Prove that $G$ is differentiable on $U$ and $G'(x)=\int^{b}_{a}D_2f(s,x)ds,\;\;x\in U$

Question

Let $U$ be open in $\Bbb{R}^n$ and \begin{align}f:[a,b]\times U\to \Bbb{R}^m\end{align} be continuous. Let \begin{align}G(x)=\int^{b}_{a}f(s,x)ds,\;\;x\in U.\end{align} We assume that $D_2 f$ is continuous on $[a,b]\times U.$ I want to prove that $G$ is differentiable on $U$ and \begin{align}G'(x)=\int^{b}_{a}D_2f(s,x)ds,\;\;x\in U.\end{align}.

The case of differentiability:

I picked $h\in U.$ Then, \begin{align}G(x+h)-G(x)=\int^{b}_{a}f(s,x+h)ds-\int^{b}_{a}f(s,x)ds,\;\;x\in U.\end{align} \begin{align}=\int^{b}_{a}\big[f(s,x+h)-f(s,x)\big]ds.\end{align} So, I got stuck here and couldn't proceed. Please, can anyone please help out?

Answer 1

Your formula for $G(x+h)-G(x)$ is wrong. The correct expression is $\int_a^{b} f(s,x+h)\, ds -\int_a^{b} f(s,x)\, ds$. To prove the two statement about $G$ it is enough restrict your attention to a compact subset of $U$. When you do this $D_2f$ becomes uniformly bounded. By MVT $f(s,x+h) - f(s,x)=hD_2f(s,t)$ for some $t$between $x$ and $x+h$. Can you now complete the argument using the fact that $D_2f(s,t) \to D_2f(s,x)$?