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Let $f$ be a function defined on $[0,1]$ by

$$f(x) = { 0, \text{ if } x = 0} $$ $$f(x) = { x \sin \frac 1 x , \text{ if } 0 < x \leq 1} $$

Prove that the curve $\{(x, f(x)) : x \in [0,1]\}$ is not rectifiable.

I'm not sure how to approach this. The general idea seems logical, we're proving that the length of the curve is infinite, but the method seems difficult to find.

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You know how to form the arclength integral, no? – J. M. Dec 12 '11 at 6:01
it's the integral of the absolute value of the derivative of any parametrization. i'm not sure where to go from there. – uwuw Dec 12 '11 at 6:11
Form the arclength integral and see if you can evaluate it. – J. M. Dec 12 '11 at 6:20

The arclength $L$ of a curve $(x,f(x))$ from $x=a$ to $x=b$ is defined as $$L=\int_a^b\sqrt{1+(f'(x))^2}dx.$$ Therefore, in this case, $f(x)=\displaystyle x\sin(\frac{1}{x})$, which implies that $$(f'(x))^2=\Big[\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})\Big]^2\geq-\frac{2}{x}\sin(\frac{1}{x})\cos(\frac{1}{x})+\frac{1}{x^2}\cos^2(\frac{1}{x}).$$ Hence, if $x\in\displaystyle[\frac{1}{2\pi n+\pi/3},\frac{1}{2\pi n}]$ where $n\in\mathbb{N}$, then $$(f'(x))^2\geq -2(2\pi n+\frac{\pi}{3})+4\pi^2 n^2\cos^2(2\pi n+\pi/3)=\pi^2n^2-4\pi n-\frac{2\pi}{3}.$$ Therefore, the arclength $L$ of $(x,f(x))$ from $x=0$ to $x=1$ can be estimated as follows: $$L=\int_0^1\sqrt{1+(f'(x))^2}dx\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+(f'(x))^2}dx$$ $$\geq \sum_{n=1}^\infty\int^{\frac{1}{2\pi n}}_{\frac{1}{2\pi n+\pi/3}}\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}dx$$ $$=\sum_{n=1}^\infty\sqrt{1+\pi^2n^2-4\pi n-\frac{2\pi}{3}}\cdot\frac{\frac{\pi}{3}}{(2\pi n)(2\pi n+\pi/3)}.$$ It's easy to see that the last series in $n$ diverges to infinity by using limite comparison test with the harmonic series $\displaystyle\sum_{n=1}^\infty\frac{1}{n}$.

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Dear Paul! I have one question. Formula that you wrote at the beginning of your post is appropriate for continuously differentiable curve but $f'(x)$ is not continuous at zero. Can you explain that? Because I have the similar problem here… – RFZ Nov 20 '15 at 12:23

The length of a curve is by definition the $\sup$ of the lengths of inscribed chord-polygons. For your curve define

$$x_0:=1,\qquad x_k:={2\over k\pi}\quad(1\leq k< N),\qquad x_N:=0$$

and consider the chord-polygon $\gamma_N$ through the points $\bigl(x_k, f(x_k)\bigr)$ $\ (0\leq k\leq N)$. As $\bigl|\sin{1\over x_k}\bigr|$ is alternatively $0$ and $1$ one has $|f(x_k)-f(x_{k-1})|\geq{1\over k}$; therefore the individual chords (apart from the first and the last one) have a length $\geq{1\over k}$. It follows that $\gamma_N$ has a total length $\geq\sum_{k=2}^{N-1}{1\over k}$. Since this sum is unbounded for $N\to\infty$ the considered curve $\gamma$ is not rectifiable.

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