# Criteria for Lipschitz continuity

Is the following statement true. Assume that $f:[0,1]\to [0,1]$ is a continuous function such that $$\sup_t\limsup_{s\to t}\frac{|f(s)-f(t)|}{|t-s|}<\infty,$$ then $f$ is Lipchitz continuous.

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This "smells" like it's related to the Dini derivatives of $f$. – kahen Oct 14 '12 at 14:09

## 1 Answer

Note that the assumption of continuity is redundant with the displayed one. Indeed, for each $t\in [0,1]$, we can find $\delta(t)>0$ such that $|f(t)-f(s)|\leq (M+1)|t-s|$ if $|t-s|<\delta$, where $M$ is the supremum involved in the hypothesis. Otherwise, we would be able to find $t\in[0,1]$ and a sequence $\{t_n\}$ converging to $t$ such that $|f(t)-f(t_n)|>(M+1)|t-t_n|$, contradicting the assumption.

The problem has been answered at math.overflow by Misha.

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How did you obtained the last inequality? – guest2939891 Oct 15 '12 at 11:19
It's actually wrong, and I will try to think more (for example if $t_k=k^{-1}$ and $s_k=t_k+k^{-2}$ we can conclude from the inequality I get assuming we don't have Lipschitzness. maybe an argument using Lebesgue covering can be performed, but I need to think (much) more. Thanks! – Davide Giraudo Oct 15 '12 at 14:01