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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 with [[spectrum|BMKATJ802A]] $\\sigma(T)$ and [[point spectrum|7FNVA0Q6YQ]] $\\sigma_p(T)$.\nFor $\\lambda \\in \\sigma(T) \\setminus \\sigma_p(T)$, the operator $T - \\lambda I$ is injective.\n\n1. The *continuous spectrum* $\\sigma_c(T)$ is the set of $\\lambda \\in \\sigma(T) \\setminus \\sigma_p(T)$ such that the [[image|0WSFKKT6FF]] of $T - \\lambda I$ is [[dense|0BSJSX7ME2]] in $X$.\n2. The *residual spectrum* $\\sigma_r(T)$ is the set of $\\lambda \\in \\sigma(T) \\setminus \\sigma_p(T)$ such that the image of $T - \\lambda I$ is not dense in $X$.\n\n## Remarks\n\nBy construction, $\\sigma(T)$ is the disjoint union $\\sigma_p(T) \\cup \\sigma_c(T) \\cup \\sigma_r(T)$.\nFor $\\lambda \\in \\sigma_c(T)$, the image of $T - \\lambda I$ is dense but not all of $X$, since a bounded bijection between Banach spaces has a bounded inverse.\n\nExample: for $D(x_n) = (x_n / n)$ on $\\ell^2$, $0 \\in \\sigma_c(D)$.\nIndeed $D$ is injective, it is not surjective, and its image contains every finitely supported sequence, a dense set.\n\nNon-example: for the right shift $S(x_1, x_2, \\ldots) = (0, x_1, x_2, \\ldots)$ on $\\ell^2$, $0$ is not in $\\sigma_c(S)$.\nThe image of $S$ is the closed subspace $\\set{x : x_1 = 0}$, which is not dense, so $0 \\in \\sigma_r(S)$.\n\nFor the right shift, $\\sigma_p(S) = \\emptyset$, $\\sigma_r(S)$ is the open unit disc and $\\sigma_c(S)$ is the unit circle.\nA self-adjoint operator on a Hilbert space has empty residual spectrum.\n",
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