Let $(H;(\cdot,\cdot))$ be a complex Hilbert space and let $A: D(A) \rightarrow H$ be a densely dened closed operator on H. Let A be an m-dissipative operator that is \begin{align} &(Ax, x) \leq 0 \text{ for all } x \in D(A) \text{ and } \\ &\lambda I - A \text{ is surjective for every } \lambda > 0. \end{align} Prove that $A$ is the innitesimal generator of a $C_0$-semigroup in $H$, that is, a semigroup $\{Q(t)\}$ satisfying the conditions of Denition 13.34(a)-(c) of Rudin's book.
Hint: show first that for each $\lambda > 0$ the operator $\lambda I - A$ has a bounded inverse with norm $\leq \frac{1}{\lambda}$.
What I have done is the following:
Let $\lambda>0$ and $x\in\mathscr{D}(A)$. Then $\lambda I - A$ is surjective. We show that $\lambda I - A$ is one-to-one so that $\lambda I - A$ is invertible. Since $(Ax,x)\leq 0$ we have $$ \lambda \|x\|^2 = (\lambda x,x) \leq ((\lambda I -A)x,x) \leq \|\lambda I -A\| \|x\|, $$ so that $$ \| (\lambda I - A)x \| \geq \lambda \|x\|. \tag{*}\label{*} $$ Now $(\lambda I -A)x=0$ implies that $\|x\|=0$ and so by 12.1 $x=0$. Hence $\mathscr{N}(\lambda I - T)=\{0\}$ and $\lambda I - A$ is one-to-one.
Let $y\in H$ then since $\lambda I - A$ is surjective we can find a $x\in\mathscr{D}(\lambda I -A)$ such that $(\lambda I - A)x=y$. Then (\ref{*}) implies $$ \|(\lambda I - A)^{-1}y\|=\|(\lambda I - A)^{-1}(\lambda I - A)x\|=\|x\| \leq \frac{1}{\lambda} \| (\lambda I - A)x \|= \frac{1}{\lambda}\|y\| $$ thus $$ \|(\lambda I - A)^{-1}\| \leq \frac{1}{\lambda}. $$ Then for positive $m$, $C=1$ and $\gamma=0$ we have $$ \|(\lambda I - A)^{-m}\| \leq \left(\frac{1}{\lambda}\right)^{-m}. $$ So by theorem 13.37 $A$ is the infinitesimal generator of a $C_0$-semigroup.
The problem for me is that I haven't used the fact that the operator is closed. So is the closeness unnecessary, or am I missing something?
