Does unbounded derivative imply No Lipschitz?

Question

We are taught that bounded derivative implies lipschitz. Also, at times, derivative may not exist, yet lipschitz could hold.

I wonder if derivative exists but is unbounded, can we directly say that it is not lipschitz?

hamam_Abdallah · Answer 1 · 2016-11-01T12:57:05.867

9

Assume $f$ is Lipschitz and differentiable at $I$ and let $a\in I$.

$\exists K>0: $

$\forall x\in I - \{a\}$

$|\frac{f(x)-f(a)}{x-a}| \leq K$

then by passage to the limit when $x\to a$, we get

$|f'(a)|\leq K$ and $f'$ is bounded at $ I$.

edited Nov 01 '16 at 12:57

answered Nov 01 '16 at 12:37

hamam_Abdallah

63,719

1

Is this answering the question? Lipschitz continuity implies bounded derivative, yes, but this is not the question... – pluton May 05 '21 at 19:21

score 4 · Accepted Answer · edited May 05 '21 at 19:46

4

If derivative is unbounded at $x$, then you can find two points near the point $x$ which will not follow the Lipschitz continuity definition for any bound $M$.

edited May 05 '21 at 19:46

pluton

1,243

answered Nov 01 '16 at 13:11

jnyan

2,525

2

Does unbounded at x mean unbounded on some neighborhood of x? – Smooth Alpert Frame Sep 15 '18 at 12:03

score -1 · Answer 3 · edited Oct 28 '20 at 02:08

-1

$$f(x) = x \sin\left(\frac1x\right) \text{ on } (0,1)$$ is a Lipschitz function with unbounded derivative

edited Oct 28 '20 at 02:08

Siong Thye Goh

153,832

answered Oct 28 '20 at 01:46

Sumit Mor

9

Do you have a proof? – pluton May 05 '21 at 19:26
This function is not Lipschitz, it is not even bounded variation, see: https://math.stackexchange.com/questions/605241/is-fx-x-sin-frac1x-with-f0-0-of-bounded-variation-on-0-1 – Connor Lockhart Aug 10 '22 at 05:01

Does unbounded derivative imply No Lipschitz?

3 Answers3