Thirteenth IMAR Mathematical Competition, 2015

a) Let $I \subset \mathbb{R}$ be a closed bounded interval, let $P(I)$ be the set of all polynomial functions $f: I \to \mathbb{R}$, and let $\|f\| = \max_{x \in I} |f(x)|$. Define $A: P(I) \to P(I)$ by $Af(x) = (f(x))^2 - \frac{1}{2}\|f\|^2$, $x \in I$. If $f$ is monic (respectively, non-constant), then so is $Af$. Further, $\|Af\| = \frac{1}{2}\|f\|^2$, so
$$\|A^n f\| = 2 \left( \frac{1}{2} \|f\| \right)^{2^n},$$
where $A^n = A \circ \dots \circ A$, $n$ times, is the $n$-th iterate of $A$. Consequently, if $\|f\| < 2$ for some $f$ in $\mathcal{P}(I)$, then $\|A^n f\| < 1$ for $n$ large enough.

b) In the notation above, let $I = [-2, 2]$, and define $B: \mathcal{P}(I) \to \mathcal{P}(I)$ by
$$Bf(x) = \frac{1}{2}f(\sqrt{x+2}) + \frac{1}{2}f(-\sqrt{x+2}), \quad x \in I.$$
If $f$ is monic, so is $Bf$. Further, $\|Bf\| \le \|f\|$, and $\deg Bf = \lfloor \frac{1}{2} \deg f \rfloor$. Consequently, if $2^n \le \deg f < 2^{n+1}$, where $n$ is a non-negative integer, then $\deg B^n f = 1$. To conclude the proof, notice that every monic polynomial function of degree 1 on $I$ has norm at least 2.

Remark. Tchebysheff's polynomials and their properties provide an alternative solution. If $n$ is a non-negative integer, and $-1 \le x \le 1$, the Tchebysheff polynomial (real-valued function) of degree $n$ is defined by ...