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.

Answer 1

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

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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.

Answer 2

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