Restriction of a lower semi-continuous functional again lower semi-continuous? Let $F: [a,b]\times \mathbb R \times \mathbb R \rightarrow \mathbb R$ be continuous, $J(u) = \int_{[a,b]} F(x, u, u') dx$ be a functional over $W^{k,p}([a,b])$. We assume that for any uniformly convergent sequence $(u_{r})_{r\in\mathbb N}$ of equi-Lipschitzian functions, $ \lim_{r\rightarrow \infty}u_{r} = u_{0}$, we have:
$$J(u_{0}) \le \liminf_{r\rightarrow\infty} J(u_{r}) $$
My question is: for $I \subset [a,b]$, is the restriction $J_{|I}(u) := \int_{I} F(x, u, u') dx$ lower semi-continous in the same sense that J is? If yes, why? If not, is this possible under addtional assumptions?
Thank you very much for your time and energy!
 A: I assume $I$ means a subinterval $[c,d]$. Let $u_r:I\to \mathbb R$ be a uniformly convergent sequence of $L$-Lipschitz functions. Extend them to $[a,b]$ via
$$U_r(x)=\begin{cases}u_r(c), \quad &x<c \\ u_r(x), &x\in [c,d] \\ u_r(d), &x>d\end{cases}\tag1$$
Clearly, $U_r$ are equi-Lipschitz and converge uniformly. Thus, $$J(  U_0)\le \liminf J(U_r)\tag2$$  The contribution of $[a,c]$ to $J$ is  $ \int_a^c F(x,u_r(c),0)\,dx$, which converges to $ \int_a^c F(x,u_0(c),0)\,dx$ due to the continuity of $F$. Similarly for the contribution of $[d,b]$. Subtract them from (2) to conclude with the desired result.

If $I$ is a general closed subset of $[a,b]$, you can extend $u_r$ in the gaps between components of $I$ by linear interpolation. This preserves $L$-Lipschitzness and uniform convergence of the sequence. The derivatives of the added linear pieces will no longer be $0$, but they will converge uniformly in all gaps of size $>\delta$, for any fixed $\delta$. The gaps of size $<\delta$ contribute little.  
