Lumer–Phillips theorem

Remarks

If $A$ generates a contraction semigroup $T(t)$, then for $\lambda > 0$ the resolvent is the Laplace transform $(\lambda I - A)^{-1}x = \int_0^\infty e^{-\lambda t} T(t)x \, dt$. This gives surjectivity of $\lambda I - A$ and the bound $\norm{(\lambda I - A)^{-1}} \le 1/\lambda$, which is dissipativity. Conversely, dissipativity and one surjective $\lambda_0 I - A$ give $\norm{(\lambda I - A)^{-1}} \le 1/\lambda$ for all $\lambda > 0$, and the Hille–Yosida theorem then produces the semigroup.

Example: on $L^2(0, 1)$, let $Af = f''$ with domain $H^2(0, 1) \cap H^1_0(0, 1)$. It is dissipative since $\ip{Af}{f} = -\int_0^1 \abs{f'}^2 \, dx$, and $I - A$ is onto because the boundary problem $f - f'' = g$, $f(0) = f(1) = 0$ has a solution for every $g \in L^2(0, 1)$. So $A$ generates the heat semigroup with Dirichlet boundary conditions, and it is a contraction semigroup.

The range condition cannot be dropped. The operator $f \mapsto f''$ on smooth functions with compact support in $(0, 1)$ is densely defined and dissipative, but it is not closed, so it is not a generator.

Going deeper

For a student

The Hille–Yosida theorem asks for the resolvent bound $\norm{(\lambda - A)^{-1}} \le 1/\lambda$ for every $\lambda > 0$. The Lumer–Phillips theorem replaces this bound by dissipativity, which is often a one-line integration by parts, plus solvability for one $\lambda_0$. For a second-order elliptic operator with Dirichlet conditions, the solvability usually comes from the Lax–Milgram theorem. For a self-adjoint operator $A \le 0$ on a Hilbert space, both conditions hold, and the semigroup is $e^{tA}$ given by the spectral theorem.