I have to proof, that
$$f(x) = \begin{cases}
x\cdot\sin(\frac{1}{x}) & x \neq 0 \\
0 & x = 0 \\
\end{cases} $$ isn't Lipschitz continuous.
I have started with
$|f(x)-f(y)|$ and tried to get to $ \leq L \cdot | x-y| $ but without any success.
$|(1+\frac{y}{x-y})\cdot\sin(\frac{1}{x}) – \frac{y}{x-y}\cdot\sin(\frac{1}{y})| \cdot |x-y|$ was best rearranging, but I don't know at this point how to simplify the $\sin$
I'm a mathematics student in first semester, so I am not allowed to use differential calculus.
Hints or solutions are welcome 🙂
Best Answer
You should try to show that the ratio $$ \frac{|f(x) - f(y)|}{|x - y|} $$ has no upper bound for $x,y \in \Bbb R$. In particular, consider taking $$ x = \frac{1}{2 \pi n + \pi/2}\\ y = \frac{1}{2 \pi n + 3\pi/2} $$ for an arbitrary integer $n$.