Find a polynomial of degree $\leq n$ such that $p(0)=1$ and $p$ has smallest supremum norm on $[1,2]$

Question

Of course, if $n=0$ then $p=1$.

I found myself that if $n=1$ then $p=1-x\frac23$, just by seeing that $p(1)$ must be $-p(2)$ cause $p$ is monotone.

What about $n>1$? I thought about sth related to Tschebyschev polynomials (scaled and translated) but it does not make sense after longer thinking.

Any hints? Thanks in advance.

Answer 1

Loosely speaking, among all polynomials whose magnitude over $[-1,1]$ is restricted, the Chebyshev polynomials $T_n(x)$ grows fastest outside $[-1,1]$. More precisely,

Let $f(x)$ be any polynomial with degree at most $n$ and $|f(x)| \le 1$ over $[-1,1]$. For all $x \in \mathbb{R}$, we have $$|f(x)| \le \max(1,|T_n(x)|)$$

For a proof, see my answer to an unrelated question about Chebyshev polynomials.

Back to the question at hand.

Let $p(x)$ be any polynomial with degree $n$ with $p(0) = 1$ and sup-norm $M= \sup\limits_{x \in [1,2]}|p(x)|$ over $[1,2]$. It is easy to see $f(x) = \frac1M p\left(\frac{3-x}{2}\right)$ is a polynomial with degree $n$ and $|f(x)| \le 1 $ over $[-1,1]$. Apply above result, we find

$$1 = p(0) = M |f(3)| \le M \max(1,|T_n(3)|) = M T_n(3)\quad\implies\quad M \ge \frac{1}{T_n(3)}$$

The lower bound $\frac{1}{T_n(3)}$ is achievable with

$$f(x) = T_n(x)\quad\iff\quad p(x) = \frac{T_n(3-2x)}{T_n(3)}$$

This means $\frac{T_n(3-2x)}{T_n(3)}$ is a polynomial with degree $n$ with smallest sup-norm over $[1,2]$.

Answer 2

Using Jyrki's conjectured form and slightly adjusting the proof of minimality from the Wikipedia page on Chebyshev polynomials:

Define $\| p \| = \sup_{x \in [1, 2]} |p(x)|$, and let $p_n = \frac{T_n(3 - 2x)}{T_n(3)}$, so $p_n$ has degree $\leq n$ and $p_n(0) = 1$. Note that we have $|p_n(x)| = \|p_n\|$ at the $n+1$ points $x_k = \frac{3}{2} - \frac{1}{2}\cos \frac{k\pi}{n}$ for $0 \leq k \leq n$ (where each $x_k \in [1, 2]$), and $p_n$ alternates in sign on $x_0, x_1, \dots, x_n$. Suppose there were some polynomial $q$ of degree $\leq n$ with $q(0) = 1$ and $\|q\| < \|p_n\|$. Then defining $f = p_n - q$, since $\|q\| < \|p_n\|$, subtracting $q$ does not change the sign of $p_n$ at the points with $|p_n(x)| = \|p_n\|$, meaning $f$ also alternates in sign on $x_0, x_1, \dots, x_n$. But by the IVT, $f$ then has at least $n$ roots on the interval $[1, 2]$ (one between $x_i$ and $x_{i+1}$ for each $0 \leq i < n$), and $f$ has another root at $x = 0$, for a total of $n+1$ roots, which is impossible as $f$ has degree $\leq n$. This means there is no such $q$, so $p_n$ minimizes $\|\cdot \|$ among all polynomials with degree $\leq n$.