Let the function $\Lambda : [0,T] \times \mathbb{R^n} \times \mathbb{R^n} \to \mathbb R$ be continuously differentiable. Assume the integral functional $I(x) = \int_{0}^{T} \Lambda (t , x(t), \dot{x} (t)) \; dt $ is finite value for all $x \in AC[0,T]$ on a neighborhood of $x_0$.
My question: Is the integral functional $I(x) = \int_{0}^{T} \Lambda (t , x(t), \dot{x} (t)) \; dt $ lipschitz on a neighborhood of $x_0$ ?
P.S: $AC[0, T]$ stands for the space of all absolutely continuous function $x: [0,T] \to \mathbb R^n$ equipped with $W^{1,1}$ norm which is $$ \| x \| := \int_{0}^{T} \|x(t)\| \; dt + \int_{0}^{T} \|x' (t)\| \; dt$$
The answer is no.
Consider the case $AC([0,1])$ and $\Lambda(t,x,y) = y\sin (y)$. This function is $C^\infty$. Set $x_n(t) = nt$ on $[0,n^{-1}]$ and $1$ for $x > n^{-1}$. These functions are uniformly bounded in $AC$.
Then $\|x_n - x_{n+1}\|_{AC} = O(n^{-1})$. But $I(x_n) = \sin n$. Clearly there is no finite Lipschitz constant.