The minimax problem We define the maximum norm of a function on $[-1,1]$ by $\|f \|_{\infty} = \underset{-1 \leq x \leq 1}{\max} |f(x)|$. Find $a,b \in \mathbb{R}$ such that $\|ax^2+x+b \|_{\infty}$ get the minimum value.
 A: A simple way to see that the minimum is at $a=b=0$ is as follows.
First, it is easily checked that for $a=b=0$, $\lVert f\rVert_{\infty}=1$.
Now, observe that for arbitrary $a,b\in\mathbb{R}$ we have
$$f(1)=a+b+1\\
f(-1)=a+b-1,$$
so that $|f(1)|>1$ whenever $a+b>0$ and $|f(-1)|>1$ whenever $a+b<0$.
Moreover, when $a+b=0$, we have $|f(1)|=|f(-1)|=1$. We see that regardless of the situation, $\lVert f \rVert_{\infty} \geq 1$, so $a=b=0$ is indeed a minimum for $\lVert f \rVert_{\infty}$.
A: Here is a cheap solution based on convexity:
Note that for a fixed $x$, the function $(a,b) \mapsto |a x^2+x+b|$ is convex,
hence it follows that $f((a,b)) = \max_{|x| \le 1} |a x^2+x+b|$ is convex.
Note that the domain of the $\max$ is symmetric about zero and compute
$f(-(a,b))$.
Use this information to compute the minimum value.

 We have $f(-(a,b)) = f((a,b))$ and so convexity gives $f((0,0)) \le {1 \over 2} (f((a,b))+ f(-(a,b))) = f((a,b))$. Hence $(a,b) = (0,0)$ is a minimiser.

Notes: 
A little work shows that that iff $|a| \le {1 \over 2}$ and $b=-a$,
then $f((a,b)) = f((0,0)) = 1$. In particular, while there is an odd function $x \mapsto 0 x^2 + 0$ that best approximates $x \mapsto x$, there is also
an even (but not odd) function $x \mapsto {1 \over 2} x^2 - {1 \over 2}$ that
best approximates $x \mapsto x$.
A similar approach can be used to show that there is a polynomial of best approximation of an odd function that has only odd powers of $x$. If $g$ is
odd,
$p_{e,o}(x) = \sum_k e_k x^{2k} + \sum_k o_k x^{2k+1}$  and $f((e,o)) = \max_{|x|\le 1} | p_{e,o}(x) - g(x)|$, then we can show that
$f((-e,o)) = f((e,o))$ and so $f((0,o)) \le f((e,o))$, so we need only consider
odd degree polynomials.
