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I posted the question stating that it was upper semicontinuous, but that was definitely wrong. I am trying to prove lower semicontinuity.

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user11314
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  • Just to clarify your question - you are asking about this function: $F(f) = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }, \mathrm{d}x$ on the space of continuously differentiable functions with sup-norm, right? That is, the norm on your space is not something like $\lVert f \rVert = \max{f(x),f'(x)}$, it does not include $f'$ at all. – Martin Sleziak Oct 07 '11 at 09:05
  • That is correct Martin. – user11314 Oct 07 '11 at 20:50

1 Answers1

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The result you're looking for can be found in the book Mathematical analysis: linear and metric structures and continuity by Mariano Giaquinta, Giuseppe Modica as Theorem 11.3, p.396.

I will copy their proof here.

The settings in which they are working is a metric space $(X,d)$.

Proposition Let $f_i:X\to\overline{\mathbb R}$, $i\in I$, be a family of lower semicontinuous functions on a metric space $X$. Then $f := \sup f_i $ is a lower semicontinuous function.

Theorem. The length functional $L(\varphi)$ is lower semicontinuous in $C^0([a,b],X)$.

Proof. Recall that we have $$L(\varphi)=\sup_{S\in\mathcal S} V_S(f),$$ where $V_S(f)=\sum_i d(f(t_i),f(t_{i+1}))$, $S=\{t_0=a<t_1<\dots<t_N=b\}$. Since the functional $f\to V_S(f)$ is continuous for every fixed subdivision $S$ of $[a,b]$, the result follows.

They also provide a simple example showing that length is not continuous: Example 6.25, p.204.

This result for functions defined on a closed interval $[0,1]$ is shown in First course in functional analysis by Casper Goffman, George Pedrick p.40.

BTW I was first trying to prove this myself. When I was unsuccessful, I tried googling for curve length semicontinuous and I found the above two references. (And you will probably find some others if you go through the search results or if you try similar searches.)

Martin Sleziak
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  • Thanks Martin! That proposition is in Willard's General Topology(where I found this problem), and it is definitely the core of the solution(along with some simple one dimensional analysis which is what I was for some reason focusing on). – user11314 Oct 08 '11 at 14:19
  • @user11314 I see, Problem 7K.5 in Willard, p.50.\BTW in my opinion it is useful to add the context of the question. Mentioning, that you are trying to solve this problem from Willard might have helped potential answerers. (The fact that this question follows the result on supremum of lsc functions can be viewed as a kind of a hint.)\ Anyway, it was an interesting question and I learned something new trying to answer it. – Martin Sleziak Oct 08 '11 at 14:34
  • Once more the motto is right: "Less means more". Looking this problem in a more general scenario removes calculus red herrings. – EA304GT Oct 04 '17 at 23:52