Let $p:[-1,1] \to \mathbb{R}$ be a polynomial. Prove that for every $\epsilon \gt0, \exists$ a polynomial $q:[-1,1] \to \mathbb{R}$ with rational coefficients s.t. $\Vert p-q \Vert_\infty \lt \epsilon$.

My overall approach is in constructing a polynomial with rational coefficients.


Let $\epsilon \gt 0$ be given and let a polynomial $p$ be given. Let $p(x) = a_0 + a_1x+ a_2x^2 +\dots+ a_nx^n$ for some $n \in \mathbb{N}$ and $a_0, a_1, \dots, a_n \in \mathbb{R}$.

Define a polynomial $q(x) = b_0 + b_1x +\dots+b_nx^n$, where each coefficient $b_i$ is defined by:

$\displaystyle b_i = \frac{l_i}{m_i}$ where $l_i,m_i \in \mathbb{N}, m_i \neq 0$, s.t. $\displaystyle \left| \frac{l_i}{m_i} -a_i \right| \lt \frac{\epsilon}{n+1}$.

Then $$ \begin{align} \displaystyle |p(x)-q(x)| & = \left| \left(a_0-\frac{l_0}{m_0}\right) + \left(a_1-\frac{l_1}{m_1}\right)x + \dots + \left(a_n-\frac{l_n}{m_n}\right)x^n\right| \\ & \leq \left| a_0-\frac{l_0}{m_0} \right| + \left| a_1-\frac{l_1}{m_1} \right| \cdot |x| + \dots + \left| a_n-\frac{l_n}{m_n} \right| \cdot |x^n| \\ & \leq \frac{\epsilon}{n+1} + \frac{\epsilon}{n+1} \cdot |x| + \dots + \frac{\epsilon}{n+1} \cdot |x^n| \\ & \leq \frac{\epsilon}{n+1} (n+1) \\ & = \epsilon . \end{align} $$

So $|p(x) - q(x)| \lt \epsilon, \ \forall x \in [-1,1]$, then $\Vert p-q \Vert_\infty \lt \epsilon$.


I would like some feedback on overall correctness, style as well as simplification if possible.

Thank you.

  • 1
    $\begingroup$ Your proof is correct. It would be notationally simpler to not specify the $b_i$ as fractions, seeing as how that is irrelevant to the matter at hand; in the sense that they key property of $\mathbb{Q}$ is its density. Indeed, your proof works for polynomials with coefficients in any dense set. $\endgroup$
    – Reveillark
    Dec 2 '19 at 5:06
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    $\begingroup$ As @Reveillark said, just say $ b_i\in \Bbb Q$ and replace each $l_i/m_i$ with $b_i$.... And for better style, before in the main display about $p(x)-q(x)$, replace "Then" with "Then for all $x\in [0,1]$ we have". Quite correct work. $\endgroup$ Dec 2 '19 at 10:53
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    $\begingroup$ My edit was to put a period after $\epsilon$ in the 2nd-last line. I commend you for your style: Grammatically complete sentences, logically related. $\endgroup$ Dec 2 '19 at 11:00
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    $\begingroup$ You could also employ the summation notation $\sum$ instead of $+...+$ but I would call this a matter of taste. But in the section about $|p(x)-q(x)|$ you should erase the terms that include $a_1,$ because they are unneeded and because it may be that $n=0.$... I have some mistakes in my other comments but it's too late to edit them. $\endgroup$ Dec 2 '19 at 11:10

A compact way of writing, maybe:

Choose $b_{i}\in\mathbb{Q}$ such that $|a_{i}-b_{i}|<\epsilon/(n+1)$, then for $x\in[-1,1]$, \begin{align*} |p(x)-q(x)|&=\left|\sum_{i=0}^{n}(a_{i}-b_{i})x^{i}\right|\\ &\leq\sum_{i=0}^{n}|a_{i}-b_{i}||x|^{i}\\ &\leq\sum_{i=0}^{n}|a_{i}-b_{i}|\\ &<\sum_{i=0}^{n}\dfrac{\epsilon}{n+1}\\ &=\epsilon, \end{align*} so $\|p-q\|\leq\epsilon$.


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