Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
8  
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
3  
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
1  
@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

2 Answers 2

$exp(L)$ 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.

share|improve this answer
    
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.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.