Question on Linear Operators - MathOverflow

Question on Linear Operators

2011-07-21T01:15:29Z

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.

Answer by Orr Shalit for Question on Linear Operators

2011-07-21T03:00:42Z

$exp(L)$ 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.

Answer by Matthew Daws for Question on Linear Operators

2011-07-21T19:40:39Z

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.