2015 Thirteenth IMAR Mathematical Competition

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 $\Vert f\Vert ={\max }_{x\in I}|f(x)|$. Define $A:P(I)\to P(I)$ by $Af(x)=(f(x){)}^2-\frac{1}{2}\Vert f{\Vert }^2$, $x\in I$. If $f$ is monic (respectively, non-constant), then so is $Af$. Further, $\Vert Af\Vert =\frac{1}{2}\Vert f{\Vert }^2$, so $$\Vert A^{n}f\Vert =2{\left(\frac{1}{2}\Vert f\Vert \right)}^{2^n},$$ where $A^n=A\circ \cdots \circ A$, $n$ times, is the $n$-th iterate of $A$. Consequently, if $\Vert f\Vert <2$ for some $f$ in $\mathcal{P}(I)$, then $\Vert A^{n}f\Vert <1$ for $n$ large enough.

b) In the notation above, let $I=[-2,2]$, and define $B:\mathcal{P}(I)\to \mathcal{P}(I)$ by $$Bf(x)=\frac{1}{2}f(\sqrt{x+2})+\frac{1}{2}f(-\sqrt{x+2}),x\in I.$$ If $f$ is monic, so is $Bf$. Further, $\Vert Bf\Vert \le \Vert f\Vert$, and $\mathrm{deg}Bf=\lfloor \frac{1}{2}\mathrm{deg}f\rfloor$. Consequently, if $2^n\le \mathrm{deg}f<2^{n+1}$, where $n$ is a non-negative integer, then $\mathrm{deg}{B}^{n}f=1$. 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 ...