I'm trying to prove that this function isn't Lipschitz continuous:
$f:[-1,1]\to \Bbb R:x\mapsto x^{2}\sin1/x^{2}$ when $x$ not equal to zero and $0$ if $x=0$.
Now I just can't find the right $x$ and $y$ to prove it.
Take $M$ from $\Bbb R^+$. Chose then $x$ and $y$ (I can't find them) and it follows that:
$$|f(x)-f(y)|....>M|x-y|$$
but without $x$ and $y$ I can't start the proof at all, can someone just help me with the $x$ and $y$, I would like to finish to prove on my own.
It is easy to see that this function is differentiable at every point. If it is Lipschitz it would follow that its derivative is bounded. But $f'(x)=2x\sin (\frac 1 {x^{2}})-(2/x) \cos (\frac 1 {x^{2}})$. This is unbounded along the sequence $(\frac 1 {\sqrt {2n \pi}})$.
If $x$ and $y$ are sufficiently close to $\frac 1 {\sqrt {2n\pi}}$ then $|f(x)-f(y)| $ would exceed $M|x-y|$.