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I was involved into this subject when I answered this question from MSE. Trying to generalize my answer, I am thinking about a following

Question. Let $X$ and $Y$ be metric spaces. When each continuous function from $X$ to $Y$ can be approximated by Lipschitz functions in the topology of uniform convergence on compact sets?

It seems that I can prove that the question has a positive answer provided a pair $(X,Y)$ satisfies the following

Condition. Each bounded (that is with finite diameter of the image) function from $A$ to $Y$ defined on a uniformly discrete subset $A$ of the space $X$ can be extended to a Lipschitz function from $X$ to $Y$. (I call a subset $A$ of a metric space $(X,d)$ uniformly discrete, provided there exists a number $\varepsilon>0$ such that $d(x,y)\ge\varepsilon$ for each distinct points $x,y\in A$.) In particular, Valentine Theorem (see, [p. 11, LR] or [V]) implies that a pair $(X,Y)$ satisfies Condition provided $X$ and $Y$ are Hilbert spaces.

We found no references on the above results and I think about a publication and a continuation of this investigation. But I am not a specialist in this domain.

So, I care about the following questions. Are the above results new, good and worthy to be published somewhere? What another related problems are worthy to be investigated? In particular, is it known which pairs $(X,Y)$ of metric spaces satisfies Condition?

Thanks.

References

[LR] Ronnie Levy, Michael David Rice. The approximation and extension of uniformly continuous Banach space valued mappings, Commentationes Mathematicae Universitatis Carolinae, 24:2 (1983), 251--265.

[V] F.A. Valentine, A Lipschitz condition preserving extension for a vector function, Amer. J. Math., 67 (1945), 83--93.

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  • $\begingroup$ There's no hypothesis about having a uniform bound on the Lipschitz numbers of the approximating functions? $\endgroup$
    – Nik Weaver
    Commented Feb 18, 2014 at 20:19
  • $\begingroup$ @NikWeaver It seems that if $f=\lim f_n$, where all $f_n$ are $L$-Lipschitz then the function $f$ is $L$-Lipschitz too. $\endgroup$
    – Alex Ravsky
    Commented Feb 19, 2014 at 0:06
  • $\begingroup$ So you're assuming there is a uniform bound on the Lipschitz numbers of the approximating functions. $\endgroup$
    – Nik Weaver
    Commented Feb 19, 2014 at 3:23
  • $\begingroup$ @NikWeaver It seems that if I assume this then the limit function will be Lipschitz and the question will be trivial. So I do not assume this. $\endgroup$
    – Alex Ravsky
    Commented Feb 19, 2014 at 9:37
  • $\begingroup$ You could have just answered "no" to my first question, but okay. I believe the result, if correct, is new, and I think it is publishable. $\endgroup$
    – Nik Weaver
    Commented Feb 19, 2014 at 17:08

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