How to prove the generator for strongly continous contraction semigroup? I have a self adjoint operator, which is also negative semi-definite, I have also a Hilbert space. 
How to show that the given operator with given X Hilbert space is  the generator for strongly continous contraction semigroup?
 A: I'm guessing you aren't struggling with showing that a negative semidefinite self-adjoint operator can generate a strongly continuous semigroup, but rather that the semigroup is contractive. Denote the negative semidefinite self-adjoint operator by $\mathcal{H}$ and the semigroup it generates by $G(t)$. Then $G'(t) = \mathcal{H}G(t)$, and
\begin{align}
\frac{d}{dt}\left\|G(t)x\right\|^{2} 
&= \left<G'(t)x, G(t)x \right> + \left<G(t)x, G'(t)x \right>\\
&= \left<\mathcal{H}G(t)x, G(t)x \right> + \left<G(t)x, \mathcal{H}G(t)x \right> \\
&= 2\left<\mathcal{H}G(t)x, G(t)x \right> \\
&\le 0.
\end{align}
By the differential form of Gronwall's inequality, we know that this implies $\left\|G(t)x\right\|^{2} \le \left\|G(0)x\right\|^{2} = \left\|x\right\|^{2} \, \forall t > 0$, so that $\left\|G(t)\right\| \le 1$.
A: Recall that $A$ is the generator of a contractive semigroup on a Banach space $X$ iff $(0,\infty)$ is in the resolvent of $A$ with
$$
               \|\lambda(\lambda I-A)^{-1}\| \le 1,\;\;\; \lambda > 0.
$$
For a densely-defined selfadjoint operator $A$ with $(Ax,x) \le 0$ for all $x \in \mathcal{D}(A)$, it is true that $(0,\infty)$ is in the resolvent. This works for a real or complex Hilbert space $X$. To see why, suppose $\lambda > 0$ is real. Then
$$
        ((\lambda I-A)x,x) = \lambda(x,x)-(Ax,x) \ge \lambda(x,x),\;\;\; x\in\mathcal{D}(A).
$$
Consequently,
$$
    \lambda\|x\|^{2} \le |((\lambda I-A)x,x)| \le \|(\lambda I-A)x\|\|x\| \\
    \lambda\|x\| \le \|(\lambda I-A)x\|, \;\;\; x \in \mathcal{D}(A).
$$
This shows that $\lambda I-A$ is invertible on its range and, because $A$ is closed, the range of $\lambda I -A$ is closed. The range is dense because
$$
                      \mathcal{R}(\lambda I-A)^{\perp} = \mathcal{N}(\lambda I-A)=\{0\}.
$$
Therefore $\lambda > 0$ is in the resolvent of $A$, and the estimate gives
$$
                             \|\lambda(\lambda I-A)^{-1}y\|\le \|y\|,\;\;\; \lambda > 0, \; y \in X.
$$
