Approximate point and compression spectrum

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $X$ be a nonzero complex Banach space and let $T\in \mathcal{B}(X)$ (A bounded linear operator between normed spaces). Write $T-\lambda$ for $T-\lambda I_X$. The approximate point spectrum of $T$ is

${\sigma }_{ap}(T):=\{\lambda \in \mathbb{C}:T-\lambda \text{ is not bounded below}\},$

bounded below meaning $\mathrm{\parallel }(T-\lambda )x\mathrm{\parallel }\ge c\mathrm{\parallel }x\mathrm{\parallel }$ for all $x$ and some real $c>0$ (A bounded operator that is bounded below), and the compression spectrum of $T$ is

${\sigma }_{cp}(T):=\{\lambda \in \mathbb{C}:\overline{\mathrm{ran}(T-\lambda )}\mathrm{\ne }X\},$

the set of $\lambda$ for which $T-\lambda$ does not have dense range.

Sequential description of the approximate point spectrum. Under Countable Choice, $\lambda \in {\sigma }_{ap}(T)$ if and only if there is a sequence $(x_n)$ of unit vectors in $X$ with $\mathrm{\parallel }(T-\lambda )x_n\mathrm{\parallel }⟶0.$ Indeed, if $T-\lambda$ is not bounded below then for each $n$ the set $\{x:\mathrm{\parallel }x\mathrm{\parallel }=1,\mathrm{\parallel }(T-\lambda )x\mathrm{\parallel }<1\mathrm{/}(n+1)\}$ is nonempty, and Countable Choice selects one unit vector $x_n$ for each $n$; the resulting sequence witnesses the failure of the bound. Conversely a sequence of unit vectors with $\mathrm{\parallel }(T-\lambda )x_n\mathrm{\parallel }\to 0$ rules out every constant $c>0$ in the estimate.

• The two sets are not spectral parts in the disjoint sense. They may overlap each other and the classical parts: an eigenvalue is in ${\sigma }_{ap}$ but may also be a compression value. A value whose range is dense but not closed is not in ${\sigma }_{cp}$ and cannot be bounded below; indeed, under Countable Choice a convergent sequence of range points has Cauchy preimages under a lower bound, and completeness then puts its limit back in the range. Thus the value lies in ${\sigma }_{ap}$. The exact relations are proved in Relations among the five spectral parts; no disjointness is claimed here.

• The normalization of approximate eigenvectors matters. The definition above demands unit vectors, so a sequence $x_n\to 0$ with $(T-\lambda )x_n\to 0$ is not evidence: without the normalization every bounded operator would qualify. The unit vectors may be chosen adaptively; the single use of Countable Choice is recorded in the display above.

• Bounded below is exactly injectivity with closed range in the Banach setting. That equivalence, proved under Dependent Choice in Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range, is what makes ${\sigma }_{ap}$ and ${\sigma }_{cp}$ cover the spectrum (Relations among the five spectral parts); the definition here does not assume it.