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] $?

Question

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 $C^2 [0 ,T] \subset AC [0 ,T]$.

Answer 1

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.