# The space of Lipschitz continuous functions is dense in that of uniformly continuous functions?

Let $$(X,d)$$ be a metric space. Then

• The space of bounded uniformly continuous functions is dense in that of bounded continuous functions w.r.t. the supremum norm. ref
• The space of bounded Lipschitz continuous functions is dense in that of bounded uniformly continuous functions w.r.t. the supremum norm. ref

Does the following relaxation hold?

• The space of uniformly continuous functions is dense in that of continuous functions w.r.t. the supremum norm.
• The space of Lipschitz continuous functions is dense in that of uniformly continuous functions w.r.t. the supremum norm. ref

Update 1: In the paper "Approximation of Continuous Functions by Lipschitz Functions", the author Radu Miculescu said that

• A continuous function $$f: X \rightarrow \mathbb{R}$$, where $$X$$ is a metric space, is a uniform limit of a sequence of locally Lipschitz maps from $$X$$ to $$\mathbb{R}$$.
• There exist continuous functions that cannot be the uniform limit of a sequence of Lipschitz functions.

Update 2: I write the counter-example of the second statement suggested by Mindlack here.

Let $$d(x,y) := \min\{|x-y|,1\}$$ be metric on $$\mathbb R$$. Assume $$f: (\mathbb R, d) \to (\mathbb R, |\cdot|)$$ is any $$L$$-Lipschitz-continuous. Then $$|f(x)-f(0)| \le Ld(x,0) \le L$$ for all $$x\in \mathbb R$$. So $$f$$ is bounded.

Let $$g: (\mathbb R, |\cdot|) \to (\mathbb R, |\cdot|)$$ be uniformly continuous and unbounded (for example, $$x \mapsto x$$). Fix $$\varepsilon>0$$. There is $$\delta<1$$ such that $$|g(x)-g(y)| < \varepsilon$$ for all $$x,y\in \mathbb R$$ such that $$|x-y|<\delta$$. Then $$|g(x)-g(y)| < \varepsilon$$ for all $$x,y\in \mathbb R$$ such that $$d(x,y)<\delta$$. This implies $$g: (\mathbb R, d) \to (\mathbb R, |\cdot|)$$ is also uniformly continuous.

So $$\sup_{x\in \mathbb R} |f(x)-g(x)| = \infty$$, let alone approximation.

• If you don’t force your functions to be bounded, there is no “supremum norm” on your functional space. Also, cross-posting on MO with so little time is a bit strange. May 12, 2022 at 8:48
• Anyway, the first point is false: any uniformly continuous function on $\mathbb{R}$ is $O(|x|)$ at infinity, so is not at finite distance from, say, $x \longmapsto x^2$. I’m not quite sure yet about the second one. May 12, 2022 at 8:58
• @Mindlack I'm sorry for not being patient enough. The point "supremum norm" is great. May 12, 2022 at 8:59
• The first result is not correct in general: on the contrary, the space $BUC(X,\mathbb R)$ of bounded uniformly continuous functions is a closed subspace of the Banach space $\big(BC(X,\mathbb R), \|\cdot\|_{\infty,X}\big)$ May 12, 2022 at 14:45
• The third quoted fact is true; more generally, I think a necessary and sufficient condition for f to be uniform limit of Lipschitz function is that it has a subadditive modulus of continuity (check the wiki article on modulus of continuity). The fourth quoted fact is not true: take X to be the integer numbers with the Euclidean distance. Then any function is uniformly continuous (the epsilon-delta test holds almost vacuously for delta<1), but e.g. $x\mapsto x^2$ has infinite uniform “distance” from any Lipschitz function. May 12, 2022 at 14:55

As written in my comment, the first point is false: any uniformly continuous function on $$\mathbb{R}$$ is $$O(|x|)$$ at infinity, so is not at finite distance from, say, $$x \longmapsto x^2$$.

The second point seems to be true for “reasonable” metric spaces. Specifically, we assume that for any small enough $$\eta>0$$, there is a constant $$C \geq 1$$ such that for any pair of points $$x,y$$ at distance at most $$N\eta$$ ($$N \geq 1$$), there is a sequence of at most $$CN$$ points with “jumps” of size at most $$\eta$$ starting from $$x$$ and arriving to $$y$$.

This happens (if I’m not mistaken) when there exists a map $$\omega: (0,1] \rightarrow \mathbb{R}$$ such that for all $$x,y \in X$$, there is a path $$[0,d(x,y)] \rightarrow X$$ from $$x$$ to $$y$$ for which $$\omega$$ is a modulus of uniform continuity. So, for instance, any convex in a normed space works, or any complete Riemannian manifold.

The practical consequence is as follows: let $$\omega$$ be the modulus of uniform continuity of $$f$$. Then for any $$\eta>0$$ small enough, there is a constant $$D < \eta$$ ($$D=\eta/C$$) such that for any $$x,y \in X$$, $$|f(x)-f(y)|$$ is at most $$\omega(\eta)$$ times the smallest integer above $$d(x,y)/D$$.

Up to considering $$f=f^+-f^-$$ we may assume $$f \geq 0$$. Then define, for every $$M \geq 0$$, $$f_M(x)=\inf_y\, f(y)+Md(x,y)$$. $$f_M$$ is clearly a positive $$M$$-Lipschitz function on $$X$$ and $$f_M \leq f$$.

Now, let $$\eta >0$$ be small, $$D$$ as above, $$M>0$$ and $$M’=M/\omega(\eta)$$. Choose $$M$$ large enough so that $$M’D >2$$.

Let $$x,y\in X$$. Suppose that $$f(y) + Md(x,y) < f(x)$$. Then $$M’d(x,y)$$ is less than the smallest integer $$t$$ above $$d(x,y)/D$$. In particular, if $$d(x,y) > D$$, then $$M’d(x,y)< t < 2d(x,y)/D$$, a contradiction, so that $$d(x,y) \leq D \leq \eta$$ and thus $$f(y)+Md(x,y) \geq f(y) \geq f(x)-\omega(\eta)$$.

Therefore, $$\|f_M-f\|_{\infty} \leq \omega(\eta)$$. This concludes.

• A typical counter-example to the second question is $(\mathbb{R},d)$ with $d(x,y)=\min(|x-y|,1)$. Because then Lipschitz-continuous functions are bounded, but the class of uniformly continuous functions is the same… May 12, 2022 at 10:57
• Could you have a check on how I formalize your counter-example in my update? May 12, 2022 at 13:24
• @Akira: looks correct to me, apart from the example of unbounded uniformly continuous function ($x \longmapsto x$ works; but it requires linear growth so $x \longmapsto e^{-x}$ does not fit the bill). May 12, 2022 at 13:49