Let's $w : \mathbb R_+ \to \mathbb R_+^*$ a continuous function (I will only be interested in the case $w : t \mapsto e^{-t}$). We use $w$ to define the Banach space $C^0_w(\mathbb R_+)$ of continuous functions $f : \mathbb R_+ \to \mathbb R$ such that $fw \xrightarrow[+\infty]{}0$, endowed with the norm $\| f \| = \|fw\|_\infty = \sup_{\mathbb R_+} |fw|$.

I want to prove that, when $w: t \mapsto e^{-t}$, the space of polynomials is dense in this Banach space. I think this is a special case of Bernstein's approximation problem, but I haven't found an explicit elementary proof with these keywords.

Because of Weierstraß's approximation theorem and a small change of variables, it is enough to show that the functions $e_d : t \mapsto e^{-dt}$ ($d \in \mathbb N$) are in the closure of the space of polynomials. It is relatively easy to show this for $e_1$, because you can show that $$ \|e_1 - P_n \| = \|e_1(e_1 - P_n)\|_\infty = O\left(\frac 1{\sqrt n}\right),\tag{$*$}$$ where $P_n$ is the degree-$n$ Taylor polynomial of $e_1$ at $0$. If we denote by $\mathscr F$ the closure of the space of polynomials in our Banach space, we have proved $e_1 \in \mathscr F$.

It seems that not much is needed to prove an equivalent result for the approximation of $e_d$. For instance, it would be enough to prove that all the $f_k : x \mapsto x^k e^{-x}$ are in $\mathscr F$ to kickstart a proof by induction: suppose all the $(f_k)$ are in $\mathscr F$ (so all the $\text{polynomial} \times e_1$ functions are, by linearity) and that $e_p$ is in $\mathscr F$ (so that we have a sequence of polynomials $(Q_n)_n$ converging to it w.r.t. the $\|\cdot\|$ norm). We then have $$\|e_1 Q_n - e_{p+1}\| = \|e_1(e_1 Q_n - e_{p+1})\|_\infty \leq \|e_1 Q_n - e_{p+1}\|_\infty = \|Q_n - e_p\| \xrightarrow[n\to\infty]{} 0.$$

However, I'm unable to prove that the $f_k$'s belong to $\mathscr F$, or more generally to make any significant progress on $e_1 \in \mathscr F$. Basically, $(*)$ doesn't give much maneuvering space...

I'm under the impression that there are much more sophisticated approaches to this problem, but I would really appreciate any help towards an elementary proof of the result.

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    $\begingroup$ I got interested in this question and I tried without success to answer it in the following way. Let $x\in \mathbb R$ be the real coordinate and embed $\mathbb R$ into the circle $\mathbb S^1\subset \mathbb R^2$ via the transformation $$\begin{cases} X= \frac{2x}{1+x^2} \\ Y= \frac{1-x^2}{1+x^2}\end{cases} \quad x=\frac{X}{1+Y} .$$ The point at infinity of $\mathbb R$ gets mapped to the south pole of $\mathbb S^1$ and the mapping $$(\mathcal{P} f)(X, Y)= f\left( \frac{X}{1+Y}\right) $$ is an isometry of $C^0_{\infty}(\mathbb R)=\left\{ f\in C^0\ :\ \lim_{|x|\to \infty} f(x)=0\right\}$... $\endgroup$ Sep 28, 2017 at 18:05
  • $\begingroup$ ... onto $C^0_\infty(\mathbb S^1)=\{ F\in C^0(\mathbb S^1) \ :\ F(0, -1)=0\}.$ The point of this construction is that, since $\mathbb S^1$ is compact, the Stone-Weierstrass theorem is available on this last space. Unfortunately this does not work as is, because the image of the vector space $$\mathcal D =\{ \exp(-|x|)p(x)\ :\ p\ \text{is a polynomial}\}$$ is not an algebra under pointwise products. This is probably a silly attempt but I wanted to write it down in the hope that some idea can be salvaged from it. $\endgroup$ Sep 28, 2017 at 18:08
  • $\begingroup$ Koosis's book on the logarithmic integral discusses the topic at great length, and your result is the Corollary on pg. 170 there. $\endgroup$
    – user138530
    Sep 30, 2017 at 2:11
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    $\begingroup$ @ChristianRemling: You should convert your comment into an answer, so it can be accepted and awarded the bounty. $\endgroup$ Oct 1, 2017 at 18:05
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    $\begingroup$ @GiuseppeNegro: Well, the OP was asking for an elementary proof, not sure if the reference fits that bill. $\endgroup$
    – user138530
    Oct 1, 2017 at 20:39


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