Showing that an operator generates a contraction semigroup Let $A$ be the infinitesimal generator of a contraction semigroup $(T(t))_{t\ge 0}$ on the Hilbert space $X$, and $D\in\mathcal{L}(X)$.
I want to show that the operator $A+D-2\|D\|I$ with domain $D(A)$ generates a contraction semigroup on $X$.
By the Lummer-Phillips Theorem, we have that if $A$ is a linear operator with domain $D(A)$ on a Hilbert space $X$, then $A$ is the infinitesimal generator of a contraction semigroup $(T(t))_{t\ge 0}$ if and only if $A$ is dissipative and $ran(I-A)=X$.
We have that a linear operator $A:D(A)\subset X\to X$ is dissipative if and only if
$$\|(\alpha I-A)x\|\ge\alpha\|x\|$$
for $x\in D(A)$, $\alpha>0$.
So for $\alpha>0$, we have
$\|(\alpha I-(A+D-2\|D\|I)x\|=\|(\alpha I-A-D+2\|D\|I)x\|$
Where do I go from here?
 A: Assuming you're working on a complex Hilbert space, the range condition and the following condition are equivalent to dissipative:
$$
                 \Re(Ax,x) \le 0,\;\;\; x \in \mathcal{D}(A).
$$
This last condition persists if you add another bounded operator $C$ satisfying the same condition. In your case, because $D$ is bounded,
$$
                     |(Dx,x)| \le \|D\|\|x\|^{2},\\
                        \Re(Dx,x) \le \|D\|\|x\|^{2}, \\
                        \Re((D-\|D\|I)x,x) \le 0.
$$
Let
$$
                  B = A+(D-\|D\|I)
$$
Because $\Re(Bx,x) \le 0$, it follows that $B$ is dissipative iff $B-\alpha I$ is surjective for some real $\alpha > 0$ because $\mathcal{R}(B-\alpha I)$ is the same for all $\alpha > 0$. Notice that
\begin{align}
    B-\alpha I & = (A-\alpha I)+(D-\|D\|I) \\
         & = \{I+(D-\|D\|I)(A-\alpha I)^{-1}\}(A-\alpha I)
\end{align}
The operator in braces is invertible for large enough $\alpha$ because
$$
        \|(D-\|D\|I)(A-\alpha I)^{-1}\| \le 2\|D\|/\alpha < 1
$$
for large enough $\alpha > 0$. Hence, $B-\alpha I$ is surjective because the operator on the right is surjective for large enough $\alpha$. And that gives you what you want: $B$ is the generator of a contractive $C_{0}$ semigroup. The same is true of $B-\|D\|I$, which is the operator you wanted to know about.
