T_t^n u – T_t u \rVert_X \leq \lVert T_t^n – T_t \rVert \lVert u \rVert_X,
$$
this would also imply $(1)$.
So my questions are:
- Is the reasoning above correct?
- If it is correct, how can we show that $( T_t^n )_{ n \in \mathbb{N} }$ is convergent?
Best Answer
You can easily show that $(T^n)$ is Cauchy: $$||T^n-T^m||=||\sum_{i=m+1}^{n} \frac{t^i L^i}{i!}||\leqslant \sum_{i=m+1}^{n} \frac{t^i}{i!} ||L||^i$$ And since $||L||$ is finite, you can make it as small as you want.