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  "body": "\n## Definition\n\nLet $X$ be a real or complex [[Banach space|06GZ956657]] and let $A$ be a linear [[operator|KRJX5RA0PW]] in $X$ with domain $D(A)$.\nThe operator $A$ is *dissipative* if\n$$\\norm{(\\lambda I - A)x} \\ge \\lambda \\norm{x} \\quad \\text{for all } \\lambda \u003e 0 \\text{ and all } x \\in D(A).$$\nA dissipative operator $A$ is *maximal dissipative*, or *m-dissipative*, if moreover $\\lambda I - A$ maps $D(A)$ onto $X$ for some $\\lambda \u003e 0$.\n\n## Remarks\n\nIf $X$ is a Hilbert space with inner product $\\ip{\\cdot}{\\cdot}$, then $A$ is dissipative if and only if $\\operatorname{Re}\\ip{Ax}{x} \\le 0$ for all $x \\in D(A)$.\nIn a general Banach space, $A$ is dissipative if and only if for each $x \\in D(A)$ there is a functional $j$ in the [[dual space|07JBSZE0V4]] with $j(x) = \\norm{x}^2 = \\norm{j}^2$ and $\\operatorname{Re} j(Ax) \\le 0$.\n\nExample: on $L^2(0, 1)$, let $Af = f''$ on smooth functions with compact support in $(0, 1)$.\nIntegration by parts gives $\\ip{Af}{f} = -\\int_0^1 \\abs{f'}^2 \\, dx \\le 0$, so $A$ is dissipative.\n\nNon-example: the identity $I$ on a nonzero Hilbert space is not dissipative, since $\\ip{x}{x} \u003e 0$ for $x \\ne 0$.\nDirectly, for $1/2 \u003c \\lambda \u003c 1$ and $x \\ne 0$, $\\norm{(\\lambda I - I)x} = (1 - \\lambda)\\norm{x} \u003c \\lambda\\norm{x}$.\n",
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