Decomposition of propagation function

Question

In the book of Jean Prüss (Evolutionary Integral Equations and Applications, 2012, pg. 111) it says in the Proposition 4.10...
Let $dc$ be completely positive, let $w(t;\tau)$ denote the associated propagation function, and let $\kappa$, $\omega$, $\alpha$ be as in (4.35). Then $$w(t; τ ) = e^{−ωτ }w_0 (t − κτ ; τ ) + e^{−ατ} e_0 (t − κτ ), t, τ > 0,$$ where $w_0(·; τ )$ is nondecreasing, left-continuous, $w_0 (t; ·)$ is right-continuous, non-increasing, and $w_0 (t; τ ) > 0$ iff $t > 0; w_0 (0+; τ ) = 0$. Here, $e_0$ is the Heavside function, $\displaystyle κ=\lim_{t\to 0+}k(t)$ where $k$ is the creep function associated, $\displaystyle \omega=\lim_{t\to \infty}k(t)/t$ and $\displaystyle \alpha=\lim_{t\to 0+}k'(t)$. On the other hand, in the page 120 we have some examples of propagation functions. One of these examples is $$\omega(t;\tau)= e^{−τ} e_0(t − τ )$$ with $dc=e^{-t}dt$ and creep function $k(t)=1+t$ (Maxwell Fluid). Comparing this example with the Proposition 4.10 we have that $\omega_0\equiv 0$, but would this be contradicing the Proposition 4.10? Because by Proposition 4.10 $\omega(t;\tau)>0$ for all $t>0$. Can somebody help me? I thought the problem would be that $dc$ wouldn't be completly positive. But we have $$\int_0^tk(t-\tau)dc(\tau)=t$$ then it is completly positive.