Showing that Lipschitz continuity is a subset of uniform continuity isn't hard.

Let's assume that it is true that $|f(x_1) - f(x_2)| \leq K(x_1-x_2)$ given an arbitrary value $\epsilon > 0$ let's define $\delta = \epsilon / K$

Then trivially if $|x_1 - x_2| \leq \delta$:

$|f(x_1) - f(x_2)| \leq K\cdot|x_1 - x_2| \leq K\cdot\delta \leq K\cdot\epsilon/K \leq \epsilon$.

Which satisfies the uniform continuity property that for every $\epsilon$ there is a $\delta$ such that $|x_1-x_2| \leq \delta \implies |f(x_1) - f(x_2)| \leq \epsilon$

However, I am having a hard time trying to understand why not all uniformly continuous functions obey this property.

I would appreciate a general explanation with at least one example of a non Lipschitz continous function that is uniformly continuous.


Although I admit I had not found the answer that is linked. That answer, although very good, still does not suffice. The author of the question provided multiple functions that are not Lipshitz continuous, but did not prove why they are not Lipshitz continuous (although he did prove that they are uniformly continuous).

My issue is not an absence of an example, but rather a lack of intuition to understand the difference between both.

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    $\begingroup$ A Lipschitz function has a linear modulus of continuity, whereas any function $\omega:[0,\infty) \rightarrow [0,\infty)$ such that $\lim\limits_{t \rightarrow 0} {\omega}(t) = {\omega}(0) = 0$ can be the modulus of continuity for some uniformly continuous function. Also, any continuous function $f:[a,b] \rightarrow \mathbb R$ is uniformly continuous (e.g. the Weierstrass nowhere nowhere differentiable function, $f:[0,1] \rightarrow \mathbb R$ defined by $f(x) = t^{\alpha}$ where $\alpha$ is a constant strictly between $0$ and $1).$ $\endgroup$ Jan 22, 2020 at 18:07
  • $\begingroup$ Also, what is $f$ or to be more precise, it's domain? Don't you need $K | x_1 - x_2 | $ as an upper bound? $\endgroup$ Jan 22, 2020 at 18:08
  • $\begingroup$ And why the tag geometric-measure-theory? $\endgroup$ Jan 22, 2020 at 18:08

2 Answers 2


Lipschitz continuous functions have "bounded derivative" (more accurately, bounded difference quotients: the slope of any line connecting two points on the graph is bounded by the Lipschitz constant). Uniformly continuous functions allow you to pick the $\delta$ in the definition of continuity independent of the $x$.

The classic example of a uniformly continuous function which is not Lipschitz is $$ f:[0;1]\rightarrow \mathbb{R}, \ f(x)=\sqrt{x}.$$ This function is uniformly continuous as every continuous function on a compact set is uniformly continuous. On the other hand its slope is not bound around the origin, so it is not Lipschitz. More rigorously, assume it was Lipschitz, then there exists $K>0$ such that for all $x\in [0;1]$ holds $$ \sqrt{x} =\vert f(x) -f(0)\vert \leq K \vert x-0\vert =K x.$$ This is equivalent to $$ \frac{1}{\sqrt{x}} \leq K$$ which is a contradiction.

In short, having steeper and steeper derivative needs not to be an issue for uniform continuity, but it kills Lipschitz continuity for sure.

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    $\begingroup$ Ah! I see so the fact that the derivative has an asymptote at x=0 breaks the requirement that the difference in the mapped space be smaller than some constant times the difference in the domain. $\endgroup$
    – Makogan
    Jan 22, 2020 at 18:19
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    $\begingroup$ @Makogan That is exactly the point. $\endgroup$ Jan 22, 2020 at 18:20
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    $\begingroup$ @DaveL.Renfro Thanks for pointing out. I added it. $\endgroup$ Jan 22, 2020 at 18:23

$$f(x)=|x|$$ is Lipschitz continious $$f(x)=\sqrt{(|x|)}$$ is continious but not Lipschitz continiuous


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