Finding the minimum of the maximum value of $|P(x)|$ 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.
$max |P(x)| =\max \{1+|p|+q,|q-\frac {p^{2}} 4|\}$ 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 $\frac {p^{2}} 4 &lt;-|p|$). ). 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 $x^{2} -\frac 1 2$.
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 $P(x)= x^2$.