ThisThe answer is quite straightforward: For the derivativeno. Indeed, let $I'(x)$$n=1$, $T=1$, and any $h\in C^1[0,T]$ we have$\Lambda(t,u,v)\equiv v^2$, so that
$$I'(x)(h)=\int_0^T (\nabla\Lambda)(t,x(t),x'(t))\cdot(0,h(t),h'(t)) \, dt, \tag{1}$$\begin{equation}
I(x)=\int_0^1 x'(t)^2\,dt.
\end{equation}
whence for allLet $x$ in any bounded neighborhood$x_0:=0$ and, for each real $U$ of$b\ge1$ and all $x_0$ we have$t\in[0,1]$,
$$|I'(x)(h)|\le M\int_0^T \|(0,h(t),h'(t))\|\,dt
\le M\Big(\int_0^T \|h(t)\|\,dt+\int_0^T \|h'(t)\|\,dt\Big)
=M\|h\|,$$\begin{equation}
y_b(t):=e^{-bt}.
\end{equation}
whereThen
$$M:=\sup_{x\in U}\|(\nabla\Lambda)(t,x(t),x'(t))\|<\infty,$$\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}
sinceso that $\Lambda$ is continuously differentiable.
So,$y_b$ in the derivativeball of radius $I'$ is locally bounded and hence$2$ centered at $I$ is locally Lipschitz$x_0$.
Details on $I'$, in response to a comment by the OP: If $I'$ is understood in the Gateaux sense (which is enough for the "locally Lipschitz" condition on $I$) However, then (1) follows from the continuous differentiability of
\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 $\Lambda$ by the dominated convergence theorem$b\to\infty$. Working just a bit harderSo, one can show that $I'$ exists in the functional Fréchet sense as well$I$ is not locally Lipschitz.
