Vector valued function in $\mathbb{R}^2$ with non-finite arc length

Question

Find a curve $\mathcal{C}$ : x = g(t) , $a \leq t \leq b$ in $\mathbb{R}^2$, where g $\in \mathcal{C}[a,b]$ such that $\mathcal{C}$ does NOT have finite arc length

Answer 1

I'm convinced the examples using a sine curve with decreasing amplitude work out fine, but I think calculating or bounding the arc length might be tedious. Let me suggest something more elementary.

Consider the curve $\gamma:[0,1]\to\mathbb R^2$ given by $$ t\mapsto \left(t, \frac{1}{\lfloor 1/t\rfloor} - \left| 2t \lceil 1/t\rceil-2-\frac{1}{\lfloor 1/t\rfloor}\right|\right). \tag{1}\label{gamma} $$

Figure: Plot of the curve $\gamma:[0,1]\to\mathbb R^2$ given by \eqref{gamma}.

All this complicated expression does is put a tent of height $\frac 1 k$ above the interval $\left[\frac 1{k+1}, \frac 1 k\right]$. It's obvious that the arc length along one tent is at least $\frac 2 k$. Thus, the length of the overall curve is at least $\sum_{k=1}^\infty \frac 2 k=\infty$.

Answer 2

Trying to follow the hint you were given.

Let $x_n = b - (b-a)/n$ for $n \in \mathbb{N}^*$, so that $x_1 = a$, $\lim_{n \to +\infty} x_n = b$ and $x_{n+1}-x_n = {1 \over n(n+1)}$.

The idea is to define a curve such that its length from $x_n$ to $x_{n+1}$ is $(n+1)$, so that its total length will be $\sum_{n=0}^{+\infty} {1 \over n} = +\infty$.

A first idea would be to simply define $g(x_1) = (0,0)$ and $g(x_{n+1}) = g(x_n) + (n+1) \, \vec{u}$ where $\vec{u}$ is some unit vector in the plane (and just interpolate by a straight line segment between $x_n$ and $x_{n+1}$). The problem is that this indeeds define a continuous curve with infinite arc length on $[a,b)$, but it does not extend continuously at $b$.

So you have to make up a slightly more complicated example, so that $g(x)$ "circles around" a point (the origin, say) when $x \to b$.¹

Of course, there could be simpler (but less "geometric") "analytic" examples, maybe $g(x) = (0, x \sin (1/x))$ works, I'm too lazy to check (this would be for $a = 0$ and $b=1$, of course it's easily adapted to the general case).

Note that your curve cannot be ${\cal C}^1$ on the whole interval $[a,b]$, otherwise it would have finite arc length.

¹ For example, $g(x)$ could travel at a constant speed (determined so that it travels a distance $n+1$) on the circle centered at the origin with radius $1 \over n$ when $x$ is between $x_n$ and $x_{n+1}$ for $n$ even, and $g(x)$ would be the straight line segment between $x_n$ and the point $(0, {1 \over n+1})$ for $n$ odd.

Answer 3

Generally, you want a curve that oscillates (or spirals) infinitely, so that you get infinite arclength, but it has to converge to some specific point, so that you get continuity at the end-point.

How about $a=0$ and $b=1$ and $g(0)=(0,0)$ and $$ g(t)=\left(t,t\sin\frac1t\right) \quad \text{for } 0 < t \le 1 $$ The arclength is infinite because it's greater than the sum of the straight-line distances between the alternating $(+/-)$ peaks, which is in turn related to $\sum\tfrac1n$. My real analysis skills are rusty from decades of neglect, but I'd guess that continuity at $t=0$ follows from the fact that $t\sin\frac1t$ approaches zero as $t \to 0$. Continuity elsewhere is obvious.