This is problem 16 chapter 23 of 'Calculus-Spivak (3rd edition)'
Consider the following 'length' describing set $$\ell(f,P)=\sum_{i=1}^{n}\sqrt{(t_{i}-t_{i-1})^2+[f(t_i)-f(t_{i-1})]^2}$$ Where $P=\{t_{1}<\dots<t_{n}\}$ is a partition of the interval being 'measured'.
Let $f(x)=x \cdot \sin( \frac{1}{x})$ for $0 < x \leq 1 $ and $f(0)=0$. Show that $f$ has an 'infinite length'on $[0,1]$ by showing the unboundness of $\ell(f,P)$ by considering the partition $$P^*=\left\{ 0, \frac{2}{(2n+1)\pi},\dots,\frac{2}{5\pi},\frac{2}{3\pi},\frac{2}{\pi},1 \right\}$$
My attempt :
Without loss of generality we can take $t_{i}$ and $t_{i-1}$ to be the terms $\frac{2}{(2k+1)\pi}$ and $\frac{2}{(2k-1)\pi}$ for some $k < n$ ( for a moment ignoring the 0 and 1 of the partition) . Then \begin{align*}f(t_{i})-f(t_{i-1}) &=t_{i} \sin\left(\frac{(2k-1)\pi}{2}\right)-t_{i-1} \sin\left(\frac{(2k+1)\pi}{2}\right) \\ &=t_{i}\sin\left(k\pi-\frac{\pi}{2}\right)-t_{i-1}\sin\left(k\pi+\frac{\pi}{2}\right) \\ &=-t_{i}-t_{i-1} \quad \text{or} \quad t_{i}+t_{i-1} \end{align*}
Now define $b_{i}=\frac{2}{(2n-2i+1)\pi}$ for $i$ in $\{0,\dots,n\}$ and let $0 < A_{n}=\sqrt{(1-\frac{2}{\pi})^2+(\sin(1)-1)^2}+\sqrt{\left(\frac{2}{(2n+1)\pi}\right)^2+\left(\frac{2}{(2n+1)\pi}\right)^2}$
then we can write \begin{align*} \ell\left(x\sin\frac{1}{x},P^*\right) &=A+\sum_{i=1}^{n+1}\sqrt{(b_{i}-b_{i-1})+[f(b_{i})-f(b_{i-1})]^2} \\ & \geq \sum_{i=1}^{n+1} \sqrt{(b_{i}-b_{i-1})^2+(b_{i}+b_{i-1})^2} \\ &= \sum_{i=1}^{n+1} \sqrt{2(b^2_{i}+b^2_{i-1})} \\ & \geq \sum_{i=0}^{n} b_i\end{align*} (According to the result preceeding the calculation).
Since $$\sum_{i=0}^{n} b_{i}=\sum_{i=0}^{n} \frac{2}{(2i+1)\pi} \geq \sum_{i=1}^{n} \frac{1}{12i}$$ With the last sum diverging by the harmonic series convergence test, we have that the set $\ell(f,P^*)$ is unbounded and thus $f$ has 'infinite length' $\blacksquare$
My question is if my proof is correct? And also if there is something i can do to make my proof look more coherent? (in the sense that it looks too be a bit all over the place.)
The proof works.
Using $(b_i-b_{i-1})^2+(b_i+b_{i-1})^2 = 2(b_i^2 + b_{i-1}^2)$ is unnecessarily clever, though -- you can get what you need simply by discarding the first term: $$ \sqrt{(b_i-b_{i-1})^2+(b_i+b_{i-1})^2} \ge \sqrt{(b_i+b_{i-1})^2} = b_i+b_{i-1} $$
As a matter of writing style I think the phrase "without loss of generality" is misplaced there. The partition is not something that someone gives you that you need to make assumptions about -- for your purpose you get to choose it, and so there is no generality to worry about losing.