Find the polynomial $f(x)$ which have the following property Find the polynomial $p(x)=x^2+px+q$ for which $\max\{\:|p(x)|\::\:x\in[-1,1]\:\}$ is minimal.
This is the 2nd exercise from a test I gave, and I didn't know how to resolve it. Any good explanations will be appreciated. Thanks!

Note that $p(0)=q, p(1)=1+p+q$ and $p(-1)=1-p+q$.
Thus 
$$p(-1)+p(1)-2p(0)=2 \,.$$
Thus
$$2= p(-1)+p(1)-2p(0) \leq 4 \max \{ p(-1), p(1), -p(0) \} \,.$$
It follows that $\max \{ p(-1), p(1), -p(0) \} \leq \frac{1}{2} (*)$ 
This proves that 
$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} \geq \frac{1}{2} \,.$$
Moreover, we can only get equality in $(*)$ if and only if $ p(-1)=\frac{1}{2}, p(1)=\frac{1}{2}, p(0)=-\frac{1}{2}$, if and only if $q=-\frac{1}{2}, p=0$.
It is now easy to check that $f(x)=x^2-\frac{1}{2}$ satisfies 
$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} = \frac{1}{2} \,.$$
This proves that
$$\max\{\:|p(x)|\::\:x\in[-1,1]\:\} \geq \frac{1}{2} \,,$$
with equality if and only if $p=0, q=-\frac{1}{2}$.
