Stone-Weierstrass theorem for $C(\mathbb{R}^n,\mathbb{R}^m)$?

Question

By the Stone-Weierstrass theorem, we know that any $X$ compact with $A\subseteq C(X,\mathbb{R})$ subalgebra, we know that $A$ is then dense. However, is it true for the case where we have $\mathbb{R}^m$ instead of $\mathbb{R}$? For example, is it true that the set of polynomials from $X\subseteq \mathbb{R}^n$ to $Y\subseteq \mathbb{R}^m$ dense in the set $C(X,Y)$?

Answer 1

Hint: There is an obvious homeomorphism

$$h : C(X,\mathbb R^m) \to C(X, \mathbb R)^m ,$$ where $C(X, \mathbb R)^m$ is the product of $m$ copies of $C(X, \mathbb R)$. Thus each dense subset $D \subset C(X, \mathbb R)$ gives a dense subset $h^{-1}(D^m)$ of $C(X,\mathbb R^m)$.

Answer 2

Like in Stone-Weierstrass’ theorem, we assume that $C(X,Y)$ has the topology of uniform convergence.

The answer is negative even when $X=Y=\Bbb R$, because we cannot approximate an exponent $e^x$, which grows faster than any polynomial. On the other hand, for natural $n$ and $m$, a set of analytic functions is uniformly dense is $C(\Bbb R^n,\Bbb R^m)$. This is a corollary of much more general result of Kurzweil, see [Ku]. For a short survey and recent advances in this topic see our paper [MR].

On the other hand, in 1885 Weierstrass showed that for each natural $n$ and each compact subset $X$ of $\Bbb R^n$, any function $f\in C(X,\Bbb R)$ can be uniformly approximated by restrictions on $X$ of polynomials on $\Bbb R^n$. By the way, this is the first known result on uniform approximation of continuous functions. We can extend it to $C(X,\Bbb R^m)$, following Paul Frost’s hint.

References

[Ku] Kurzweil J. On approximation in real Banach spaces, Studia Mathematica 14 (1954), 214–231.

[MR] Mytrofanov M., Ravsky A. A note on approximation of continuous functions on normed spaces, Carpathian mathematical publications (2020) 12:1 107–110.