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

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.

Proof:

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 $$\Vert p-q \Vert_\infty \lt \epsilon$$.

$$\Box$$

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

Thank you.

• 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. Dec 2 '19 at 5:06
• 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. Dec 2 '19 at 10:53
• 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. Dec 2 '19 at 11:00
• 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. Dec 2 '19 at 11:10

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 $$\|p-q\|\leq\epsilon$$.