Is Sin 1 x Continuous at 0

2. Local Lipschitz continuity of x⋅sin(1/x) at $0$

Local Lipschitz continuity of x⋅sin(1/x) at $0$

Solution 1

If $f $ were Lipschitz on $(-\delta,\delta) $ for some $\delta>0$, you would have $$|f (x)-f (y)|<k\,|x-y|$$ for all $x,y\in (-\delta,\delta) $. This implies that $|f'(x)|\leq k $ for all $x $ in $(0,\delta) $. But $f'$ is not bounded on $(0,\delta) $.

Solution 2

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 $n$.

Solution 3

You could use this criterion:

A function $f:D \subseteq \mathbb{R}^k \to \mathbb{R}^d$ is not locally lipschitz if and only if there exist two sequences $x_n,y_n \in D$ with $x_n \neq y_n$ such that $x_n \to p \leftarrow y_n$ and such that $\frac{\|f(x_n)-f(y_n)\|}{\|x_n-y_n\|} \to + \infty$.

Then take

$$ x_n = \frac{1}{2 \pi n}\\ y_n = \frac{1}{2 \pi n + \pi/2} $$

and note that:

$$\frac{\|f(x_n)-f(y_n)\|}{\|x_n-y_n\|} = \frac{\frac 1 {2n \pi + \pi/2}}{\frac{1}{2 \pi n + 8 n^2 \pi}} = \frac{2\pi n + 8 n^2 \pi}{2 n \pi + \pi/2} \to + \infty$$

Note that the sequences where already suggested in the linked question, the conceptual contribution of this answer is an explicit criterion to solve this kind of question.

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Comments

• 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).

  • 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?

  • 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.

  • @MarvinF. that post talks about global Lipschitz continuity. I know it isn't global Lipschitz. My question regards about it being locally Lipschitz specifically at the origin - for open balls (intervals) small enough contaning that point. It can be locally Lipschitz without being globally.

• So if a function is Lipschitz, it implies the function has a bounded derivative? I didn't know the converse was true. Do you prove it by MVT as well? Also, the derivative of $f$ is $sin(1/x) - cos(1/x)/x$. How can I prove local Lipschitz continuity for other points? I know that for zero $1/x$ goes to infinity, but what about other points?

• If the derivative exists (and it does in this case) and the function is Lipschitz, then the derivative is bounded because the Lipschitz condition tells you precisely that all Newton quotients are bounded by the same constant. And you don't have to prove Lipschitz for anything, I don't know where you are going with that.

• It's not on the original question, I was just trying to grasp the discussion more...

• Oh, ok. I don't really know any canonical way to show that a function is Lipschitz. An obvious way would be: if the derivative is bounded on an interval, then the function will be Lipschitz with that bound on the interval. But the whole point of considering Lipschitz is to study functions that are more or less nice but not differentiable.

• "But the whole point of considering Lipschitz is to study functions that are more or less nice but not differentiable." In what sense?

• It is a condition that is weaker than being differentiable, but that gives a useful property about the function. Lipschitz implies uniformly continuous, so it is somewhere in between uniform continuity and differentiability.

• $|f(x_n) - f(y_n)| \neq 1$

• @Javier You are right. I have edited the answer.

Recents

Related

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Source: https://9to5science.com/local-lipschitz-continuity-of-x-sin-1-x-at-0

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