On the resolvent set of an unbounded operator 
Suppose $A$ is the infinitesimal generator of a $C_0$ semigroup $S(t)$ on an Hilbert space $X$. If
  $$\langle Ax, x\rangle \leq \omega \|x \|^2 \ \ \ \forall x \in \mathfrak{D}(A)$$
  then
  $$\|S(t)\| \leq e^{\omega t} \ \ \forall t \geq 0$$

How do I prove this? Notice that $A$ is unbounded, closed and densely defined, but we don't know anything else.
My idea until now is to prove that we are in the hypotheses of the Hille-Yosida theorem, namely that:
$$\rho(A) \supset (\omega, \infty) \ \ \ \mathrm{and} \ \ \ \|R_{\lambda}\| \leq \frac{1}{\lambda - \omega} \ \ \forall \lambda > \omega$$
In particular, it is easy to see that if $\omega < \lambda \in \rho(A)$ then the second hypothesis is immediately satisfied. Therefore, I only need to show that $(\omega, \infty) \subset \rho(A)$. Now, if by contradiction we assume that $\lambda > \omega$ is not in $\rho(A)$, then the image $\mathfrak{R}(\lambda I- A) \not = X$. Therefore, there exists a $y \not = 0$ such that 
$$\langle x - Ax, y \rangle = 0 \ \ \ \forall x \in \mathfrak{D}(A)$$
I would like to show that $y = 0$, but I don't know how to do it. Notice, for instance, that $y \in \mathfrak{D}(A^*)$, but we don't know if $A^*$ is dissipative. At the same time, we don't know if $y \in \mathfrak{D}(A)$.
How can I approach this problem?
 A: Take $B=A-\omega$. The inequality above implies that $B$ is dissipative. You need $\lambda- B$ to be surjective for some $\lambda>0$ to conclude. This follows easily since $A$ is a generator of $C_0$-semigroup. Corollary 3.20 in the book of Engel & Nagel, implies that the semigroup $T(t)=e^{-\omega t} S(t)$ of $B$ is a contraction.
A: Because $\frac{d}{dt}(S(t)x=AS(t)x$ for all $t > 0$ (and as a right derivative at $0$,) then
$$
         \frac{d}{dt}\|S(t)x\|^2=\frac{d}{dt}\langle S(t)x,S(t)x\rangle \\
   = \langle AS(t)x,S(t)x\rangle+\langle S(t)x,AS(t)x\rangle \\
   \le 2w\langle S(t)x,S(t)x\rangle = 2w\|S(t)x\|^2
$$
Therefore, for all $t \ge 0$, $x\in X$,
$$
                   \frac{d}{dt}\|S(t)x\|^2-2w\|S(t)x\|^2 \le 0 \\
                        \frac{d}{dt}(e^{-2wt}\|S(t)x\|^2) \le 0 \\
                        e^{-2wr}\|S(r)x\|^2|_{r=0}^{r=t} \le 0 \\
                         e^{-2wt}\|S(t)x\|^2 \le \|S(0)x\|^2 \\
                          \|S(t)x\|^2 \le e^{2wt}\|x\|^2 \\
                            \|S(t)x\| \le e^{wt}\|x\|.
$$
Because this holds for all $x$, it follows that
$$
                               \|S(t)\| \le e^{wt},\;\;\; t \ge 0.
$$
