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.


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 $u_0 = 1$)

  • $\begingroup$ I see. Can we at least obtain $||v_0||_{\infty}=||w_0||_{\infty}$, where $w_0(s):=\int_{c}^{s}u_0(\xi)d\xi-u_0(c)$, $s\in (0,1)$ for some convenient choice of $c\in (0,1)$? In your example we can take $c=0$, and both $\infty$-norms are equal to ${1\over 2}$. Is this true in the general case? I would settle for that. We can add zero boundary conditions (as you did), if it helps. Thanks. $\endgroup$ – Andrija Jan 5 '18 at 7:53
  • $\begingroup$ No, I don't think so. You could modify my example with $\int_0^1 v_n'\,\mathrm{d}x = -42$. Then, $\|w_0\|_\infty$ will always be very big. $\endgroup$ – gerw Jan 5 '18 at 8:56
  • $\begingroup$ Correction: given $v_0$, $w_0$ should be defined as $w_0(s):=\int_c^su_0(\xi)d\xi-A(v_0)$, where $A=A(v_0)$ is some constant which is allowed to depend on $v_0$, and $c\in (0,1)$ is some chosen point, say $c=0$. Typically, if $||v_0||_{\infty}$ is large, $A(v_0)$ is also large. In my previos comment the formula for $w_0$ is too restrictive since in that case we have $||w_0||_{\infty}\leq 2$. The point here is that I want to achieve $w_0'=u_0$ and $||w_0||_{\infty}=||v_0||_{\infty}$. $\endgroup$ – Andrija Jan 5 '18 at 8:59
  • $\begingroup$ I think you ment, in your last comment, that $||v_0||_{\infty}$ will always be very big. I agree with that, this is why I modified the formula for $w_0$. Now it stands a chance. Or does it? $\endgroup$ – Andrija Jan 5 '18 at 9:05
  • $\begingroup$ Ofcourse, we can allow the constant $A$ to depend also on $u_0$, so we can write $A=A(v_0,u_0)$. $\endgroup$ – Andrija Jan 5 '18 at 9:18

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