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I've got a question about a topic that has already been discussed here: In $C[0,1]$ prove that the subset of Lipschitz functions is dense

We have proved that every function in C[a,b] is Lipschitz using the mean value theorem. So if I get that right, there is no function in C[0,1] that is not Lipschitz?

Then I am confused how to proof, that Lipschitz functions are dense in C[0,1]

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Moala
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    Are you possibly confusing the fact that every $C^1[a,b]$ function (i.e. every continuously differentiable function on a bounded closed interval) is Lipshitz with the (incorrect) claim, that every $C$ function (i.e. every continuous function) is Lipshitz? – Thomas May 17 '16 at 12:04
  • Ahh, yes, that was the problem, just mistaking the two notations then...thank you! – Moala May 17 '16 at 12:11

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$x\mapsto \sqrt x$ is in $C[0,1]$, but not Lipschitz.

Hagen von Eitzen
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