Could you tell me what is the name and/or reference for the following theorem:

Let $M$ be a metric space. Then any continuous function $f:M\to\mathbb R$ can be a be uniformly approximated by a locally Lipschitz functions.

  • $\begingroup$ On any metric space? Why do you think this is true? $\endgroup$
    – Will Jagy
    Oct 26, 2010 at 3:46
  • $\begingroup$ I believe I have a proof :) $\endgroup$
    – ε-δ
    Oct 26, 2010 at 4:32
  • $\begingroup$ @WillJagy is there a counterexample? $\endgroup$ Jul 26, 2018 at 16:42
  • $\begingroup$ @AryehKontorovich I believe that I was just reacting to the placement of a mathematical statement and calling it a theorem, without any indication of the source. I see, the person posting it had a likely proof. $\endgroup$
    – Will Jagy
    Jul 26, 2018 at 16:51
  • 1
    $\begingroup$ I see, thank you for the clarification. $\endgroup$ Jul 26, 2018 at 17:14

3 Answers 3


Actually the uniform density of locally Lipschitz functions is quite an immediate consequence of the paracompactness of metric spaces (Stone's theorem), and of the fact that, of course, metric spaces admit locally Lipschitz partitions of unity. Note that this way you also have the general result for Banach-valued functions, that is, with a given Banach space as a codomain.

A close result is that uniformly continuous ($\mathbb{R}$-valued) functions on a convex set of a normed space can be uniformly approximated by (uniformly) Lipschitz functions. In this case, an explicit approximation for a function $f$ is obtained just taking $f_k:=$ the infimum of all $k$-Lipschitz functions above $f.$ Then $f_k$ is k-Lipschitz and $f_k\to f$ uniformly as $k\to \infty$ (moreover, the uniform distance of $f$ and $f_k$ can be evaluated in terms of the modulus of continuity of $f$), without need of Stone's theorem. I think that variant of this construction should work for locally Lipschitz approximation of continuous functions (always in the scalar-valued case).

  • $\begingroup$ So you meant the space of Lipschitz continuous functions is dense in that of uniformly continuous functions w.r.t. the supremum norm. Is the space of uniformly continuous functions dense in that of continuous functions w.r.t. the supremum norm? Thank you so much for your elaboration! Please see my question here? $\endgroup$
    – Akira
    May 11, 2022 at 21:17
  • 1
    $\begingroup$ In general, no: Uniformly continuous functions are a close set w.r.to the uniform norm. $\endgroup$ May 12, 2022 at 13:58

oh what a good question it is ! i think this question is from the following paper :

see : https://projecteuclid.org/download/pdf_1/euclid.rae/1230939175

and your question is the same with theorem 2 in this paper !

is it helpful ? thank you !


In that case, take a look at:

Lipschitz-type functions on metric spaces

M. Isabel Garrido, Jesús A. Jaramillo

Journal of Mathematical Analysis and Applications

Volume 340, Issue 1, 1 April 2008, Pages 282-290


In order to find metric spaces X for which the algebra Lip*(X) of bounded Lipschitz functions on X determines the Lipschitz structure of X, we introduce the class of small-determined spaces. We show that this class includes precompact and quasi-convex metric spaces. We obtain several metric characterizations of this property, as well as some other characterizations given in terms of the uniform approximation and the extension of uniformly continuous functions. In particular we show that X is small-determined if and only if every uniformly continuous real function on X can be uniformly approximated by Lipschitz functions.

Keywords: Lipschitz functions; Banach–Stone theorem; Uniform approximation

  • 2
    $\begingroup$ I think you may have overlooked the distinction between Lipschitz and locally Lipschitz functions. $\endgroup$ Oct 26, 2010 at 14:18
  • $\begingroup$ Mark, you make a good point. I see Pietro left a complete answer while I was asleep, good. $$ $$ I did not rememeber anything substantial about this material, so all I did was Google phrases (including Locally Lipschitz) the OP used, the great majority of items gave very specific conditions on the metric space. After he said he felt he had a proof, I put the one that appeared to consider all metric spaces. So it is not so much that I overlooked something, it is that I found something that might have been useful that did not fit in a comment window. $\endgroup$
    – Will Jagy
    Oct 26, 2010 at 16:34

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