Au = u′ (in this case a weak derivative) with domain D(A) equal to those functions u in the Sobolev space with u(1) = 0. D(A) is dense in L2([0, 1]; R). Moreover, for every u in D(A), using integration by parts, Hence, A is a dissipative operator. Furthermore, since there is a solution (almost everywhere) in D to for any f in H, the operator A is maximally dissipative. Note that in a case of infinite dimensionality like this, the range can be the whole Banach space even though the domain is only a proper subspace thereof. Consider H = H02(Ω; R) (see Sobolev space) for an open and connected domain Ω ⊆ R'''n and let A'' = Δ, the Laplace operator, defined on the dense subspace of compactly supported smooth functions on Ω. Then, using integration by parts, so the Laplacian is a dissipative operator. Notes References (Definition 12.25) Operator theory Linear operators