Question on Linear Operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:56:22Z http://mathoverflow.net/feeds/question/70866 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70866/question-on-linear-operators Question on Linear Operators Najdorf 2011-07-21T01:15:29Z 2011-07-21T20:37:07Z <p>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.</p> http://mathoverflow.net/questions/70866/question-on-linear-operators/70872#70872 Answer by Orr Shalit for Question on Linear Operators Orr Shalit 2011-07-21T03:00:42Z 2011-07-21T03:00:42Z <p>$exp(L)$ 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.</p> http://mathoverflow.net/questions/70866/question-on-linear-operators/70925#70925 Answer by Matthew Daws for Question on Linear Operators Matthew Daws 2011-07-21T19:40:39Z 2011-07-21T19:40:39Z <p>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.</p>