---
tag: YG46P1T47G
kind: definition
name: Dissipative operator
msc: [47B44, 47D06]
defines: 3S7HJ5GQRF
uses: [06GZ956657, KRJX5RA0PW]
sources:
  - url: https://en.wikipedia.org/wiki/Dissipative_operator
---

## Definition

Let $X$ be a real or complex [[Banach space|06GZ956657]] and let $A$ be a linear [[operator|KRJX5RA0PW]] in $X$ with domain $D(A)$.
The operator $A$ is *dissipative* if
$$\norm{(\lambda I - A)x} \ge \lambda \norm{x} \quad \text{for all } \lambda > 0 \text{ and all } x \in D(A).$$
A dissipative operator $A$ is *maximal dissipative*, or *m-dissipative*, if moreover $\lambda I - A$ maps $D(A)$ onto $X$ for some $\lambda > 0$.

## Remarks

If $X$ 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 $x \in D(A)$.
In 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 $\operatorname{Re} j(Ax) \le 0$.

Example: on $L^2(0, 1)$, let $Af = f''$ on smooth functions with compact support in $(0, 1)$.
Integration by parts gives $\ip{Af}{f} = -\int_0^1 \abs{f'}^2 \, dx \le 0$, so $A$ is dissipative.

Non-example: the identity $I$ on a nonzero Hilbert space is not dissipative, since $\ip{x}{x} > 0$ for $x \ne 0$.
Directly, for $1/2 < \lambda < 1$ and $x \ne 0$, $\norm{(\lambda I - I)x} = (1 - \lambda)\norm{x} < \lambda\norm{x}$.