Let $H$ be a Hilbert space, let $T \in B(H)$, and let $\lambda \in \mathbb{C}$. Suppose there is a sequence $\{x_n\}^{\infty}_{n=1}$ in $H$ such that $\|x_n\| = 1$ for all $n$ and $\|(T - \lambda I)x_n\| \to 0$. Show that $\lambda \in \sigma(T)$.
I am new to Operator Algebra and stuck with this problem. Here is what I've tried. $$ \|(T - \lambda I)x_n\| \leq |T - \lambda I| \|x_n\| = |T - \lambda I| \to 0 $$ It follows that $\lambda I - T = 0$. Namely, $\lambda I - T$ is not invertible. Hence, $\lambda \in \sigma(T)$.
Please tell me is it alright? Any help would be appreciated!
$|T-\lambda I| \to 0$ makes no sense.
Suppose, if possible, $\lambda \notin \sigma(T)$. Then $(T-\lambda I)$ has a bounded inverse. Now $1=\|x_n\|=\|(T-\lambda I)^{-1}(T-\lambda I)x_n\| \leq \|(T-\lambda I)^{-1}\|\|(T-\lambda I)x_n\| \to 0$ which is a contradiction.