On weak-star convergence of functions and almost everywhere convergence of derivatives

Question

Suppose that a sequence $(v_n)$ of ${\rm H}^2(0,1)$ functions satisfies: $v_n\rightarrow{}v_0$ weakly-star in ${\rm L}^{\infty}(0,1)$ as $n\rightarrow+\infty$, $v_n'\rightarrow u_0$ almost everywhere on $(0,1)$ as $n\rightarrow+\infty$, where $u_0\in {\rm BV}((0,1);\{-1,1\})$. Is it true that it follows that we have $v_0'=u_0$? Remark. If it were true that almost everywhere convergence of $(v_n)$ implies convergence in the sense of distributions (i.e. in ${\cal D}'(0,1)$), then the claim follows. But I do not if this is true. Still, I think my initial assertion is true, but I do not know how to prove it. The point here is that $u_0$ is BV-function which takes values $-1$ and $1$ only, so I guess $v_0$ can not be constant. How to proceed? Thanks.

Answer 1

Take a functions $v_n' \in H^1(0,1)$, $n \ge 3$, with $v_n' = 1$ on $(0,1/2-1/n) \cup (1/2+1/n,1)$ and $\int_0^1 v_n' \, \mathrm{d}x = 0$. Then, we set $v_n(t) = \int_0^t v_n' \, \mathrm{d}x$.

Now, we can check that $v_n' \to u_0 = 1$ a.e. and $v_n \to v_0$ weak-* in $L^\infty(0,1)$ with $$v_0(t) = \begin{cases} t & \text{for } t < 1/2, \\ t - 1 & \text{for } t > 1/2.\end{cases}$$ Further, $v_0$ is not (weakly) differentiable.

(However, we still have that $v_0$ is differentiable a.e. and this derivative is $u_0 = 1$)