Exercise 3.D.19

Proof. _(a)_ Let $P=P_m(ℝ)$ for some positive integer $m$. Suppose $p\in P$. We have $m\ge \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->p\ge \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->\mathit{Tp}$. Therefore $\mathit{Tp}\in P$. $T$ restricted to $P$ is an operator on $P$ and, since $T$ is injective, $T$ restricted to $P$ is also injective and, thus, surjective. This means that $T$ maps to every polynomial with a degree less or equal to $m$. Because $m$ was arbitrary, it follows that $T$ is surjective.

_(b)_ Suppose there exists $p\in P_m(ℝ)$ such that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->p=m$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->\mathit{Tp}<\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->p$. Let $n=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->\mathit{Tp}$. $T$ restricted to $P_n(ℝ)$ is surjective, hence there exists $q\in P_n(ℝ)$ such that $\mathit{Tq}=\mathit{Tp}$. Since $m>n$, it follows that $p-q\ne 0$, but

This is a contradiction to the fact that $T$ is injective, therefore we cannot have $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->\mathit{Tp}<\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->p$. □