I'm trying to understand intuitively the notion of Lipschitz function.

I can't understand why bounded function doesn't imply Lipschitz function.
enter image description here

I need a counterexample or an intuitive idea to clarify my notion of Lipschitz function.

I need help

Thanks a lot

  • $\begingroup$ also assume continuity? $\endgroup$
    – Halil Duru
    Apr 6, 2013 at 20:38

5 Answers 5


A Lipschitz function is such that $$|f(x)-f(y)|\leq \alpha |x-y|$$ for any points you pick. Writing this as $$\left|\frac{f(x)-f(y)}{x-y} \right|\leq \alpha $$

what we're saying is that the slope of the secant line joining $(x,f(x))$ and $(y,f(y))$ is always bounded above by $\alpha$. An example of a Lipschitz function is $\sin x$, or $x$. An example of a function which is not Lipschitz but is bounded is $$\sin (x^2)$$ over $\Bbb R$. This is because as we go further towards $+\infty$, the oscillation becomes faster, and thus the slope of the secant lines get nearer and nearer to vertical ones.

An example of a function that is not Lipschitz nor bounded is $\sqrt x$ over $\Bbb R_{>0}$. This is because if we fix $x=0$ and make $y$ very close to $0$, the slope of the secant line grows without bound.

Finally, we can give an example of a function which is Lipschitz but isn't bounded: $x+\sin x$ over $\Bbb R$. Its slope will never get larger than $1$.

  • $\begingroup$ Excellent explanation! +1 $\endgroup$
    – Clayton
    Apr 6, 2013 at 20:33
  • $\begingroup$ If a function has a derivative that is bounded, must it be a Lipschitz function? $\endgroup$ Apr 7, 2013 at 0:28
  • $\begingroup$ @PeterOlson Assume $f$ has bounded derivative. By the Mean Value Theorem, for each $x,y\in\operatorname{dom}f$, what can you equate $$\frac{f(x)-f(y)}{x-y}$$ to? $\endgroup$
    – Pedro
    Apr 7, 2013 at 0:34
  • $\begingroup$ In case of root x, near zero, isn't the slope becoming zero? Can we still say the slope is increasing without bound? ( though I know it is not lipschitz, only trying to understand it intuitively). $\endgroup$
    – aarbee
    Nov 3, 2016 at 6:01
  • $\begingroup$ Hi Pedro Tamaroff, can you guide us with the next please math.stackexchange.com/questions/2440926/… $\endgroup$
    – mathd
    Sep 25, 2017 at 5:04

It means that your function does not explode at some point - made mathematically rigorous.

This idea can also be shown geometrically like so (picture from wikipedia):

enter image description here

The Lipschitz condition (or Lipschitz continuity) ensures that your function always remains entirely outside the white cone, so it cannot e.g. become infinitely steep at one point.

Concerning your picture: Think of the function $sin(1/x)$ (source W|A):

enter image description here

You see that it is becoming infinitely steep at the origin (and would therefore run into the white cone!) but it is still bounded.

This condition is e.g. used in the theory of differential equations where it guarantees the existence and uniqueness of solutions to certain types of differential equations.

  • $\begingroup$ yes, but in every bounded function I can draw this kind of cone (see the picture of my question). $\endgroup$
    – user42912
    Jun 16, 2015 at 13:50
  • $\begingroup$ @user42912: No, you cannot: See my edit for clarification. $\endgroup$
    – vonjd
    Jun 16, 2015 at 14:18
  • $\begingroup$ @user42912: Does this help you? $\endgroup$
    – vonjd
    Jun 16, 2015 at 16:40
  • 1
    $\begingroup$ yes, now I got it! thank you very much! $\endgroup$
    – user42912
    Jun 16, 2015 at 17:09
  • $\begingroup$ It's a nice way of thinking. Thanks. But I am not able to relate root x with this line of thinking. Any help pls? $\endgroup$
    – aarbee
    Oct 29, 2016 at 5:09

Look at the square root function on $[0,1]$ and its behaviour at 0.


Look at the unitary circle, $f(x)=\sqrt{1-x^2}$ how is the tangent line in $x=1$?


With this version (just assuming boundedness) , any discontinuous bounded function forms a counter-example .[why?]

But if you also assume continuity , see the examples below/above.

Furthermore , you may even have differentiability and boundedness without Lipschitzness......

  • $\begingroup$ How could a function be differentiable without being lipschitz? $\endgroup$
    – treble
    Apr 6, 2013 at 21:01
  • $\begingroup$ Look at Tamaroff's example $sin[x^2]$ ..Isn't it differentiable? $\endgroup$
    – Halil Duru
    Apr 6, 2013 at 21:19
  • $\begingroup$ OK, I wasn't thinking about the derivative going to infinity. $\endgroup$
    – treble
    Apr 6, 2013 at 22:28

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