Recall that a left topological zero divisor in a Banach algebra $A$ is an element $a\in A$ such that there exists a sequence of unit vectors $(a_{n})$ in $A$ with $\lim_{n\rightarrow\infty}aa_{n}=0$. Let $u\in B(H)$ for a Hilbert space $H$.
I want to show that $u$ is a left topological zero divisor in $B(H)$ if and only if $u$ is not bounded below. I've managed to show $\Rightarrow$ but I'm having some trouble with $\Leftarrow$ because assuming that $u$ is not bounded below, I can get a sequence $(x_{n})$ in $H$ such that $u(x_{n})\rightarrow 0$ but in the definition of left topological zero divisor, I need a sequence of operators in $B(H)$.
Update: The part above has been solved with the help of the comment below by Prahlad Vaidyanathan.
The approximate point spectrum $\sigma_{ap}(u)$ is defined to be $\{\lambda\in\mathbb{C}:u-\lambda\;\text{is not bounded below}\}$. I know that $\partial\sigma(u)\subset\sigma_{ap}(u)\subset\sigma(u)$. I also have that $u$ is bounded below if and only if $u$ is left invertible in $B(H)$. I want to show that if $u$ is normal, then $\sigma_{ap}(u)=\sigma(u)$.
See this paper, theorem 5 and its corollary. $\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}\phantom{}$