The minimizer of $\sup \{ |x^3+ax^2+bx+c| | x \in [0, 4] \}$ for $a,b,c \in \mathbb{R}$

Question

Question. Let $f(x) = x^3+ax^2+bx+c$

Find the maximum of $M$ such that $$ \forall a,b,c \in \mathbb{R} \; \exists x_0 \in [0,4]: |f(x_0)| \geq M$$

I saw a brief "solution," saying that:

By symmetricity, the polynomial $f(x)$ must have a symmetric point at $(2,0)$. And we must let the (positive) local maximum be equal to the endpoint maximum in order to "squeeze" $f(x)$ in the interval. And thus $f(x)=(x-2)^3-3(x-2)$ with $M=2$.

The answer indeed is $M=2$. Playing around with the graph of $(x) = x^3+ax^2+bx+c$ (with fixed $a=c=0$), I get the intuition of this (so-called) solution. However, the person who provided this solution could not write down a solid and rigorous solution. I also tried doing this with a calculus approach, but it did not simplify the question much.

How can I rigorously solve this question?

Answer 1

Let $P$ be the collection of monic polynomials in $x$ over $[0,4]$. For each $p \in P$, let $$\|p\|_\infty = \max\{ |p(x)| : x \in [0,4] \}$$

This is usually called the uniform norm (or sup norm) of $p$.

The problem at hand is equivalent to finding the maximum $M$ such that for all $p \in P$, $\|p\|_\infty \ge M$. $M$ is given by the infimum of $\|p\|_\infty$ as $p$ varies over $P$. ie.

$$M = \inf \{ \|p\|_\infty : p \in P \}$$

For the polynomial $f(x)=(x−2)^3−3(x−2)$, $f \in P$ and satisfies

• $|f(x)| \le 2$ for $x \in [0,4]$
• $f(0) = f(3) = -2$ and $f(1) = f(4) = 2$.

This means $\|f\|_\infty = 2$. Being the infimum of all such sup-norms, we have $M \le 2$.

It is impossible for $M < 2$. Otherwise, we can find a $g \in P$ such that $\|g\|_\infty < 2$. For such a $g$, $|g(x)| < 2$ for all $x \in [0,4]$ and hence the difference $h = g - f$ is positive at $x = 0, 3$ and negative at $x = 1,4$. This implies $h$ has at least 3 roots, one in each of the three open intervals $(0,1)$, $(1,3)$ and $(3,4)$. Now $h$ is a polynomial with degree at most $2$, this forces $h$ to be identical zero and $g = f$. Contradict with the condition $\|g\|_\infty < 2$.

Combine these, we can deduce $M = 2$.

What you see here is really a property of Chebyshev's polynomials.
Quoting wiki:

For any given $n \ge 1$, among the polynomials of degree n with leading coefficient 1 (monic polynomials), $$f(x) = \frac1{2^{n-1}} T_n (x)$$ is the one of which the maximal absolute value on the interval $[−1, 1]$ is minimal.

This maximal absolute value is $\displaystyle\;\frac{1}{2^{n-1}}$ and $|f(x)|$ reaches this maximum exactly $n + 1$ times at $$x = \cos \frac{k\pi}{n}\quad\text{ for }\quad 0 \le k \le n.$$

Your $f(x)$ is simply $2 T_3\left(\frac{x-2}{2}\right)$.