$\int_{-1}^1\frac1f=\infty$ iff $\int_{-1}^1(u_n')^2f\to0$ Let $f$ be a continuous function on $[-1,1]$ such that $f(x\neq0)>0,f(0)=0$.
How can I show that $\int_{-1}^1\frac{1}{f(t)}dt=\infty$ iff there exists a sequence of functions $u_n$, $C^1$ on $[-1,1]$ such that $\int_{-1}^1 (u_n'(t))^2f(t)dt\underset{{n\infty}}\to0$ and $(u_n)$ converges pointwise to the sign function on $[-1,1]$ ?
The sign function is defined by $sgn(x<0)=-1,sgn(0)=0,sgn(x>0)=1$.
By $C^1$ I mean derivable with a continuous derivative.

Neither way of the equivalence seems easy, and other than having confirmed that the above works both ways on easy cases (such as $f$ being $|x|$), I haven't really made any progress.
What is more since the convergence is only pointwise in the reverse way, it tells us very little on the derivative of $u_n$.
 A: One direction is easy.
If $u_n$ is close to $1$ at some  points and close to -1 at other points then $\int |u_n'|>1$. Writing $u_n=(u_nf^{1/2})f^{-1/2}$, the Cauchy-Schwarz inequality shows that $$
1<\left(\int|u_n'|\right)^2\le\int(u_n')^2f\int1/f,$$so if $\int1/f<\infty$ then $\int (u_n')^2f$ cannot tend to $0$.
Ah. For the other direction I'm going to pretend the problem's posed on $[0,1]$ and let you add the other half. (So we want $u_n(0)=0$ and $u_n(t)\to1$ for $0<t\le 1$.) Suppose $\int1/f=\infty$. Then we can find $a_n$ and $b_n$ with $0<a_n<b_n$, $b_n\to0$, and $$\int_{a_n}^{b_n}1/f\to\infty.$$
Let $v_n$ satisfy $v_n(0)=0$ and $$v_n'=\frac cf\chi_{[a_n,b_n]},$$where $c$ is chosen so that $v_n=1$ on $[b_n,1]$. (Hence $v_n(t)\to1$ for $t\in(0,1]$.) This implies that $c=\left(\int_{a_n}^{b_n}1/f\right)^{-1}$, and hence that $$\int(v_n')^2f=\left(\int_{a_n}^{b_n}1/f\right)^{-1}\to0.$$
Of course $v_n$ is not quite $C^1$. It's piecewise $C^1$; let $u_n$ be $v_n$ except smoothed out a little at the corners.
