Finding the minimum of the maximum value of $|P(x)|$

Question

In the book "Problem Solving Strategies", I found the following problem

Find the polynomial $p(x) = x^2 + px + q$ for which $\max\limits_{x\in[-1,1]}|P(x)|$ is minimal.

I have no idea how to deal with this problem. Any help will be highly appreciated.

Answer 1

$max |P(x)| =\max \{1+|p|+q,|q-\frac {p^{2}} 4|\}$ since the maximum value is attaiined at one of the points $-1,1,-\frac p 2$. Claim: either $1+|p|+q \geq \frac 1 2$ or $|q-\frac {p^{2}} 4| \geq \frac 1 2$. I leave it to you to prove this by contradiction. (It is quite easy: if both inequalities fail you will get the contradiction $\frac {p^{2}} 4 <-|p|$). ). It follows that $max |P(x)| \geq \frac 1 2$ always. This value is attained when $p=0$ and $q =-\frac 1 2$ so the required polynomial is $x^{2} -\frac 1 2$.

Answer 2

I presume that you mean "$P(x)= x^2+ px+ q$", not "p(x)". That is a parabola opening upward. It takes on its maximum value on any interval at one of the endpoints of the interval. So the maximum of P(x) on [-1, 1] is either 1-p+ q or 1+ p+ q. The smallest value of either of those occurs when p= q= 0 or $P(x)= x^2$.

Answer 3

I don't know how to formally prove this, but a little geometric consideration could help. Since the problem is asking for the minimal value of the maximum of the modulus of $p(x)$, we need to set up the parabola such that the minimum and the maximum of $p(x)$ on the interval $[-1,1]$ are symmetric with respect to 0. Then, it is the best option to keep the modulus of the derivative of $p(x)$ small on this interval, such that $\max_{x\in[-1,1]} p(x) - \min_{x \in [-1,1]}$ is as small as possible. Since $p'(x) = 2x + p$, this happens exactly at $p=0$, i.e. when the minimum is at $x=0$. Setting $q=-1/2$ then makes sure that $\max_{x \in [-1,1]} p(x) = - \min_{x \in [-1,1]} p(x) \equiv 1/2$.