If $x\mapsto F(x)=\int^{x}_{a}f(s)ds$, then $F$ is $C^1$ and $F'(x)=f(x),\;\;\forall\;x\in[a,b]$

Question

Assuming that \begin{align}f:[a,b]\to\Bbb{R}^n\end{align} is continuous, and \begin{align}F:[a,b]\to\Bbb{R}^n\end{align} \begin{align}x\mapsto F(x)=\int^{x}_{a}f(s)ds.\end{align} I want to prove that $F$ is $C^1$ and $F'(x)=f(x),\;\;\forall\;x\in[a,b].$

MY WORK

To prove differentiablity, it suffices to prove that \begin{align}\Vert F(x_0+h)- F(x_0)-hf(x_0)\Vert \leq\Vert h \Vert\epsilon(h)\end{align}

Since, $f:[a,b]\to\Bbb{R}^n$, then it is uniformly continuous.

\begin{align}\Vert F(x_0+h)- F(x_0)-hf(x_0) \Vert\end{align} \begin{align}=\Vert \int^{x_0+h}_{a}f(s)ds- \int^{x_0}_{a}f(s)ds-hf(x_0) \Vert\end{align} \begin{align}=\Vert \int^{x_0+h}_{x_0}f(s)ds-hf(x_0) \Vert\end{align} I'm stuck at this point, can anyone help me out?

Answer 1

Use that the integral over a constant is the constant times the interval length, in reverse \begin{align} \left\Vert \int^{x_0+h}_{x_0}f(s)\,ds-hf(x_0) \right\Vert &=\left\Vert \int^{x_0+h}_{x_0}(f(s)-f(x_0))\,ds \right\Vert\\ &\le \int^{x_0+h}_{x_0}\Vert f(s)-f(x_0) \Vert\,ds \end{align}