Is a continuous and locally Lipschitz function Lipschitz?

Question

Let $f:\Omega \to \mathbb R$ be locally Lipschitz and continuous on $\Omega$, where $\Omega$ is open, bounded subset of $\mathbb R^n.$ From these two conditions can we say that $f$ is Lipschitz?

Edit: It is not true while we are taking $\Omega$ as open , bounded ,but will it be true when $\Omega$ is compact?

Please someone help.Thank you.

I think $\Omega$ needs to be compact for the result to be true — AmorFati, Nov 19 '17 at 11:14
"Locally Lipschitz and continuous" - doesn't locally Lipschitz already mean continuous? — Mathemagical, Nov 19 '17 at 11:51
Yeah. Now I think that it should have been piecewise Lipschitz and continuous. — Mini_me, Nov 19 '17 at 11:58

José Carlos Santos · Accepted Answer · 2017-11-19T12:29:11.677

4

No. Take $f\colon(0,1)\longrightarrow\mathbb R$ defined by $f(x)=\frac1x$.

About the new question that you added (you shouldn't; asking a new question is the right way of acting), I could provide an answer, but someone has already done that.

edited Nov 19 '17 at 12:29

answered Nov 19 '17 at 11:14

José Carlos Santos

440,053

If we take $\Omega$ to be compact then is this true? – Mini_me Nov 19 '17 at 11:16
@mathiu_lady Yes, but then $\Omega$ will not be open. – José Carlos Santos Nov 19 '17 at 11:17
Yeah of course. Can you give some hints. to prove that – Mini_me Nov 19 '17 at 11:20
@mathiu_lady Post it as a question. – José Carlos Santos Nov 19 '17 at 11:21
Okay. I'm editing my question. Please see it. – Mini_me Nov 19 '17 at 11:23

score 2 · Answer 2 · edited Nov 19 '17 at 11:17

2

Try this: $f:\mathbb{R}_+ \to \mathbb{R}$, defined by $f(x) =\frac{1}{x} .$

edited Nov 19 '17 at 11:17

AmorFati

1,989

answered Nov 19 '17 at 11:13

user3010768

164

I'd advise not using $\mathbb{R}_+$ as there is no concensus on if the notation includes 0 or not. – JPhy Nov 19 '17 at 11:57
2

@JamalS I have never seen it used to include 0 personally. – Wojowu Nov 19 '17 at 16:16

Is a continuous and locally Lipschitz function Lipschitz?

2 Answers2