Proposition 2.6.4.label Let $E$ be a Banach space and $A: E \to E$ be a densely defined, closed operator, then $A$ generates a strongly continuous contraction semigroup if and only if:

  1. (1)

    $A$ is dissipative.

  2. (2)

    For each $\lambda > 0$, $(\lambda - A)$ is surjective.

Proof. First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$. By the Hahn-Banach theorem, there exists $\phi \in E^{*}$ with $\norm{\phi}_{E^*}= \norm{x}$, $\dpb{x, \phi}{E}= \norm{x}_{E}^{2}$.

Since $\bracs{T_t|t \ge 0}$ is a contraction semigroup,

\[\dpb{T_tx, \phi}{E}\le \norm{\phi}_{E^*}\norm{x}_{E} \le \norm{x}^{2} = \dpb{x, \phi}{E}\]

At $t = 0$,

\[0 \ge \frac{d}{dt}\dpb{T_tx, \phi}{E}= \dpb{Ax, \phi}{E}\]

Now suppose that $A$ is a dissipative operator with $(\lambda - A)$ being surjective for all $\lambda > 0$. For any $x \in D(A)$, let $\phi \in E^{*}$ with $\norm{\phi}_{E^*}= \norm{x}_{E}$, $\dpb{x, \phi}{E}= \norm{x}_{E}^{2}$, and $\dpb{Ax, \phi}{E}\le 0$, then

\begin{align*}\lambda \norm{x}_{E}^{2}&\le \lambda \dpb{x, \phi}{E}- \dpb{Ax, \phi}{E}\\&= \dpb{(\lambda - A)x, \phi}{E}\le \norm{(\lambda - A)x}_{E}\norm{\phi}_{E^*}\end{align*}

so $(\lambda - A)$ is coercive with $\normn{(\lambda - A)^{-1}}_{L(E; E)}\le \lambda^{-1}$.$\square$