TITLE: Find a polynomial of degree $\leq n$ such that $p(0)=1$ and $p$ has smallest supremum norm on $[1,2]$
QUESTION [10 upvotes]: 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.

REPLY [7 votes]: 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 $[1,2]$.