Show that if $h \in C^1([a,b]\times[-ε,ε])$ with $ε>0$, then the function $s \in [-ε,ε] \mapsto \int_a^b h(t,s) \,dt$ is $C^1$

Question

Show that if $h \in C^1([a,b]\times[-ε,ε])$ with $ε>0$, then the function $$s \in [-ε,ε] \mapsto \int_a^b h(t,s) \,dt$$ is $C^1$ and $${d\over ds}\int_a^b h(t,s) \,dt=\int_a^b {\partial h \over \partial s}(t,s)\,ds$$

Any ideas how to show these parts?any help would be appreciated.

Answer 1

Whenever possible, replace derivatives with integrals instead. For example, \begin{align} \int_{-\epsilon}^{u}\int_{a}^{b}\frac{\partial}{\partial s}h(t,s)dtds & = \int_{a}^{b}\int_{-\epsilon}^{u}\frac{\partial}{\partial s}h(t,s)dsdt \\ & = \int_{a}^{b}h(t,u)-h(t,-\epsilon)dt \\ & = \int_{a}^{b}h(t,u)dt - C \end{align} The left side is differentiable in $u$, which means that the right side must also be. And, by the Fundamental Theorem of Calculus, $$ \int_{a}^{b}\frac{\partial}{\partial u}h(t,u)dt = \frac{d}{du}\int_{a}^{b}h(t,u)du. $$ You do have to argue that $\int_{a}^{b}\frac{\partial}{\partial s}h(t,s)dt$ is a continuous function of $s$ in order to apply the Fundamental Theorem, and you have to argue that the interchange of orders of integration is justified by the joint continuity in $t$, $u$ of the integrand, but that's not so tough. In general, it's simpler to try to turn differentiation problems into integration and to interchange orders of integration than to try interchanging differentiation and integration.