Limit of the lengths $=$ lengths of the limit

Question

This question is motivated by this other: Equal perimeters of squares and right angled isosceles triangles

Intuitively, it seems that if a sequence $\alpha_n$ of piecewise twice differentiable paths converges uniformly to a path $\alpha$ and each $\alpha''_n$ has constant sign where it is defined, $$\ell(\alpha_n)\to\ell(\alpha)$$ where $\ell(\gamma)$ denotes the length of the piecewise differentiable path $\gamma$.

A picewise differetiable path is a continuous, piecewise differentiable function $\gamma:[0,1]\to\Bbb R^n$. Its length is $$\ell(\gamma)=\int_0^1\|\gamma'(t)\|dt$$

I'm not really sure of that this is true, and if it is, I don't know the name of the theorem, if it has any. I have googled with no result so far.

If it is true, I don't need a proof, only a confirmation, and if it is possible, a link to some document with the theorem.

Answer 1

False. You need both $\alpha_n \rightarrow \alpha$ and $\alpha'_n \rightarrow \alpha'$, because the length is defined through $\alpha'$.

Otherwise, the attached image shows you how you could prove that any segment is equal to $\pi$ times its length.

The problem on the figure is that, when you take a close look at the path when divided in a large number of half-circle, you see that the derivative (i.e. the direction) of the sequence is every where distinct of the straight line.

So all what you can say about position ($\alpha_n \rightarrow \alpha$) passes to the limit, but what ever you can say which involve the derivative (length, angle, volume,...) do not.

Answer 2

It can not be true since in the following video : https://www.youtube.com/watch?v=D2xYjiL8yyE suppose this theorem true implies that $\pi=4$.

The problem is in "piece wise": each $\alpha_n$ is piece wise differentiable, but you do not control the number of parts of $\alpha_n$. So you have be more restrictive: for example $(\alpha_n)$ is $C^1$, converge uniformly and $(\alpha_n')$ is bounded may be sufficient.