How to prove $|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|\geq 1/8$ for some $x\in[-1, 1]$

Question

I am trying to prove $|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|\geq 1/8$ for some $x\in[-1, 1].$

I found one case where $\max|x^{4}+ ax^{3}+ bx^{2}+ cx+ d|= 1/8$ for some $x\in[-1, 1],$ which is $$x^{4}- x^{2}+ \frac{1}{8}$$

Answer 1

Let $T_4$ be the Chebyshev polynomial of degree $4$. Then $p(x)=\tfrac18 T_4(x) = x^4-x^2 + \tfrac18$ is monic and oscillates between $-\tfrac18$ and $ \tfrac18$ on $[-1,1]$. In particular, there are numbers $$-1 = x_0 < x_1 < x_2 < x_3 < x_4 = 1$$ such that $$p(x_0)=p(x_2)=p(x_4)=\tfrac18$$ and $$ p(x_1)=p(x_3)=-\tfrac18.$$ Now suppose that $q(x)$ is a polynomial of degree $4$ such that $\lvert q(x) \rvert < \tfrac18$ on $[-1, 1]$. Then $q\neq p$ and $p - q$ must have a root in each of the four intervals $(x_0, x_1), \ldots, (x_3, x_4)$ since it switches sign on each of these. In particular $p-q$ must be of degree $4$ and therefore $q$ cannot be monic.

In other words for any monic $q$ of degree $4$ there is some $x \in [-1, 1]$ such that $\lvert q(x) \rvert \geq \tfrac18$.

In the same way (using $T_n$) it follows that this bound is $2^{1-n}$ for monic polynomials of degree $n \geq 1$.