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Today I came across the following definition:

Definition: A function $f:[a,b] \to \Bbb C$ is Lipschitz to the right of $t_0 \in [a,b]$ if exists $L>0$ such that $|f(s+t_0)-f(t_0^+)| <Ls$ for all $s >0$ sufficiently small.

In the same way we define what it is to $f$ being Lipschitz to the left of $t_0$. It is quite clear that if $f$ is Lipschitz at both right and left of $t_0$, then $f$ is continuous at $t_0$. And talking to a friend, a question arose:

Is there a function Lipschitz on the right of every point, but everywhere discontinuous?

We can guess that $f$ can never be Lipschitz at the left of any point.

Ivo Terek
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    I can't quite tell whether your function is right continuous. If it is, then actually the set of discontinuities is countable, so the answer is a definitive "no". – Ian Jun 23 '15 at 13:51
  • Have a look at one of the two answers to this question: http://math.stackexchange.com/questions/65941/if-f-mathbbr-to-mathbbr-is-a-left-continuous-function-can-the-set-of-disc – PhoemueX Jun 23 '15 at 14:29
  • Extended comment: You seem to assume that the right-sided limit $f(t_0^+) = \lim_{t \downarrow t_0} f(t)$ exists for every $t_0$ (otherwise, your definition of being Lipschitz to the right of $t_0$ makes no sense). By the answer of MikeF to the question I linked above, this implies that $f$ has at most countably many discontinuities. In particular, $f$ can not be discontinuous everywhere. – PhoemueX Jun 23 '15 at 20:56

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