Relationship between reflexivity and dissipative operator

Question

This seems like a simple question, but trying is not trivial Question: Let H be a reflective space. Show that a semigroup is one of contractions over H if and only if A is dissipative.

Answer 1

Well, without any additional assumptions on the generator $A$ this is generally not true. However, making use of the reflexivity one can proof the following equivalence:

$(T(t))_{t \geq 0}$ is a contraction semigroup on $H$ if and only if $A$ is dissipative, i.e., $\lVert (\lambda - A) x \rVert \geq \lambda \lVert x \rVert$ for all $\lambda > 0$, and there exists $\lambda > 0$ such that $(\lambda - A) H = H$.

$\Leftarrow$: This is a well-known corollary of the Lumer-Phillips theorem (see Corollary II.3.20 in "One-Parameter semigroups" by Engel and Nagel).

$\Rightarrow$: Since $(T(t))_{t \geq 0}$ is a contraction semigroup we know that $\{\lambda \in \mathbb C : \operatorname{Re} \lambda > 0 \} \subseteq \rho(A)$ since the spectral bound $\mathrm s(A)$ is always less or equal to the growth bound $\omega_0(T) = 0$ of the semigroup. In particular, one has $(0, \infty) \subseteq \rho(A)$ and thus $(\lambda - A)$ is surjective for all $\lambda > 0$. Furthermore, we know that $$\lVert R(\lambda, A) \rVert \leq \frac 1 \lambda \qquad (\lambda > 0)$$ since $A$ is the generator of a contractive semigroup. Now let $x \in D(A)$ and set $y := (\lambda - A) x$. Plugging this into the inequality above we get $$\lVert x \rVert = \lVert R(\lambda, A) (\lambda - A) x \rVert = \lVert R(\lambda, A) y \rVert \leq \frac 1 \lambda \lVert y \rVert = \frac 1 \lambda \lVert (\lambda - A) x \rVert$$ and, thus, $\lVert (\lambda - A) x \rVert \geq \lambda \lVert x \rVert$ for all $\lambda > 0$ and $x \in D(A)$. So $A$ is dissipative.