0
$\begingroup$

I'm looking for a version of Weierstrass's approximation theorem that works for a continuous function $f:D \to \mathbb{R}^d$.

Versions that I know of

Multivariate Weierstrass theorem? uses a generalization of Weierstrass for multivariate domain based on the Stone's generalization of the theorem.

Wikipedia gives several versions but none (as far as i read) works with a multivariate image.

What I need it for

I need it to properly justify Proof of Peano's existence theorem for ODEs or Proving Peano's Existence Theorem by approximating with $C^{\infty}$ functions using Weierstrass' Theorem.

Seen the above application that I'm seeking I think that perhaps I can do it component-wise. (????)

Statement of the (possible) theorem

Let $D \subseteq \mathbb{R} \times \mathbb{R}^d$ be an open set and $f:D \to \mathbb{R}^d$ a continuous function.

$\exists.p_n:D \to \mathbb{R}^d$ a sequence of polynomials such that $p_n \stackrel{\|\cdot\|_{\infty}}{\to} f$ on a compact set of the form $[a,b] \times \overline{B}(x_0,b) \subseteq D$.

$\endgroup$
  • $\begingroup$ acadpubl.eu/monographs/201301/1012732acadpubl.201301.pdf seems a good place to start, althougth i would like the theorem restated in my easy conditions, perhaps theorem 4.3.1 $\endgroup$ – Javier Apr 8 '18 at 14:27
  • $\begingroup$ State the version you would like to hold. $\endgroup$ – zhw. Apr 8 '18 at 15:03
  • $\begingroup$ Why not just use Stone Weierstrass on each coordinate function of $f?$ $\endgroup$ – zhw. Apr 8 '18 at 15:31
1
$\begingroup$

Perhaps this simple case will help: Suppose $f=(f_1,\dots,f_k):\mathbb R^j\to \mathbb R^k$ is continuous. By Stone-Weierstrass, for each $m$ there exist polynomials $p_{m1}, \cdots, p_{mk}$ such that

$$|p_{mi}-f_i| < 1/m\,\, \text { on } \overline {B(0,m)},\, i=1,\dots, k.$$

Set $p_m = (p_{m1},\dots , p_{mk}).$ Then $p_m\to f$ uniformly on compact subsets of $\mathbb R^j.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.