IMC 1995 · Day 2 · P10

a) Let $n$ be such that $(1 - \varepsilon^2)^n \le \varepsilon$. Then $|x (1 - x^2)^n| < \varepsilon$ for every $x \in [-1,1]$. Thus one can set $\lambda_k = (-1)^{k+1} \binom{n}{k}$ because then $$x - \sum_{k=1}^{n} \lambda_k x^{2k+1} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{2k+1} = x (1 - x^2)^n.$$

b) From the Weierstrass theorem there is a polynomial, say $p \in \Pi_m$, such that $$\max_{x \in [-1,1]} |f(x) - p(x)| < \frac{\varepsilon}{2}.$$ Set $q(x) = \frac{1}{2} \{ p(x) - p(-x) \}$. Then $$f(x) - q(x) = \frac{1}{2} \{ f(x) - p(x) \} - \frac{1}{2} \{ f(-x) - p(-x) \}$$ and $$\tag{1} \max_{|x| \le 1} |f(x) - q(x)| \le \frac{1}{2} \max_{|x| \le 1} |f(x) - p(x)| + \frac{1}{2} \max_{|x| \le 1} |f(-x) - p(-x)| < \frac{\varepsilon}{2}.$$ But $q$ is an odd polynomial in $\Pi_m$ and it can be written as $$q(x) = \sum_{k=0}^{m} b_k x^{2k+1} = b_0 x + \sum_{k=1}^{m} b_k x^{2k+1}.$$ If $b_0 = 0$ then (1) proves b). If $b_0 \ne 0$ then one applies a) with $\dfrac{\varepsilon}{2 |b_0|}$ instead of $\varepsilon$ to get $$\tag{2} \max_{|x| \le 1} \left| b_0 x - b_0 \sum_{k=1}^{n} \lambda_k x^{2k+1} \right| < \frac{\varepsilon}{2}$$ for appropriate $n$ and $\lambda_1, \lambda_2, \dots, \lambda_n$. Now b) follows from (1) and (2) with $\max\{n, m\}$ instead of $n$.