Example of a continuous non-lipschitz function with domain $[0,1]$ and co-domain $\mathbb R$ I would like an example of a function which is continuous with domain $[0,1]$ but is not Lipschitz continuous. Is this possible? I know a continuous function with domain $[0,1]$ is uniformly continuous, is this also true for Lipschitz continuity?
Thank you in advance, regards.
 A: Ah there should be a fair number of them; for example consider 
"https://en.wikipedia.org/wiki/Cantor_function"
Cantor's Function is not absolutely continuous and not lips-chitz continuous,
$$[0,1] \to [0,1]$$
 But it is uniformly continuous and an even 'Holder Continuous' which is stronger again on that domain and range
$[0,1] \to [0,1]$ 
also see https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function
Minkowskis ? question function F, is even more peculiar: it is even 'odd', 'symmetric', with some other linear like properties, associated with lipschitz functions, and strictly increasing, continuous.
$$[0,1] \to [0,1]$$ 
This function agrees $F(x)=x$, a linear (lip-schitz exponent 1) function on at  least five places (five points where F(x)=x); five fixed points, 
$$F(1)=1\,,F(0.5)=0.5\,\, ,F(0)=0$$
and on at least two other elements of the domain.
I believe that 'F' biject-ive, and may possess a stronger form of holder continuity again. But F is not absolutely continuous on that interval and thus not Lip-schitz continuous
