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  "body": "\n## Definition\n\nLet $X$ be a complex Banach space with identity map $I$, and let $T \\colon X \\to X$ be a bounded operator.\nThe *approximate point spectrum* of $T$ is the set $\\sigma_{ap}(T)$ of $\\lambda \\in \\C$ for which there is a sequence $(x_n)$ in $X$ with\n$$\\norm{x_n} = 1 \\text{ for all } n \\quad \\text{and} \\quad \\norm{(T - \\lambda I) x_n} \\to 0.$$\n\n## Remarks\n\nEquivalently, $\\lambda \\in \\sigma_{ap}(T)$ if and only if $T - \\lambda I$ is not bounded below, that is, there is no $c \u003e 0$ with $\\norm{(T - \\lambda I)x} \\ge c \\norm{x}$ for all $x$.\nIt satisfies $\\sigma_p(T) \\cup \\sigma_c(T) \\subseteq \\sigma_{ap}(T) \\subseteq \\sigma(T)$, with the [[point spectrum|7FNVA0Q6YQ]], the [[continuous spectrum|YVH149HCAG]] and the [[spectrum|BMKATJ802A]].\nIt is closed, and it contains the boundary of $\\sigma(T)$, so it is nonempty when $X \\ne \\set{0}$.\n\nExample: for $D(x_n) = (x_n / n)$ on $\\ell^2$, the unit vectors $e_n$ satisfy $\\norm{D e_n} = 1/n \\to 0$, so $0 \\in \\sigma_{ap}(D)$.\n\nNon-example: for the right shift $S$ on $\\ell^2$, $\\norm{Sx} = \\norm{x}$ for all $x$, so $S$ is bounded below and $0 \\notin \\sigma_{ap}(S)$, although $0 \\in \\sigma(S)$.\nMore generally $\\norm{(S - \\lambda I)x} \\ge (1 - \\abs{\\lambda}) \\norm{x}$, so $\\sigma_{ap}(S)$ is the unit circle.\n\nFor a normal operator on a Hilbert space, $\\sigma_{ap}(T) = \\sigma(T)$.\n",
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