Lower semicontinuous functional Consider the space $A=(C^1([0,1],\mathbb R),\|.\|_{L^\infty})$ norm and look at the functional
$$
\mathcal F: A\to\mathbb R_+, f\mapsto \int_{0}^1\left|f'(t)\right|~\mathrm dt.
$$
This functional is not continuous. My question: Is it lower semi-continuous?
And could there be a meaningful extension of $\mathcal F$ to the closure of $A$ under the $L^\infty$-norm?
 A: Your functional is the "length" functional. It is lower-semicontinuous. It can be extended to all continuous function as the usual "length" of a curve. You can extend to all the space $L^\infty$ by letting its value be $+\infty$ on non rectifiable curves.
Some more details. 
Consider the functional
$$
\ell(f) = \sup \left\{\sum |f(x_{i+1})-f(x_{i})|\colon 0=x_0\le x_1<x_2<\dots<x_N=1\right\}.
$$
One can prove that if $f\in C^1$ then
$$
\ell(f) = \int |f'(t)|\, dt.
$$
Moreover if $f\in C^0$ then $\ell(f)<+\infty$. The space of $f$ such that $\ell(f)<+\infty$ is called $BV$ (functions of Bounded Variation).
Clearly the functional $\ell$ is lower-semicontinous because it is the $\sup$ of a family of continuous functionals.
A: This functional is DEFINITELY continuous, when defined on $C^1[0,1]$, with its standard norm.
Let $f\in C^1[0,1]$, then
$$
\lvert Af\rvert=\Big\lvert\int_0^1 f'(t)\,dt\Big\rvert\le \int_0^1 \lvert f'\rvert\,dt
\le \int_0^1 \| f'\|_\infty\,dt=\| f'\|_\infty\le \| f\|_\infty+\| f'\|_\infty=\|f\|_{C^1[0,1]}.
$$
Is it lower semi-continuous, w.r.t. to $\|\cdot\|_\infty$-norm?
Let's be reminded of the definition: $A$ is lower semi-continuous at $f=f_0$, if for every $\varepsilon>0$, there exists an open $U\subset C^1[0,1]$ (open w.r.t. to the $\|\cdot\|_\infty$-norm), with $f_0\in U$, such that
$$
\int_0^1 \lvert f'(x)\rvert\,dx\le\int_0^1 \lvert f'_0(x)\rvert\,dx+\varepsilon, 
$$
for every $f\in U$.
Apparently this is NOT possible, as if $U$ is open, then $B_\delta(f_0)\subset U$, for some $\delta>0$, and 
$$
f_n=f_0+\frac{\delta}{2}\sin nx\in B_\delta(f_0)\subset U,
$$
and 
$$
\lim_{n\to\infty} \lvert Af_n\rvert=\infty.
$$
