Welcome to MathOverflow

Question

-2

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.

edited Jul 21 2011 at 16:15

Yemon Choi
8,437●1●19●44

asked Jul 21 2011 at 1:15

Najdorf
494●3●8

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 2011 at 1:31

Should we assume the series converges? – Ricky Demer Jul 21 2011 at 2:41

3

Given your tag for the question, don't you want $L$ to be unbounded? – Yemon Choi Jul 21 2011 at 16:14

In what sense is the series assumed or supposed to converge? In the operator norm, in the strong topology, etc. ? – Mark Schwarzmann Jul 21 2011 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 2011 at 20:38

Orr Shalit · Answer 1 · 2011-07-21 03:00:42Z UTC

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.

answered Jul 21 2011 at 3:00

Orr Shalit
208●1●9

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 2011 at 15:39

Matthew Daws · Answer 2 · 2011-07-21 19:40:39Z UTC

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.

Question on Linear Operators

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.