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!
 A: Here's an informal argument that doesn't use calculus. Notice that $p(x)$ is congruent to $y = x^2$ (for example, simply complete the square). Now suppose that we chose our values for the coefficients $p,q$ carefully, and it resulted in producing the minimal value of $m$. Hence, we can think of the problem instead like this:

By changing the vertex of $y=x^2$, what is the minimal value of $m$ such that for all $x\in [-1,1], -m \le p(x) \le m$?

By symmetry, there are only two cases to consider (based on the location of the vertex).
Case 1: Suppose the vertex is at $(1,-m)$ and that the parabola extends to the top left and passes through the point $(-1,m)$. Using vertex form, we have $p(x)=(x-1)^2-m$ and plugging in the second point yields $m=(-1-1)^2-m \iff 2m=4 \iff m = 2$.
Case 2: Suppose that the vertex is at $(0, -m)$ and that the parabola extends to the top left and passes through the point $(-1,m)$. Using vertex form, we have $p(x)=x^2-m$ and plugging in the second point yields $m=(-1)^2-m \iff 2m = 1 \iff m = 1/2$.
Since $m=1/2$ is smaller, we conclude that $\boxed{p(x)=x^2-\dfrac{1}{2}}$ is the desired polynomial.
A: An answer that uses more advanced mathematics (which is probably not what you want) ...
The monic polynomial of degree $n$ that has minimal sup-norm on $[-1,1]$ is $T_n(x)/2^{n-1}$, where $T_n(x)$ is the Chebyshev polynomial of degree $n$. In fact, this minimum norm property can serve as the defining property of Chebyshev polynomials.
In our case, $n=2$, so the desired polynomial is $T_2(x)/2$ = $(2x^2 - 1)/2 = x^2 - \tfrac12$.
More about Chebyshev polynomials here.
A: 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 $p=0, q=-\frac{1}{2}$.
A: Here are some hints:


*

*$p''>0$ implies that $\max\{p(x):x\in[-1,1]\}=\max\{p(1),p(-1)\}$,

*$p(1)=1+p+q$ and $p(-1)=1-p+q$,

*Figure out which of those two values is bigger for each $p$ and $q$.


EDIT: I misread the question, so that the first point is wrong but can be fixed looking at $p(v_x)$ where $v_x$ is the $x$-coordinate of the vertex $v_x=\frac{-p}{2}$.
A: Here's a start, but I must sleep now.
Maxima can occur at $-1$, $1$, and points where the polynomial has derivative $0$.
At $-1$, the polynomial has absolute value $|1 - p + q|$.
At $1$, it has absolute value $|1 + p + q|$.
$p'(x) = 2x + p$, which vanishes at $x = -p/2$ (assuming this is in range), at which point the absolute value is $|p ^ 2/4 - p^2/2 + q|$ = $|-p^2/4+q|$
