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Is there a countable subset of polynomials $C$ with the property that every continuous function on $[a, b]$ can be uniformly approximated by polynomials from $C$?

This is problem from Abbott Understanding Analysis. I know that there exists a sequence of polynomials $p_n(x)$ that converges to $f$ uniformly on $[a, b]$. Moreover the set of polynomials $C$ is a countable set if it means polynomial with rational coefficients.

Hash Nuke
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  • You mean $C$ is countable, nt uncountable – Hagen von Eitzen Oct 02 '18 at 15:09
  • The set of polynomials with rational coefficients is countable. Unfortunately in the usual proof of the Weierstrass approximation theorem the Bernstein polynomials have rational coefficients but you need to multiply them by reals to get the approximation.. – Ross Millikan Oct 02 '18 at 15:10
  • @RossMillikan there is different proof on the book. It approximates $f$ by a polygonal function $\phi$ which is then approximated by polynomial with rational coefficients. – Hash Nuke Oct 02 '18 at 15:27
  • @HagenvonEitzen Yup does $C$ mean polynomial with real coefficients? – Hash Nuke Oct 02 '18 at 15:27

2 Answers2

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Let $p \in \mathbb{R}[X]$ and $\varepsilon > 0$. Now, if $p = a_nX^n + \dots + a_0$, for any $b_0, \dots, b_n \in \mathbb{Q}$ the polynomial $q = b_nX^n + \dots + b_0$ satisfies, given $x \in [a,b]$,

$$ |p(x)-q(x)| \leq \sum_{i=0}^n|x|^i|a_i-b_i|. $$

Since the interval and the polynomial (in particular its degree) are fixed, we can define

$$ \eta_j := \max_{x \in [a,b]}|x|^j $$

which only depend on $p$ and $[a,b]$. Thus, by density of the rationals in the reals, we can take each $b_i$ so that $|a_i-b_i| < \frac{\varepsilon}{2n_i}$ and then,

$$ |p(x)-q(x)| \leq \frac{\varepsilon}{2} < \varepsilon \quad (\forall x \in [a,b]). $$

This proves that the set of polynomials with rational coefficients is dense on the polynomials with real coefficients on $[a,b]$, and so it is dense in the continuous functions $C([a,b])$. Moreover, it is countable, because

$$ \begin{align*} & \qquad \qquad \mathbb{Q}[X] \longrightarrow \bigcup_{n \geq 1} \mathbb{Q}^n \\ &b_nX^n + \dots + b_0 \longmapsto(b_0,\dots, b_n) \end{align*} $$

is an injection.

qualcuno
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  • Thanks for the nice proof, can I ask why you alternate between notation $x$ and $X$? – Micah Smith May 09 '19 at 00:11
  • @MicahSmith in retrospective, there were a couple equalities in which I have not been consistent. Generally, I prefer writing uppercase letters for polynomials to remember they are formal expressions, and to distinguish them from the function they induce (for which I use lowercase letters). It's just a convention. – qualcuno May 09 '19 at 01:31
  • @qualcuno Shouldn't the requirement on $b_i$ be that $|a_i-b_i| < \frac{\varepsilon}{2 \eta_i \times(n+1)}$ rather than just $|a_i-b_i| < \frac{\varepsilon}{2 \eta_i}$? – TryingHardToBecomeAGoodPrSlvr Mar 23 '22 at 14:13
  • Yes, right (also the $n_i$ should be an $\eta_i$). Or we can say that $p$ and $q$ are at distance $(n+1)\varepsilon/2$ and that still proves the claim, since. $\varepsilon$ is arbitrary and $n$ only depends on $p$. – qualcuno Mar 23 '22 at 14:28
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    @qualcuno Thanks for the confirmation. Also, thank you for answering this question. It's a nice proof! – TryingHardToBecomeAGoodPrSlvr Mar 23 '22 at 15:09
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The set of polynomials with rational coefficients satisfies your statement.

TheSilverDoe
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