Weak $L^p$ convergence for passing to the limit in piecewise linear approximation of sign function?

Question

Consider $$ S_\epsilon(\xi) = \begin{cases} 1 & \text{ if } \xi > \epsilon \\ \xi/\epsilon &\text{ if } |\xi| < \epsilon \\ -1 &\text{ if } \xi < - \epsilon \end{cases}$$ which is a smoothed version of the $\mathrm{sign}$ function.

Suppose that $u_n \to u$ weakly in $L^p([0,1])$ for all $p \in [1,\infty]$ as $n \to \infty$. Is it true that $S_\epsilon(u_n-1) \to S_\epsilon(u-1)$ weakly in some $L^p$?

Answer 1

Suppose $\epsilon \le 1$. On $[0,1]$, let $$ u_n(x) = \cases{ 4 & if $x \in \left[\tfrac{2j}{2n},\tfrac{2j+1}{2n}\right)$\\ 0 & if $x \in \left[\tfrac{2j+1}{2n},\tfrac{2j+2}{2n}\right)$. } $$ Then $u_n \rightharpoonup 2$ in $L^p([0,1])$ for $1 \le p < \infty$, but $S_\epsilon(u_n-1) \rightharpoonup 0 \ne \epsilon = S_\epsilon(2-1)$.

Not sure about $p = \infty$, but I doubt this counterexample works.