Prove that $f(x)=x\sin(1/x)$ for $x\ne0$, $f(0)=0$, is not Lipschitz on $[0,1]$

Question

Prove that $f(x)=\cases{0& if $x=0$\\x\sin(1/x)& otherwise,}$ is not Lipschitz on $[0,1]$

MY TRIAL

My idea is to show that $f$ does not have a bounded derivative.

So, suppose for contradiction that there exists $K>0$ such that $$\left| f'(x)-f'(0)\right|=\left| \sin\left(\frac{1}{x}\right)-\frac{1}{x}\cos\left(\frac{1}{x}\right)\right|\leq K |x|,\;\forall\;x\in [0,1].$$ As $x\to 0$, \begin{align}\lim\limits_{x\to 0}\left| f'(x)-f'(0)\right|&=\lim\limits_{x\to 0}\left| \sin\left(\frac{1}{x}\right)-\frac{1}{x}\cos\left(\frac{1}{x}\right)\right|\leq K\lim\limits_{x\to 0} |x|,\end{align} which implies that \begin{align}+\infty=\lim\limits_{x\to 0}\left| \sin\left(\frac{1}{x}\right)-\frac{1}{x}\cos\left(\frac{1}{x}\right)\right|\leq K\lim\limits_{x\to 0} |x|=0,\;\;\text{contradiction.}\end{align} Please, I'm I right? Any other way of showing this is also accepted.

Answer 1

$f\left(\frac{2}{(2n+1)\pi}\right)=\frac{2}{(2n+1)\pi}(-1)^n.$ So, for any $C>0$ $\left|f\left(\frac{2}{(2n+1)\pi}\right)-f\left(\frac{2}{(2n-1)\pi}\right)\right|=\frac{2}{(2n-1)\pi}+\frac{2}{(2n+1)\pi}=\frac{8n}{(2n+1)(2n-1)\pi}> C\frac{4}{(2n+1)(2n-1)\pi}=C\left|\frac{2}{(2n-1)\pi}-\frac{2}{(2n+1)\pi}\right|$

for $n>C.$ So, $f$ can't be Lipschitz.

Answer 2

$f'(0)$ does not exist so your approach won't work. But the idea is correct:

You want to show that $\frac{|f(x) - f(y)|}{|x - y|}$ is unbounded in $[0,1].$

Now, it is a question of choosing judiciously a sequence for which the quotient blows up. It appears that the problem is at $0$ so it makes sense to try a sequence $(x_n)$ such that $x_n\to 0.$ But in such a way that the numerator is bounded away from $0$ while the denominator gets small. So, looking at $\frac{|x\sin(1/x)-y\sin(1/y)|}{|x - y|}$, we might try $x_n = \frac{1}{2 \pi n + \pi/2}\\$ because then the sines are equal to $1$. I'll leave it to you to find $y_n\in [0,1]$ that gives us what we want.