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Background

One of my friends told me the following story: A child must walk from his home at point A = (1,0) to his school at point B = (0,1). The laws in his country state that you can only walk parallel to the horizontal and vertical axis. No matter how he tries to get to school, he finds that he must walk at least 2 miles. He is very frustrated that he cannot walk diagonally. He doesn't want to get in trouble, so he puts up with this silly law until his senior year of high school. At this point he takes calculus, and learns about limits, and so he decides that each day he will walk a new "zig-zag" path around the line, in such a way that his sequence of paths approaches the line from A to B uniformly. On the 30th day he is pulled over by a policeman for walking diagonally.

The point of the story is to get you to think about the notion of the length of a curve. Here we have an instance of a sequence of polygonal paths which approach a curve uniformly, but the lengths of the polygonal approximations do not converge to the length of the curve.

Our usual definition of the arclength of a curve involves approximating it with polygonal secant segments. My question is why this is the "best" definition. Or to make this precise, how do we know that any other sequence of polygonal approximations to the curve will not have a shorter limit? EDIT: In the case of a straight line, this is clear, but my precise question below is about answering this for more general curves.

Precise Question:

Let $C:[0,1] \to \mathbb{R}^n$ be a rectifiable curve (or feel free to add as many smoothness requirements as you like), and let $P_n: [0,1] \to \mathbb{R}^n$ be a sequence of piecewise linear curves which converge uniformly to $C$. Is it true that limsup{Length($P_n$)} $\geq$ Arclength(C)? EDIT I goofed: I meant to ask whether or not liminf{Length($P_n$)} $\geq$ Arclength(C).

Hopefully this question is not too elementary; my analysis skills are almost pitifully weak. I strongly suspect that the answer to this precise question is "Yes", because otherwise I think that the usual definition of arc length is incorrect. I have enough faith in mathematics to believe that we have found the right definition, but I would still like to see a proof of this fact.

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I had some trouble with LaTeXing the inequality in the penultimate paragraph. How do you get { and } to show up properly in "math mode"? –  Steven Gubkin Nov 22 '10 at 13:53
    
You may try typing \\{ and \\} in the math mode. –  Andrey Rekalo Nov 22 '10 at 14:00
    
The usual definition of length is the one that is universal with respect to the triangle inequality (that is, it is the smallest number consistent with the triangle inequality), so in that sense it is the "best" definition. –  Qiaochu Yuan Nov 22 '10 at 14:08
    
Even if this statement were false, I would not necessarily take it as evidence that the definition of arc length is wrong; instead I would take it as evidence that uniform convergence is too weak for the purpose of thinking about arc length. –  Qiaochu Yuan Nov 22 '10 at 14:15
    
@Qiaochu: Sorry if my question was not clear. I am not interested only in the length of straight line segments, but in general rectifiable curves. I accept the length of line segments as a starting point, but I am curious why we approximate curves with secant segments, and not some other piecewise linear curves. I hope that the usual definition is minimal. –  Steven Gubkin Nov 22 '10 at 14:16

2 Answers 2

up vote 5 down vote accepted

Yes it is true and quite elementary too. The definition of arclength you gave is also known as the (classical) total variation, and you can define it for any curve $\gamma:[a,b]\to M$ in any metric space $(M,d)$ as $\mathrm{V}(\gamma,[a,b]):=\sup_P \mathrm{V}(\gamma, P)$ where the supremum is taken over all subdivisions à la Riemann $P:=\{a=t_0 < t_1 < \dots < t_n=b\}$ of the interval $[a,b],$ and where we denote $\mathrm{V}(\gamma, P):=\sum_{i=0}^{n-1}d\left(\gamma(t_{i+1}),\gamma(t_i)\right),$ the variation w.r.to $P$. Clearly, for any $P$, the variation wrto $P$ is a continuous functional on curves, even wrto point-wise convergence. The total variation over $[a,b]$ is a supremum of continuous functionals, and as such, what one can say is: it is (sequentially) upper semicontinuous, that is, the inequality you wrote holds for any sequence $\gamma_n$ point-wise convergent to $\gamma$ : $$\limsup_{n\to\infty}V(\gamma_n,[a,b])\geq V(\gamma, [a,b]).$$

(just because for all $P$ we have $\limsup_{n\to\infty}V(\gamma_n,[a,b])\geq \limsup_{n\to\infty}V(\gamma_n,P)= V(\gamma, P)$: so we have it taking the $\sup_P$ over all $P$.).

