consists of a single element. Using this notation, A is dissipative if and only if for all x ∈ D(A) there exists a x' ∈ J(x) such that In the case of Hilbert spaces, this becomes for all x in D(A). Since this is non-positive, we have Since I−A has an inverse, this implies that is a contraction, and more generally, is a contraction for any positive λ. The utility of this formulation is that if this operator is a contraction for some positive λ then A is dissipative. It is not necessary to show that it is a contraction for all positive λ (though this is true), in contrast to (λI−A)−1 which must be proved to be a contraction for all positive values of λ. Examples For a simple finite-dimensional example, consider n-dimensional Euclidean space Rn with its usual dot product. If A denotes the negative of the identity operator, defined on all of Rn, then so A is a dissipative operator. So long as the domain of an operator A (a matrix) is the whole Euclidean space, then it is dissipative if and only if A+A* (the sum of A and its adjoint) does not have any positive eigenvalue, and (consequently) all such operators are maximally dissipative. This criterion follows from the fact that the real part of which must be nonpositive for any x, is The eigenvalues of this quadratic form must therefore be nonpositive. (The fact that the real part of must be nonpositive implies that the real parts of the eigenvalues of A must be nonpositive, but this is not sufficient. For example, if then its eigenvalues are negative, but the eigenvalues of A+A* are −5 and 1, so A is not dissipative.) An equivalent condition is that for some (and hence any) positive has an inverse and the operator is a contraction (that is, it either diminishes or leaves unchanged the norm of its operand). If the time derivative of a point x in the space is given by Ax, then the time evolution is governed by a contraction semigroup that constantly decreases the norm (or at least doesn't allow it to increase). (Note however that if the domain of A is a proper subspace, then A cannot be maximally dissipative because the range will not have a high enough dimensionality.) Consider H = L2([0, 1]; R) with its usual inner product, and let