Proving the arc length of a certain function is $\infty$

Question

I'm having trouble with a homework problem. We're supposed to prove the following curve has arc length $\infty$ on the interval $[0,1]$:

\begin{gather*} f: \mathbb{R} \to \mathbb{R^2}: t \to \begin{cases} (0,0) & \text{when t= 0}\\ (t, t\sin(\frac{\pi}{t})) & \text{otherwise} \end{cases} \end{gather*}

I was told by the lecturer that we are obliged to do this using the following formula:

\begin{gather*} L(f|_{[0,1]})= \sup\limits_{\mathcal{N} net}L(f|_{[0,1]}, \mathcal{N}) \end{gather*}

The net I came up with is $t_{j} = \frac{2}{3+2j}$. This seems right to me intuitively, but I can't seem to find a lower limit for it. I'd greatly appreciate some help with that.

Answer 1

Let $t_n = \frac{2}{2n+1} > 0 $. This sequence is strictly decreasing and $t_1 = \frac{2}{3} < 1$. We have $$\sin(\frac{\pi}{t_n}) = \sin(n\pi + \frac{\pi}{2}) = (-1)^n = \begin{cases} 1 & n \text{ even} \\ -1 & n \text{ odd} \end{cases},$$ $$t_n + t_{n+1} = \frac{2}{2n+1} + \frac{2}{2n+3} > \frac{1}{n} .$$

For each $k \in \mathbb N$ let $$\mathcal N_k = \{s_0 = 0,s_1 = t_k, s_2 = t_{k-1}, \ldots, s_k = t_1, s_{k+1} = 1\}. $$ Then $$L(f|_{[0,1]}, \mathcal{N}_k) = \sum_{j=0}^k \lVert f((s_{j+1}) - f(s_j)) \rVert > \sum_{j=1}^{k-1} \lVert f((s_{j+1}) - f(s_j)) \rVert = \sum_{i=1}^{k-1} \lVert f(t_i) - f(t_{i+1}) \rVert .$$ For $i = 1,\ldots,k-1$ we have $$f(t_i) - f(t_{i+1}) = (t_i,t_i\sin(\frac{\pi}{t_i})) - (t_{i+1},t_{i+1}\sin(\frac{\pi}{t_{i+1}})) = (t_i,t_i(-1)^i) - (t_{i+1},t_{i+1}(-1)^{{i+1}})) = (t_i -t_{i+1},(-1)^i(t_i + t_{i+1})) ,$$ thus $$\lVert f(t_i) - f(t_{i+1}) \rVert = \sqrt{(t_i -t_{i+1})^2 + ((-1)^i(t_i + t_{i+1}))^2} = \sqrt{(t_i -t_{i+1})^2 + (t_i + t_{i+1})^2} > \sqrt{(t_i + t_{i+1})^2} = t_i + t_{i+1} > \frac{1}{i} .$$ Therefore $$L(f|_{[0,1]}, \mathcal{N}_k) > \sum_{i=1}^{k-1}\frac{1}{i} .$$ But it is well-known that $\sum_{i=1}^{k-1}\frac{1}{i} \to \infty$ as $k \to \infty$ (harmonic series).