93 Proof

As in frame 48, the solution is determined by

$$X(t)={\int }_0^t\exp (((t-v)A)BU(v)dv$$

whcih a convolution type integral of functions in $(E)$, since

$$\parallel \exp (tA)\parallel \le M_1{e}^{{\beta }_{1}t},\parallel U(t)\parallel \le M_2{e}^{{\beta }_{2}t}$$

so with $M=M_1{M}_2\parallel B\parallel$ and $\beta =\max \{{\beta }_1,{\beta }_2\}$, we have

$$\parallel \exp (((t-v)A)BU(v)\parallel \le M{e}^{\beta t}$$

so

$$\parallel X(t)\parallel \le tM{e}^{\beta t}\le M{e}^{(\beta +1)t}$$

and so $X$ satisfies $(E)$ and has a Laplace transform. From the differential equation, $dX/dt$ also satisfies $(E)$ and has a Laplace transform, and likewise $Y$ satisfies $(E)$.