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I came across the following simple definition

A path $\gamma$ in $\mathbb{R}^n$ that connects the point $a \in \mathbb{R}^n$ to the point $b \in \mathbb{R}^n$, is a continuous $\gamma : [0, 1] \to \mathbb{R}^n$ such that $\gamma(0) = a$ and $\gamma(1) = b$. We denote by $\ell(\gamma)$ the (Euclidean) length of $\gamma$. $\ell(\gamma)$ is always defined and is either a non-negative realnumber or $\infty$.

However, I cannot seem to think of a path, defined in this manner (specifically, where the domain is compact), the length of which is infinite. Can anyone provide an example ?

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I wonder whether the intention was to look at the situation for which the length $\ell(t)$ of the curve is defined and finite for [0,t] and tends to $\infty$ as $t$ tends to 1. – Mark Bennet Aug 27 '12 at 9:48
up vote 5 down vote accepted

Let me start with an example of the infinite-length path. Consider any nowhere-differentiable curve that connects points $a$ and $b$. For example, let $a = (0,f(0))$ and $b = (1,f(1))$ where $f$ is the Weierstrass function. Consider $\gamma(t) = (t,f(t))$ for $t\in[0,1]$. Then the length of $\gamma$ is infinite.

Note, however, that any $\gamma$ compact (as you mentioned) and hence bounded in $\Bbb R^n$. Due to this reason, the only cause for the infinite-length can come from local behaviour of the path, not from the global one - i.e. for $\gamma$ it is not possible to go "too far away" from points $a$ and $b$.

Finally, note that path of the infinite length are "more probable" than those of the finite length in the sense that Brownian motion (which, btw, ranges over the space of continuous curves) is nowhere-differentiable with probability $1$.

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I don't think this works as stated. For example, consider the graph of $|x|$ for $-1\leq x\leq 1$, which is a non-differentiable curve of length $2\sqrt{2}$ connecting $(-1,1)$ and $(1,1)$. – Alex Becker Aug 27 '12 at 8:18
@AlexBecker: sorry for the confusion, I meant a function which is nowhere-differentiable. To avoid this confusion, I provided an explicit example now. – Ilya Aug 27 '12 at 8:19
@AlexBecker Thanks. Do we have to resort to such special functions ? Would the answer be different in case we only considered (say) differentiable paths ? – Teddy Aug 27 '12 at 8:28
@Teddy: I've just added a comment on the importance of the local behaviour. For the case you've mentioned: it holds that $$ \ell(\gamma) = \int\limits_0^1 \sqrt{\sum\limits_{i=1}^n \dot x^2_i}\mathrm dt <\infty $$ as all coordinate functions $\dot x_i$ are continuous and the integration is taken over an interval $[0,1]$. – Ilya Aug 27 '12 at 8:30
@Teddy As Ilya says, non-differentiability is key. The example I would of used is a space-filling curve, which is arguably even stranger. – Alex Becker Aug 27 '12 at 8:32

$\ell(\gamma)$ can still be infinite even if $\gamma$ is differentiable. Take any differentiable function $f:[0,1]->\mathbb R$ that is of unbounded variation; for instance, $f(x) = x^2\sin(1/x^2)$ (with $f(0) = 0$). Now just define $\gamma(t) = (t,f(t))$.

If $\gamma$ is continuously differentiable, on the other hand, then $d\gamma/dt$ is bounded on $[0,1]$, so $\ell(\gamma) = \int_0^1|d\gamma/dt|dt$ is finite.

Updated to show unbounded arc length:

$f$ crosses the $x$-axis at $x_n=(\pi n)^{-1/2}$ for $n = 1,2,3,...$ For large $n$, the arc length of the curve between $x_n$ and $x_{n+1}$ is about $2x_n^2 = 2/\pi n$, and is certainly greater than $1/\pi n$. So the total arc length is greater than $\sum_{n=1}^\infty1/\pi n$, which is unbounded.

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Your $f$ is not a function, or not differentiable at $x=0$. – akkkk Aug 27 '12 at 9:57
@Auke: Fixed now. – TonyK Aug 27 '12 at 10:01
TonyK, your current arc length is bounded on $[0,1]$, according to mathematica. – akkkk Aug 27 '12 at 10:02
You are right, I am a Mathematica noob and entered the integral incorrectly. – akkkk Aug 27 '12 at 10:16

Yes, for instance Koch snowflake is a such example. Let's do the following construction:

  1. Start with the segment $A_0 = [0,1]$.
  2. Subdivise $A_0$ in three equal pieces.
  3. Replace the middle third by an equilateral triangle with base $[\frac13, \frac23]$.
  4. Suppress the base of that triangle. You get the path $A_1$ something which looks like a saw teeth.
  5. By induction, to construct $A_{n+1}$ replace each of the $4^n$ segments of $A_n$ by an equilateral triangle of base this segment, and then remove the segment.
  6. The limit object is a path (whose image is compact set) joining $0$ and $1$.

The length of $A_n$ is $\left( \frac43 \right)^n$ tends to $+\infty$ as $n \to +\infty$.

The Hausdorff dimensions is $\frac{\ln 4}{\ln 3} \approx 1.26$

The following picture summarize the construction.

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