Suppose D∈L(P(R),P(R)) is such that degDp=(degp)−1 for every nonconstant polynomial p∈P(R). Prove that D is surjective.
Consider p∈P(R) we must show there exists a q∈P(R) such that Dq=p.
Let n=(degp)+1 and consider D as a map from Pn(R) to itself.
Since deg(Dp)=(degp)−1 we have null D=P0(R) (the space of constants) meaning dimnull D=1. Therefor by 3.22
dimrange D=dimPn(R)−dimnull D=(n+1)−1=dimPn−1(R)
Thus there exists a q∈Pn such that Dq=p for p∈Pn−1(R).
This shows any differential-like operator which takes constants to zero is surjective! I can see how this could be useful in differential equation theory! See 27 for a basic example.