Boundary of spectrum lies in approximate point spectrum

Since $\lambda$ is a boundary point, for each $n\ge 1$ the disc $\{\mathrm{\mid }z-\lambda \mathrm{\mid }<1\mathrm{/}n\}$ meets $\rho (T)$; Countable Choice selects ${\lambda }_n\in \rho (T)$ with $\mathrm{\mid }{\lambda }_n-\lambda \mathrm{\mid }<1\mathrm{/}n$ for every $n$; in particular ${\lambda }_n\to \lambda$.

First suppose that the resolvent norms are bounded along this sequence, say $\mathrm{\parallel }R({\lambda }_n)\mathrm{\parallel }\le M$ for all $n$, and suppose $M\ge 1$. For $n$ with $\mathrm{\mid }{\lambda }_n-\lambda \mathrm{\mid }<1\mathrm{/}(2M)$ the operator $T-\lambda =(T-{\lambda }_n)+({\lambda }_n-\lambda )=(T-{\lambda }_n)(1-({\lambda }_n-\lambda )R({\lambda }_n))$ is a product of invertible factors: the displayed identity holds because $(T-{\lambda }_n)R({\lambda }_n)=-1$ by [L2], and the second factor is invertible by the Neumann series since $\mathrm{\parallel }({\lambda }_n-\lambda )R({\lambda }_n)\mathrm{\parallel }\le 1\mathrm{/}2<1$. Hence $T-\lambda$ would be invertible and $\lambda \in \rho (T)$, contradicting $\lambda \in \sigma (T)$.

Consequently the norms $\mathrm{\parallel }R({\lambda }_n)\mathrm{\parallel }$ are unbounded; passing to a subsequence, which we relabel, we may assume $\mathrm{\parallel }R({\lambda }_n)\mathrm{\parallel }\to \mathrm{\infty }$.

For each $n$ the set of unit vectors $x$ with $\mathrm{\parallel }R({\lambda }_n)x\mathrm{\parallel }\ge \frac{1}{2}\mathrm{\parallel }R({\lambda }_n)\mathrm{\parallel }$ is nonempty, because the operator norm is the supremum of $\mathrm{\parallel }R({\lambda }_n)x\mathrm{\parallel }$ over the unit sphere; Countable Choice selects such a unit vector $x_n$ for every $n$, and we set $y_n:=R({\lambda }_n)x_n\mathrm{/}\mathrm{\parallel }R({\lambda }_n)x_n\mathrm{\parallel }$, a unit vector.

Then $\mathrm{\parallel }(T-{\lambda }_n)y_n\mathrm{\parallel }=\mathrm{\parallel }{x}_n\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }R({\lambda }_n)x_n\mathrm{\parallel }\le 2\mathrm{/}\mathrm{\parallel }R({\lambda }_n)\mathrm{\parallel }\to 0$ and $\mathrm{\mid }{\lambda }_n-\lambda \mathrm{\mid }\to 0$, so $\mathrm{\parallel }(T-\lambda )y_n\mathrm{\parallel }\le \mathrm{\parallel }(T-{\lambda }_n)y_n\mathrm{\parallel }+\mathrm{\mid }{\lambda }_n-\lambda \mathrm{\mid }\to 0$ along unit vectors.

By [L4] such a sequence rules out $T-\lambda$ being bounded below with any constant $c>0$; hence $\lambda \in {\sigma }_{ap}(T)$. Since $\lambda \in \mathrm{\partial }\sigma (T)$ was arbitrary, $\mathrm{\partial }\sigma (T)\subseteq {\sigma }_{ap}(T)$.