# Continuity of mild solution

This is a question on the proof of Theorem 6.1.2, Pazy's book Semigroups of Linear Operators and Applications to Partial Differential Equations. Also the title might not be too accurate, as my question is the continuity of the element mapped by $$F$$ defined below.

$$X$$: a Banach space
$$f:[t_0,T]\times X\to X$$, continuous in $$t$$ on $$[t_0,T]$$ and uniformly Lipschitz continuous (with constant $$L$$) on $$X$$.
$$T(t)$$, $$t\ge 0$$: $$C_0$$ semigroup on $$X$$

It is (implicitly) claimed that the mapping $$(Fu)(t)=T\left(t-t_{0}\right)u_{0}+\int_{t_{0}}^{t}T(t-s)f(s,u(s))ds,\quad t_{0}\leq t\leq T,$$ takes elements in $$C([t_0,T]:X)$$ to $$C([t_0,T]:X)$$.

But I do not see why $$Fu\in C([t_0,T]:X)$$.

The continuity of the first term of the integral above is clear, but I do not see why $$\int_{t_{0}}^{t}T(t-s)f(s,u(s))ds$$ is continuous. In the book, for the case where $$T(t)$$ is analytic, various decompositions of $$\int_{t_{0}}^{t}T(t-s)f(s,u(s))ds$$ are considered but we cannot use the properties of analytic semigroups here.
I have been considering $$\|(Fu)(t_1)-(Fu)(t_2)\|_X$$, but the $$t$$-dependence of "$$T(t-s)$$" is making things hard.
If $$T$$ is analytic on the positive real axis, say, then for $$t>0$$ we can use the uniformly continuity (in time) of $$T$$ in the operator norm on positive intervals, but Theorem 6.1.2 does not say anything about the continuity of $$T$$ in the operator norm.

$$f$$ is continuous, $$u$$ is continuous, thus the composition $$t\mapsto f(t,u(t))$$ is continuous. Elementary calculus then shows that $$\|f(t,u(t))\|$$ is bounded over the compact interval $$[t_0,T]$$. The semi-group map $$T$$ is also continuous, thus likewise bounded. This implies that that $$t\mapsto\int_{t_0}^tT(t-s)f(s,u(s))\,ds$$ is even Lipschitz continuous with the product of both bounds as the Lipschitz constant. $$\|(Fu)(t_2)-(Fu)t_1|\le \|T(t_2-t_0)-T(t_1-t_0)\|\,\|u_0\|+\|T\|_{[0,T-t_0]}\,\|f\circ(id,u)\|_{[t_0,T]}\, |t_2-t_1|$$
• Ah, ok, thank you. But also sorry but I do not follow the last sentence. Using the boundedness trying to show the Lipschitzness is exactly what I did, but since I have $t$ in the integrand $T(t-s)$ I did not find this straightforward. – user41467 Jan 15 at 16:25
• I'm not seeing that, as $t-s\in[0,t-t_0]\subset [0,T-t_0]$, so the same global bounds apply. I only see thing becoming more involved if you want to prove some uniform property for some class of functions $u$. – LutzL Jan 15 at 16:35
• I do not understand how you got the second term $\|T\|_{[0,T-t_0]}\,\|f\circ(id,u)\|_{[t_0,T]}\, |t_2-t_1|$ (how did you get $|t_2-t_1|$?). I have \begin{align} &\int_{t_0}^{t_2}T(t_2-s)f(s,u(s))\,ds-\int_{t_0}^{t_1}T(t_1-s)f(s,u(s))\,ds\\ &=\int_{t_0}^{t_1}T(t_2-s)f(s,u(s))\,ds+\int_{t_1}^{t_2}T(t_2-s)f(s,u(s))\,ds-\int_{t_0}^{t_1}T(t_1-s)f(s,u(s))\,ds, \end{align} and the Lipschitzness of the middle one is fine. – user41467 Jan 15 at 17:06
• For the first and third, we have \begin{align} &\int_{t_0}^{t_1}\|(T(t_2-s)-T(t_1-s))x(s)\|\,ds \\&\leq\int_{t_0}^{t_1} \|T(t_1-s)\|\|T(t_2-t_1)x(s)-x(s)\|_X \,ds\leq \|T\|_{[t_0,T]} \int_{t_0}^{T} \|T(t_2-t_1)x(s)-x(s)\|_X\,ds, \end{align} but $t\mapsto \|T(t)\|$ is not continuous and I don't know how to proceed nor how you got there. – user41467 Jan 15 at 17:06