<s>[INST] Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$ 
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$,  what is the $\max$ value of $|a|+|b|+|c|+|d|$?

My try:


*

*Put $x=0$, we get $p(0)=d$,

*Similarly put $x=1$, we get $p(1)=a+b+c+d$,

*similarly put $x=-1$, we get $p(-1)=-a+b-c+d$,

*similarly put $\displaystyle x=\frac{1}{2}$, we get $\displaystyle p\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d$


So, given that $|p(x)|\leq 1\forall |x|\leq 1$, we get $|d|\leq 1$.
Similarly $$\displaystyle |b|=\left|\frac{p(1)+p(-1)}{2}-p(0)\right|\leq \left|\frac{p(1)}{2}\right|+\left|\frac{p(1)}{2}\right|+|p(0)|\leq 2$$
Now I do  not understand how can I calculate the  $\max$ of $|a|$ and $|c|$.
 [/INST] I'll assume that $x$ is real.
Consider system of four equations with four variables. 
$$
a + b + c + d = p(1),
$$
$$
-a+b-c+d = p(-1),
$$
$$
\frac{a}{8}+\frac{b}{4} + \frac{c}{2} + d = p(1/2),
$$
$$
-\frac{a}{8}+\frac{b}{4} - \frac{c}{2} + d = p(-1/2).
$$
You can just solve it and find values of $a,b,c,d$.
From this equations we get
$$
a+c = \frac{p(1)-p(-1)}{2},
$$
$$
\frac{a}{4} + c = p(1/2) - p(-1/2)
$$
Thus
$$
\frac{3a}{4} = \frac{p(1)-p(-1)}{2} - (p(1/2) - p(-1/2)),
$$
and $|a|\le4$. In the same way we get 
$$
3c = 4(p(1/2) - p(-1/2)) - \frac{p(1)-p(-1)}{2},
$$
and $|c|\le 3$.
The example $T(x) = 4x^3 - 3x$ shows existence of polynomial with $a=4$ and $c=3$.
So, $|a|+|b|+|c|+|d|\le 1 + 2 +3 +4 = 10$, though I don't know if such a polynomial (with $|a|+|b|+|c|+|d|\le 10$ and $p(x)\le 1$) exists. 
</s>