Let $\Bbbk$ be a field and let us consider the $\Bbbk$-algebra $\Bbbk[X]$ of polynomials in one indeterminate with its natural Hopf algebra structure, i.e. $$\Delta(X)=X\otimes 1+1\otimes X, \quad s(X)=-X\quad \text{and} \quad \varepsilon(X)=0 .$$ On its linear dual $\Bbbk[X]^*$ we may define the convolution product by $$(f*g)(X^n)=\sum_{i+j=n} \binom{n}{i}f(X^i)g(X^j)$$ for all $f,g\in\Bbbk[X]^*$ and $n\in\mathbb{N}$. Let $$\Bbbk[X]^\circ=\left\{f\in\Bbbk[X]^*\mid f\left(X^kp(X)\right)=0\textrm{ for all }k\in\mathbb{N}\textrm{ and some }p(X)\in\Bbbk[X]\right\}$$ be the finite dual of $\Bbbk[X]$. In different but equivalent words, $\Bbbk[X]^\circ$ is the $\Bbbk$-subalgebra of $\Bbbk[X]^*$ of all those $f$ which vanishes on a finite-codimensional ideal of $\Bbbk[X]$.
Question: Can we claim that if $f*g\in\Bbbk[X]^\circ$ and $f\in\Bbbk[X]^\circ$ then $g\in\Bbbk[X]^\circ$ as well?
Equivalently: If a product $f∗g$ and one of its factors $f$ vanish on an ideal of finite codimension, is it necessarily true that the other factor $g$ also vanishes on an ideal of finite codimension?
P.S.: the question is strictly related with this other question on MO, but with a less technical taste (I hope). My thanks to KonKan for helping in clarifying the question.
A negative answer finally has been given to this question. I'm grateful to JJR because his/her answer to the following related question solved the problem once for all.
Let $\xi:\Bbbk[X]\to\Bbbk$ be given by $\xi(X^n)=\delta_{n,1}$ (in some sense, $\xi$ is the dual of the basis element $X$), and consider $g:\Bbbk[X]\to \Bbbk$ be given by $g(X^n)=\frac{1}{n+1}$. Then $$(\xi*g)(X^n)=\sum_{k=0}^{n}\binom{n}{k}\xi(X^k)g(X^{n-k})=\cases{ng(X^{n-1}) =1\quad\,\, n\geq 1 \\ 0 \qquad\qquad\qquad\quad n=0}$$ It follows that $\langle X^2-X\rangle\subseteq \ker{(\xi*g)}$ and $\langle X^2\rangle\subseteq \ker(\xi)$, so both $\xi$ and $\xi*g$ belong to $\Bbbk[X]^\circ$, but $g$ does not (the associated sequence is not linearly recursive, as it is the harmonic sequence).