Prove that $f: \mathbb{R} \rightarrow \mathbb{R}$, with $$f(x) = \begin{cases} x\sin(\frac{1}{x}) & x \neq 0 \\ 0 & x = 0 \\ \end{cases} $$ is not Lipschitz continuous in any interval containing zero. Since a function is local Lipschitz (L.L.) when, given $x_0 \in (a,b)$, there is an open ball $B$ centered at $x_o$ with radius $r>0$ such that for all $x,y \in B$, $$|f(x) – f(y)| \leq K_{x_0}|x-y|.$$
Being so, the job is to show $f$ isn't L.L. in any open ball centered at $0$ (isn't L.L. at the origin).
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What I've tried is to bound $|f(x) – f(y)|$ by factoring out $|x – y|$ and using the triangular inequality for adding the terms that subtract at $(x-y)(sin(\frac{1}{x}) – sin(\frac{1}{y}))$ but are not on $sin(\frac{1}{x}) – sin(\frac{1}{y})$, which would get me an unbounded term of $|x||sin(\frac{1}{y})|$, which then would have no bound $K_{x_0}$. But that doesn't guarantee there isn't such a $K$ between $|f(x) – f(y)|$ and the upper bound we got, does it?
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Apart from that, I know that every function with a bounded derivative in an interval (which can be unbounded) is Lipschitz and a continuously differentiable function in the neighborhood of a point is L.L. at that point. Any tip that touches on those results are welcome, as well as intuitive advice.
I've been looking for an answer on the trails of the one on this post:
Lipschitz continuity of $\frac{1}{x}$ and $x^2$ (the $1/x$ part of the accepted answer, using the definition).
Edit: not duplicate of Lipschitz continuity of $x\cdot\sin(1/x)$ , that regards about global Lipschitz continuity of the same function. The question addressed here is about local Lipschitz continuity at a specific point.
Best Answer
Let $$ x_n=\frac{1}{2\,n\,\pi},\quad y_n=\frac{1}{2\,n\,\pi+\pi/2}. $$ Then $$|f(x_n)-f(y_n)|=\frac{1}{2\,n\,\pi+\pi/2}\sim\frac{C}{n}$$ and $$ x_n-y_n=\frac{\pi}{2}\,\frac{1}{2\,n\,\pi\,(2\,n\,\pi+\pi/2)}\sim\frac{C}{n^2}, $$ so that it is impossible to have $|f(x_n)-f(y_n)|\le L\,|x_n-y_n|$ for any constant $L$ and all $n$.