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

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).

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 $$g(D)(f)=k!a_k\ne0$$.
• can you explain why it is $k! a_k$? I am not getting it. Apr 22, 2020 at 5:22