Rmk. The relevance of the notion of upper semi-continuity is linked to existence results for minima. For instance, it is true that in any compact metric space, if two points are joined by a curve of finite length, then there is also a curve of minimal length connecting them. It's a consequence of the Ascoli-Arzelà compactness theorem.

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This is true, but as per the edit to the original question, and to get a stronger result by the same argument, I think you want $\liminf$ instead of $\limsup$. –  Noah Stein Nov 22 '10 at 14:52
    
Thanks for this. –  Steven Gubkin Nov 22 '10 at 15:00
    
I think I will wait a day before accepting any answers. –  Steven Gubkin Nov 22 '10 at 16:25
    
Noah: yes but using $\liminf$ or $\limsup$ gives actually equivalent conditions in general, when stated for every sequence converging to $\gamma.$ –  Pietro Majer Nov 22 '10 at 16:34
    
Steven, I suggest much more than one day! ;) –  Pietro Majer Nov 22 '10 at 17:02

Here's a somewhat rigorous geometric argument (perhaps harking back to another of today's questions). Suppose that C is a continuous curve in $\mathbb{R}^n$ and that P is a polygonal path approximating C in that every point of C has a point of P within $\epsilon$ of it. I claim that it is approximately longer than some polygonal path of secants to C.

The idea is to modify P by performing rotations and translations (thus, isometries) on the segments to try to put them into secant position. Sometimes, we need to cut to finish the job, but (almost) never stretch. Using the parameterization of S, I will freely refer to the "beginning" and "end" of a segment on P without becoming confusing.

First, consider the first segment S of P, and denote by x the beginning of C. The beginning of S is within $\epsilon$ of some point of S by hypothesis, so for simplicity, let's assume it's within $\epsilon$ of x. (This assumption is harmless since we would otherwise have to add only up to the length of S, which goes to zero in the limit anyway.) Then by the triangle inequality we can move it to x without adding more than length $\epsilon$ to P. Assume that we have done so.

Now, assume inductively that we have a segment S of P and its beginning lies on C. We rotate S around this beginning until its end is also on C (this part requires continuity of C, if you are keeping track). In doing so, we keep the next segment attached to its end, allowing it to translate and rotate as necessary to keep its end within $\epsilon$ of C.

Suppose this latter goal is impossible: one cannot bring the end of the next segment T to within $\epsilon$ of C after transforming S. Then we can put some interior point y of T on C, and form a polygonal path P' of secants to C from x to y. Everything on P after y is extra length above that of P' (to within $\epsilon$, from the first step).

Conversely, if we can do this, then by adding at most another $\epsilon$ to the length of the last segment of P we can put its end at the end y of C and get P' again, which is at most $2\epsilon$ longer than P.

Either way, the liminf (as $\epsilon \to 0$) of lengths of all possible paths P is greater than or equal to that of paths P', as desired.

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I like this idea. I started to write something like it (only rougher and just in the plane) but then I wondered about this situation: start with a path $P(x)$ which increases, decreases and crosses the $x$ axis at the same places as $\sin x$ (say alternately slope 1000 and -1000 starting at (0,0) and turning at $(2k+1)\pi/2$ stopping at $(100\pi,0)$. Now let $C(x)=P(x)+\epsilon \sin x$. Then we have to stretch every segment of $P$ in order to get all endpoints on $C$. This isn't over [0,1] but rescale to make it so. Is this a problem? –  Aaron Meyerowitz Nov 22 '10 at 16:05
    
@Aaron: Indeed, I do seem to have overlooked the possibility that P could be substantially shorter than C. Clearly in your example I am not going to be able to find a more efficient secant path since, in fact, P is not that good an approximation from the wrong direction. I will think about this. –  Ryan Reich Nov 22 '10 at 17:46
    
I doubt that it is very much shorter. But that attractive type of argument gets messier, if it is possible. –  Aaron Meyerowitz Nov 22 '10 at 23:36
    
@Aaron: It is much shorter as compared to $\epsilon$. You may be right that this argument isn't going to get more correct while staying so neat. –  Ryan Reich Nov 23 '10 at 12:39

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