What is the derivative of $F(x)= f(x(0) , x(1))$?

Question

Let $f: \mathbb {R^n \times R^n } \to \mathbb R$ be continuously differentiable function. Define the mapping $F: AC[0,1] \to \mathbb R $ with $F(x)= f(x(0) , x(1)). $

My question : Is $F$ continuously differentiable in Frechet sense ? If so then what is the derivative of $F$?

P.S: $AC[0, 1]$ stands for the space of all absolutely continuous function $x: [0,1] \to \mathbb R^n$ equipped with $W^{1,1}$ norm which is $$ \| x \| := \int_{0}^{1} \|x(t)\| \; dt + \int_{0}^{1} \|x' (t)\| \; dt$$

Answer 1

To show continuity of the map $x\mapsto x(t)$ for $t\in[0,1]$ fixed, notice that for any $u\in[0,1]$ $$x(t)=x(u)+\int^t_ux'(s)ds$$ Hence $$|x(t)|\leq |x(u)|+\int^1_0|x'(s)|\,ds$$ Integrating over $[0,1]$ gives $$|x(t)|\leq \int^1_0|x(u)|\,du +\int^1_0|x'(s)|\,ds=\|x\|$$

Answer 2

Use the chain run. The map $\lambda:x\mapsto (x(0),x(1))$ from AC to $\mathbb{R}^2$ is linear. The function $F=f\circ\lambda$.