Let the function $\Lambda : [0,T] \times \mathbb{R^n} \times \mathbb{R^n} \to \mathbb R$ be continuously differentiable. Then the integral functional $I(x) = \int\_{0}^{T} \Lambda (t , x(t), \dot{x} (t)) \; dt $ is finite value for all $x \in C^{2}[0 ,2]$ . 


My question: Is the integral functional $I(x) = \int\_{0}^{T} \Lambda (t , x(t), \dot{x} (t)) \; dt $ lipschitz on a neighborhood of $x\_0$ on the space $C^2 [0,T]$ equipped with the norm $W^{1,1}$ ?


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 $$ \| x \| := \int\_{0}^{T} \|x(t)\| \; dt + \int\_{0}^{T} \|x' (t)\| \; dt$$ clearly $C^2 [0 ,T] \subset AC [0 ,T]$.



Is the integral functional $I(x) = \int_{0}^{T} \Lambda (t , x(t), \dot{x} (t)) \; dt $ locally lipschitz on the space $C^2 [0 ,T] $?
The answer is no. Indeed, let $n=1$, $T=1$, and $\Lambda(t,u,v)\equiv v^2$, so that 
\begin{equation}
 I(x)=\int\_0^1 x'(t)^2\,dt. 
\end{equation}
Let $x\_0:=0$ and, for each real $b\ge1$ and all $t\in[0,1]$,
\begin{equation}
 y\_b(t):=e^{-bt}. 
\end{equation}
Then 
\begin{equation}
 \|y\_b-x\_0\|=\|y\_b\|=\int\_0^1 |y\_b(t)|\,dt+\int\_0^1 |y\_b'(t)|\,dt\le1/b+1\le2,
\end{equation}
so that $y\_b$ is in the ball of radius $2$ centered at $x\_0$. 


However, 
\begin{equation}
 I(y\_b)-I(x\_0)=I(y\_b)=\int\_0^1 y\_b'(t)^2\,dt\sim b/2\to\infty
\end{equation}
as $b\to\infty$. So, the functional $I$ is not locally Lipschitz. 


