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Let $V$ be a normed infinite dimensional vector space. Let $L: V \longrightarrow V$ be a bounded linear operator. Moreover assume that $L$ is 'locally nilpotent' that is: $$ \forall v \in V \quad \exists n \in \mathbf{N}: L^n (v) = 0. $$ Now my question is if the linear operator: $$ \exp (L) = \sum_{n=0}^{\infty} \frac{L^n}{n!} $$ is bounded or not.

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Am I missing something? If $L$ has norm $|L|$, then doesn't $\text{exp}(L)$ trivially have norm at most $\exp(|L|)$? – Qiaochu Yuan Jul 21 '11 at 1:31
Should we assume the series converges? – Ricky Demer Jul 21 '11 at 2:41
Given your tag for the question, don't you want $L$ to be unbounded? – Yemon Choi Jul 21 '11 at 16:14
In what sense is the series assumed or supposed to converge? In the operator norm, in the strong topology, etc. ? – Mark Jul 21 '11 at 19:44
@Mark: if $L$ is locally nilpotent, $\text{exp}(L) v$ is well-defined for any $v$ without making any use of topological structure, and it is not hard to see that it is linear etc. – Qiaochu Yuan Jul 21 '11 at 20:38

$exp(L)$ is bounded, regardless of the local nilpotentcy, since $\|L^n\|\leq \|L\|^n$. On the other hand, if you wanted to ask the question about unbounded $L$ (say, for all $v$ in the domain), then the answer is no.

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If $L$ is a closed operator (but not bounded), then $\exp(L)$ as you define it is also closed. $L$ can be closed but not bounded since $V$ is not complete... – Gerald Edgar Jul 21 '11 at 15:39

Let $V$ be the vector space of all sequences which are eventually zero. Let $L$ be the backwards shift-- this is obviously "locally nilpotent". Given $V$ the norm $$ \| (x_n) \| = \sum_n a_n x_n, $$ where $(a_n)$ is some sequence of positive numbers. Let $e_n$ be the vector which is 1 in the $n$th place, and zero elsewhere. Then $$ \exp(L)(e_n) = (\cdots,1/2,1,1,0,\cdots), $$ where the final 1 is in the $n$th place. So $$ \|\exp(L)(e_n)\| / \|e_n\| \geq (a_{n-1}+a_n)/a_n = 1 + a_{n-1}/a_n. $$ Hence just choose $(a_n)$ so that $( a_{n-1}/a_n )$ is an unbounded sequence, and then $\exp(L)$ will be unbounded. E.g. $(a_n)=(1,2,1,3,1,4,1,\cdots)$ will work.

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Of course in your example $L$ is not bounded, so this is not surprising. – Jack Huizenga Jul 21 '11 at 23:38
@Jack-- But if $L$ is bounded, then its very very simple to see that $\exp(L)$ is bounded (as Orr explained in his answer)... So I don't understand your comment? – Matthew Daws Jul 22 '11 at 9:56
Merely that $L$ bounded was assumed in the original post. – Jack Huizenga Jul 22 '11 at 17:09
@Jack-- I was following Yemon, and assuming that this was a mistake (given that the question has the trivial answer "yes" otherwise!) – Matthew Daws Jul 22 '11 at 18:48

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