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 $|c|$.
 A: Let $p(1)=u,$ $p(-1)=v$, $p\left(\frac{1}{2}\right)=w$ and $p\left(-\frac{1}{2}\right)=t$.
Thus, we have the following system:
$$a+b+c+d=u,$$
$$-a+b-c+d=v,$$
$$\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d=w$$ and
$$-\frac{a}{8}+\frac{b}{4}-\frac{c}{2}+d=t,$$ which gives
$$a=\frac{2u-2v-4w+4t}{3},$$
$$b=\frac{2u+2v-2w-2t}{3},$$
$$c=\frac{-u+v+8w-8t}{6}$$ and
$$d=\frac{-u-v+4w+4t}{6}.$$
Now, $$a+b+c+d=u\leq1,$$
$$a+b+c-d=\frac{4u+v-4w-4t}{3}\leq\frac{13}{3},$$
$$a+b-c+d=\frac{4u-v-8w+8t}{3}\leq7,$$
$$a+b-c-d=\frac{5u-12w+4t}{3}\leq7,$$
$$a-b+c+d=\frac{-u-4v+4w+4t}{3}\leq\frac{13}{3},$$
$$a-b+c-d=-v\leq1,$$
$$a-b-c+d=\frac{-5v-4w+12t}{3}\leq7$$ and
$$a-b-c-d=\frac{u-4v-8w+8t}{3}\leq7,$$
which gives
$$|a|+|b|+|c|+|d|\leq7.$$
But for $p(x)=4x^3-3x$ the equality occurs, which says that $7$ is a maximal value.
A: A few simulations made me wonder if the $10$ bound is tight, and I tried to find other ones, as it seems that one could jointly bound the coefficients. I am not able yet to push them to the end, but I propose the first steps. I hope I am not completely wrong.
A  result by V. A. Markov states that:

If $p(x)$ is a polynomial of degree at most $m$ and if $|p(x)| \le 1$
  whenever $-1 \le x \le 1$, then $|p^{(k)}(x)| \le T^{(k)}_m (1)$
  whenever $-1 \le x \le 1$ and $1 \le k \le m$, with $T_m$ the $m$-th
  Chebyshev polynomial.

This can be found for instance in Markov's Inequality for Polynomials on Normed Linear Spaces, L. A. Harris, 2002. From the explicit formula, $T^{(3)}_m (1) = 24$, $T^{(2)}_m (1) = 24$, $T^{(1)}_m (1) = 9$.
We have $p'(x) = 3ax^2+2bx+c$, $p''(x) = 6ax+2b$ and $p'''(x) = 6a$.
For $k=3$,  we get $|6a|\le 24$, hence we recover  $|a|\le 4$. 
For $k=2$,  $|6ax+2b| \le 24$. This produces a simplex, along the maxima attained at $x=1$ and  $x=-1$, that reduces the original bounds, since for instance we should have $3a+b\le 12$.
For  $k=1$, this becomes a little more tedious: $|3ax^2+2bx+c|$ reaches its maxima either at $x=1$ (value $|3a+2b+c|$),  $x=-1$  (value $|3a-2b+c|$) or at the apex of the parabola $x=-\frac{b}{3a}$ (value $|c-\frac{b^2}{3a}|$). All these three quantities should be lower than $9$.
These coupled inequalities, together with those given by @Virtuoz, define a volume (non empty, since it contains the Chebyshev  polynomial $T_3(x) =4x^3−3x$) of admissible $(a,b,c)$ triplets.
Courageous wills are now welcome to plug $d$ in and find tighter bounds.
A: 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. 
A: Let $A=\max(|a|,|c|),C=\min(|a|,|c|),B=\max(|b|,|d|),D=\min(|b|,|d|).$ Then $$|A|+|B|+|C|+|D|=|a|+|b|+|c|+|d|.$$
For $|x|\le1$, $|Ax^2-C|\le|ax^2+c|$ and $|Bx^2-D|\le|bx^2+d|$. Then
$$|(A+B)x^3-(C+D)x|\le |Ax^3-Cx|+|Bx^2-D|\le |ax^3+cx|+|bx^2+d|\le |p(x)|\text{ or } |p(-x)|.$$
Therefore we only need consider $p(x)=ax^3-cx$ with $a\ge c\ge0$. 
At $x=1$ we need $a\le c+1$ and at $x=\frac{1}{2}$ we need $|a-4c|\le8$. The required maximum is attained for $a=4,c=3$.
