Prove there is no nonzero polynomial g(t) such that g(D)=0

Question

Let $V=P(R)$(a set of polynomials having infinite degrees) and this is an infinite dimensional vector space. Let $D:V \rightarrow V$ be the linear operator defined by $D(f(x))=f'(x)$. Prove there is no nonzero polynomial g(t) such that g(D)=0

I want to show that linear operators on infinite dimensional vector spaces do not always have minimal polynomials. That means I want to show that there exists no g(D) such that g(D)=0. But I am not sure how to argue this. Any help is appreciated. I am wondering does Cayley-Hamilton theorem help with this (but that is for finite dimensional).

Answer 1

Assume $g(D)=0$ where $g(x)=\sum_{k=0}^n a_kx^k$. Let $k$ be minimal with $a_k\ne 0$. Then let $f(x)=x^k$ and observe that $g(D)(f)=k!a_k\ne0$.