Suppose that $X$ is a metric space. Is the family of all real-valued uniformly continuous functions on $X$ dense in the space of all continuous functions with respect to the topology of uniform convergence on compact sets?
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3 Answers
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Yes, and even more is true. The argument is as follows: let $f\colon X \to \mathbb R$ be a continuous function and let $K\subset X$ be a compact set. Then $f|_K$ is uniformly continuous; let $\omega$ be its nondecreasing subadditive modulus of continuity. By McShane-Whitney's extension formula $f|_K$ admits a uniformly continuous extension to $X$ with the same modulus - more concretely, $$F(x)=\inf\{f(k)+\omega(d(k,x))\colon k\in K\}$$ is this extension which is uniformly continuous on the whole $X$.
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$\begingroup$ I'm glad we used the same argument with almost the same wording --too bad I was 33'' delayed :) $\endgroup$ Commented May 23, 2018 at 10:13
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$\begingroup$ Yes. Notice that some effort still has to be done to show that such a sub-additive modulus can be really found. Effort which we both elegantly avoided ;) $\endgroup$ Commented May 23, 2018 at 15:13
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$\begingroup$ "let K⊂X be a compact." Is there a missing word "set" here? $\endgroup$ Commented May 23, 2018 at 17:02
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1$\begingroup$ @Tony: Yes: on the existence of special moduli of continuity, the wiki article treats this issue too (it's me who wrote it :) ) $\endgroup$ Commented May 23, 2018 at 17:38
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Yes. For any compact subset $K\subset X$ consider a uniformly continuous extension of $f_{|K}$ (there is such an extension even with the same subadditive modulus of continuity, see here). This shows that the uniformly continuous functions on $X$ are a dense subspace of $C(X)$ in the topology of uniform convergence on compacta.
answered May 23, 2018 at 9:55
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Let $C(X)$ be the algebra of continuous functions endowed with the compact-open topology.
Notice that the functions $d_x : X \to [0, \infty)$ given by $d_x (y) = \min( d(x,y), 1)$ separate the points of $X$: if $d_x (p) = d_x (q)$ for all $x \in X$, taking $x = p$ gives $d_p(q) = 0$, whence $d(p,q) = 0$.
The functions $d_x$ are Lipschitz (with Lipschitz constant $1$). There are 3 cases to examine:
if $d(x,y) \le 1$ and $d(x,z) \le 1$, then $| d_x(y) - d_x(z) | = |d(x,y) - d(x,z)| \le d(y,z)$ by the triangle inequality;
if $d(x,y) \le 1$ and $d(x,z) > 1$ then
$$| d_x(y) - d_x(z) | = 1 - d(x,y) \le d(x,z) - d(x,y) \le d(y,z)$$
again by the triangle inequality; similarly for $d(x,y) > 1$ and $d(x,z) \le 1$;
if $d(x,y) > 1$ and $d(x,z) > 1$ then $| d_x(y) - d_x(z) | = | 1 - 1 | = 0 \le d(y,z)$.
Remember now that the sum and product with scalars of Lipschitz functions are again Lipschitz, and that the product of bounded Lipschitz functions (and the $d_x$ are bounded) is again Lipschitz (of course, the Lipschitz constant changes).
Consider the subalgebra $A$ of $C(X)$ generated by the family $(d_x)_{x \in X}$ and the constant function $1$. By the preceding paragraph it will be made only of Lipschitz functions. Using the version of the Stone-Weierstrass theorem for the compact-open topology (Stephen Willard, "General Topology", 1970, p. 293, par. 44B.3), the closure of $A$ is $C(X)$, so you get that not only uniformly continuous, but even Lipschitz functions are dense in $C(X)$
