Show that a subspace is closed in Hilbert space $H$

Question

Let $u\in B(H)$ , $\lambda < 0$. Also we have $\|(u-\lambda)x\|\geq |\lambda|\|x\|$. So $u-\lambda$ is bounded below. To show $(u-\lambda)(H)$ is closed in $H$, suppose $\{(u-\lambda)x_n\}$ be Cauchy in $H$, then for $\epsilon >0$, we have $$|\lambda|\|x_n-x_m\|\leq \|(u-\lambda)(x_n-x_m)\| < \epsilon$$ Put $x:=\lim x_n$, show that $(u-\lambda)x=\lim(u-\lambda)x_n$. For this we have $$\|(u-\lambda)(x-x_n)\|\leq (\|u\| + |\lambda|)\|x-x_n\|\to 0$$ Please check my process. Thanks for your help.

Answer 1

The process is good. Maybe some wording could be added. For example, the first displayed inequality only holds for $m$ and $n$ large enough. The existence of the limit of the sequence $(x_n)$ has to be justified (by completeness of a Hilbert space).