Dissipative operator

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 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 $x \ne 0$, $\norm{(\lambda I - I)x} = (1 - \lambda)\norm{x} < \lambda\norm{x}$.