I want to show that the inequality
(1) $\int_{0}^{\infty}\vert h(t,x) \vert \varphi(x)\leq e^{-\mu(t-y_{0})} \int_{0}^{\infty}\vert h(y_{0},x)\vert \varphi (x) dx < \infty$
with
(2) $ h(t,x)=\tilde{n}(t,x)-\int_{0}^{\infty}n^{0}(y)\phi(y)dyN(x)$
is equal to
(3) $\int_{0}^{\infty}\vert mN(x)-\tilde{n}(t,x)\vert \varphi(x)dx\leq C_{0}e^{-\mu(t-y_{0})}\int_{0}^{\infty}\vert mN(x)-\tilde{n}(y_{0},x)\vert \varphi(x)dx dx$
with
$m=\int_{0}^{\infty}n^{0}(x)\phi(x)dx.$
My idea was just du insert (2) into (1), but then I don't know where the $C_0$ and the second dx gets from. And I'm not sure if the second dx is a mistake or there on purpose.
Given that $$m=\int_{0}^{\infty}n^{0}(x)\phi(x)\ \mathrm dx$$ We have $$h(t,x)=\tilde{n}(t,x)-mN(x)$$ For the left hand side, we have $$\int_{0}^{\infty}\left|h(t,x)\right|\varphi(x)\ \mathrm dx$$ $$=\int_{0}^{\infty}\left|\tilde{n}(t,x)-mN(x)\right|\varphi(x)\ \mathrm dx$$ For the right hand side, we have $$e^{-\mu\left(t-y_0\right)}\int_{0}^{\infty}\left|h(y_0,x)\right|\varphi(x)\ \mathrm dx$$ $$=e^{-\mu\left(t-y_0\right)}\int_{0}^{\infty}\left|\tilde{n}(y_0,x)-mN(x)\right|\varphi(x)\ \mathrm dx$$ The extra $\mathrm dx$ on the right hand side of your post is a typo and $C_0$ is most likely a positive constant. Without more information, we cannot tell where $C_0$ actually comes from. You were pretty much on the right track.