Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$

Question

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 $|c|$.

Answer 1

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.

Answer 2

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.