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Im reading Chapter12 of Carothers' Real Analysis, 1ed. Here is a reading material of Lip(X) which denotes the set of all Lipschitz functions on a compact set X, enter image description here

How to show that the set of all Lipschitz functions on a compact set X is dense in C(X)?

I want to show $Ball_ε$(g) ∩ Lip(X) ≠ empty for every continuous function g ∈ $C$(X) and every ε>0. But I got stuck here cos I need to find a function f in Lip(X) such that $||f - g||_∞$<ε.

Bear and bunny
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    What you have copied is a proof that Lip(X) is dense in C(X) by the Stone Weierstrass theorem. http://mathworld.wolfram.com/Stone-WeierstrassTheorem.html – Chris Janjigian Feb 06 '14 at 04:19
  • @ChrisJanjigian: Ok. So Lip(X) is a subalgebra of C(X) and separates points in X and vanishes at no point of X, right? – Bear and bunny Feb 06 '14 at 04:24
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    It's a subalgebra containing the constant functions which separates points. Those three conditions imply that it is dense. – Chris Janjigian Feb 06 '14 at 04:25
  • @ChrisJanjigian: Can I claim that Lip(X) contains all polygonal functions on X, the colletction of which is also dense in C(X)? – Bear and bunny Feb 06 '14 at 04:28
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    @Frank_W What is a "polygonal function on $X$"? From the context it seems that $X$ is a metric space, not necessarily a subset of $\mathbb R^n$. For example, the elements of $X$ can be former presidents of the United States, the distance between two different presidents being $1$ if they belong to the same party and $3$ if they do not. What is your idea of a polygonal function on the set of presidents? – user125646 Feb 06 '14 at 05:43
  • @user125646: Yes, you are right. I made a mistake here. – Bear and bunny Feb 06 '14 at 20:41

1 Answers1

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As Chris Janjigian said, the passage you quoted is a proof that Lipschitz functions are dense; perhaps one should say at the end

Thus, if $X$ is compact, then $\operatorname{Lip}X$ is dense in $C(X)$ by the Stone-Weierstrass theorem.

But to directly answer the question posed in the title: a constructive self-contained proof (without Stone-Weierstrass) goes as follows. Given a continuous function $g$ and a number $\epsilon>0$, pick $\delta>0$ such that $|g(x)-g(y)|<\epsilon$ whenever $d(x,y)<\delta$ — this is possible by uniform continuity of $g$. Also let $M=\sup_X |g|$. Define $$f(x) = \sup_{y\in X} \left(g(y)- 2M \delta^{-1}d(x,y)\right)\tag{1}$$ I claim that $f$ is Lipschitz and $\sup_X |f-g|\le \epsilon$.

  1. $f(x_1)-f(x_2)\le 2M \delta^{-1}d(x_1,x_2)$ by the triangle inequality. Reversing the roles of $x_1,x_2$, we see that $f$ is Lipschitz with constant $2M \delta^{-1}$.
  2. We have $f(x)\ge g(x)$, because the expression under the supremum in (1) turns to $f(x)$ when $y=x$.
  3. If $d(x,y)\ge \delta$, then the expression under the supremum in (1) is at most $-M$, which is less than $g(x)$.
  4. If $d(x,y)< \delta$, then the expression under the supremum in (1) is at most $g(x)+\epsilon$, by the choice of $\delta$.
  5. Combine the items 2-3-4 to obtain $g(x)\le f(x)\le g(x)+\epsilon$.
user125646
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  • Why do you need #3? Isn't it enough just to show #2 and #4? – Kashif Feb 06 '14 at 08:08
  • @Glassjawed I want to prove $f(x)\le g(x)+\epsilon$. This means the supremum in (1) must be estimated from above. The supremum is taken over all $y\in X$. Therefore, the quantity $g(y)-2M\delta^{-1}d(x,y)$ must be estimated from above for all $y\in X$. This is done in #3 for $y$ far from $x$, and in #4 for $y$ close to $x$. – user125646 Feb 07 '14 at 04:44