124 Proof

Let $V(t)=⟨KX(t),X(t)⟩$, so $V(t)\ge 0$ for all $t\ge 0$, and use the differential equation to find

$$\frac{dV}{dt}=⟨K\frac{dX}{dt},X(t)⟩+⟨KX,\frac{dX}{dt}⟩$$

$$=⟨KAX(t),X(t)⟩+⟨KX(t),AX(t)⟩$$

$$=⟨KAX(t),X(t)⟩+⟨A^{†}KX(t),X(t)⟩$$

$$=-⟨-(A^{†}K+KA)X(t),X(t)⟩\le 0.$$

Hence $V(t)$ is decreasing on $(0,\mathrm{\infty })$. Since $K$ is positive definite, the eigenvalues of $K$ are ${\kappa }_1\ge {\kappa }_2\ge \mathrm{\dots }\ge {\kappa }_n$, where ${\kappa }_n>0$; so by W3.3

$$0\le {\kappa }_n⟨X(t),X(t)⟩\le ⟨KX(t),X(t)⟩\le ⟨KX(0),X(0)⟩,$$

and so $\parallel X(t)\parallel \le {(⟨K{X}_0,X_0⟩/{\kappa }_n)}^{1/2}$ for all $t\ge 0$.