2
$\begingroup$

Let $X=\mathbb{R}$, and $\mathcal{A}:=\mathbb{R}[x]$ be the subalgebra (of $C(X)$) of univariate polynomials.

Given $\varphi\in C_b(X)$ and $K\subset X$ compact, we know from Stone-Weierstrass that

$$\tag{1} \forall\,\varepsilon>0 \, : \ \exists\, p_\varepsilon\in\mathcal{A} \quad\text{such that}\quad \sup\nolimits_{x\in K}\!|\varphi(x) - p_\varepsilon(x)|\leq\varepsilon.$$

(Here, $C_b(X)$ is the space of all bounded continuous $\mathbb{R}$-valued functions on $X$.)

Question: Can the polynomials in $(1)$ be chosen such that $p_\varepsilon \leq \varphi$ pointwise on $X$?

If the answer is affirmative, can we generalise this to the case where $X$ is a $\sigma$-compact and bounded closed subset of a Banach space and $\mathcal{A}$ is a point-separating and pointwise non-vanishing subalgebra of $C_b(X)$?

(Note for the general case that the algebra $\mathcal{A}$ is dense in $C_b(X)$ wrt. the strict topology.)

$\endgroup$
3
  • 2
    $\begingroup$ $\mathbb{R}[x]$ is not a subalgebra of $C_b(\mathbb{R})$. The "b" means "bounded". $\endgroup$
    – Nik Weaver
    Commented Jul 29, 2023 at 18:44
  • 2
    $\begingroup$ No. For example $\varphi(x)=-e^x$ tends faster to $-\infty$ than any polynomial. $\endgroup$ Commented Jul 29, 2023 at 19:11
  • $\begingroup$ @JochenWengenroth Thanks for the comment and apologies, I deleted one "b" too many (in response to Nik Weaver's comment); the original question is now restored accordingly. $\endgroup$
    – fsp-b
    Commented Jul 29, 2023 at 19:16

1 Answer 1

4
+50
$\begingroup$

Yes, this works. Let's say $K\subseteq [-1,1]$. Start out by finding a polynomial $q$ such that $\varphi-\epsilon/2 \le q \le \varphi-\epsilon/4$ on $[-3,3]$. We can then take $p(x)=q(x)-(\epsilon/4)(x/2)^{2N}$. This will approximate $\varphi$ on $K$ with the desired accuracy for any choice of $N$, and also $p\le q\le\varphi$ on $[-3,3]$ for any $N$. Finally, taking $N$ large enough will make sure that also $p\le\varphi$ outside $[-3,3]$.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer to the case $X=\mathbb{R}$. Would you have any ideas for the general case as well? $\endgroup$
    – fsp-b
    Commented Jul 29, 2023 at 20:46
  • 1
    $\begingroup$ Nothing substantial, but if I had to guess, I'm a bit skeptical about this, even when $X=\mathbb R$ because maybe in general it could be possible for $\mathcal{A}$ to only contain functions that diverge to $+\infty$ at one end. $\endgroup$ Commented Jul 29, 2023 at 20:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .