# Functional is weak lower semicontinuous but not weak continuous

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?

• 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. – MaoWao Nov 25 '18 at 16:54
• 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? – mathstu Nov 26 '18 at 8:57
• 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). – MaoWao Nov 26 '18 at 9:36
• 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 $F$. – mathstu Nov 26 '18 at 15:12
• I was completely wrong. Of course $F$ is convex even if the square root isn't a convex function. Thanks for your help! – mathstu Nov 26 '18 at 16:41