I want to show that the functional $L(u)=\int_0^1 \sqrt{1+(u'(x))^2} dx$ is lower semicontinuous in terms of weak convergence in $W^{1,p}(0,1), p\in(1,\infty)$ but not continuous. Our definition of weak lower semicontinuous is: A function $F:X \rightarrow \mathbb{R}$ is weak lower semicontinuous, if $u_k \rightarrow u$ weakly in $X$, then $F(u) \leq \liminf\limits_{k \rightarrow \infty} F(u_k)$. Hint: For a counterexample to the weak continuity, try to approximate a constant function with spike functions.

I don't exactly know how to show this. In class we've seen that for a bounded from below, smooth, convex function, the functional is weak lower semicontinuous. But here $u$ doesn't have to be smooth or convex. I've also found an approach with an epigraph argument here on stackexchange, but we haven't seen this argument in class so I suppose I should show this with the definition of weak lower semicontinuous and some inequalities. I thought of Poincare inequality but I still wasn't able to show the assumption.

Could someone give me a hint on how to show the lower semicontinuity of this functional?

  • $\begingroup$ The functional $L(u)=\int F(u')\,dx$ is weakly lower semicontinuous if $F$ is bounded from below, smooth and convex (I guess that's the result you are referring to). You can apply this criterion directly here. $\endgroup$ – MaoWao Nov 25 '18 at 16:54
  • $\begingroup$ But we don't know if $F$ is smooth or not because we only know that $u' \in L^p$, but $u'$ doesn't have to be continuous in $(0,1)$ or am I wrong? $\endgroup$ – mathstu Nov 26 '18 at 8:57
  • $\begingroup$ Maybe my notation was confusing, the function $F$ appearing in the integrand in my comment is not the function $F$ from your question. Anyway, smoothness in this criterion for weak lower semicontinuity has nothing to do with the regularity of $u$. You should try to properly understand this result from class before you try to apply it (it's applicable here). $\endgroup$ – MaoWao Nov 26 '18 at 9:36
  • $\begingroup$ Ok, I think I understand why $F$ is smooth: $F$ is the square root function and $1+(u'(x))^2 > 0$, so $F$ is smooth. But now I have another question: the square root function is a concave function on (0,1) or isn't it? Because then we have a problem with the assumption of the convexity of $F$. $\endgroup$ – mathstu Nov 26 '18 at 15:12
  • $\begingroup$ I was completely wrong. Of course $F$ is convex even if the square root isn't a convex function. Thanks for your help! $\endgroup$ – mathstu Nov 26 '18 at 16:41

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