The $\epsilon$-pseudospectrum of a linear operator $A$ is given by $$\Lambda_{\epsilon}(A) = \{\lambda\in\mathbb{C}: ||(\lambda\mathbb{I}-A)^{-1}||\geqslant\epsilon^{-1}\}.$$
Let $\Lambda(A) = \{\lambda\in\mathbb{C}:(\lambda\mathbb{I}-A) \mathrm{\,\,is\,\, not\,\, invertible}\}$ be the spectrum of $A$.
Is it true that $$\lim_{\epsilon\rightarrow 0}\Lambda_{\epsilon}(A) = \Lambda(A)?$$
I know that $\Lambda(A)$ is contained in $\Lambda_{\epsilon}(A)$, and intuitively it feels like the above must be true in at least some cases. However, I have not found a proof of this anywhere.
Yes, this is always true. As $\Lambda_{\varepsilon}$ is decreasing, we have $$\lim\limits_{\varepsilon \to 0} \, \Lambda_{\varepsilon}(A) = \bigcap\limits_{\epsilon > 0} \Lambda_{\varepsilon}(A).$$ By definition, $\Lambda(A) \subseteq \bigcap\limits_{\epsilon > 0} \Lambda_{\varepsilon}(A)$. Conversely, assume $\lambda \in \bigcap\limits_{\epsilon > 0} \Lambda_{\varepsilon}(A)$. Then $||(\lambda\mathbb{I}-A)^{-1}||\geqslant\varepsilon^{-1}$ for all $\varepsilon > 0$. Hence the inverse of $\lambda\mathbb{I}-A$ is not bounded and so $\lambda \in \Lambda(A)$.