43 Conclusion of proof of Corollary

$$\parallel \exp (tA)\parallel \le \parallel S\parallel \parallel S^{-1}\parallel \sum _{j=1}^r\parallel \exp (t{J}_{k_j}({\lambda }_j))\parallel$$

$$\le \parallel S\parallel \parallel S^{-1}\parallel \sum _{j=1}^r{e}^{t\mathrm{\Re }{\lambda }_j}\sum _{\mathrm{\ell }=0}^{k_j-1}\frac{t^{\mathrm{\ell }}\parallel N_{k_j}^{\mathrm{\ell }}\parallel }{\mathrm{\ell }!}.$$

Now choose $\epsilon >0$ such that . Observe that $t^{\mathrm{\ell }}{e}^{-\epsilon t}$ is bounded for all $t>0$, also $e^{t\mathrm{\Re }{\lambda }_j}\le e^{\beta t}{e}^{-\epsilon t}$, so

$$M=\underset{t>0}{sup}\left(e^{-\epsilon t}\parallel S\parallel \parallel S^{-1}\parallel \sum _{j=1}^r\sum _{\mathrm{\ell }=0}^{k_j-1}\frac{t^{\mathrm{\ell }}\parallel N_{k_j}^{\mathrm{\ell }}\parallel }{\mathrm{\ell }!}\right)$$

is finite. Hence

$$\parallel \exp (tA)\parallel \le M{e}^{\beta t}\mathit{\hspace{1em}\hspace{1em}}(t>0).$$