Let $\Bbbk[X]$ be the polynomial algebra in one indeterminate over a field $\Bbbk$, endowed with the primitive-like bialgebra structure, i.e. $\Delta(X)=X\otimes1+1\otimes X$ and $\varepsilon(X)=0$.

If we consider its linear dual $\Bbbk[X]^*$, then it turns out to be an augmented $\Bbbk$-algebra with the convolution product $(f*g)(x)=\sum_{(p)} f\left(x_{(1)}\right)g\left(x_{(2)}\right)$ (in Sweedler Sigma Notation), with the unit $\varepsilon$, and with the augmentation $\varepsilon_*:\Bbbk[X]^*\to\Bbbk$ given by $\varepsilon_*(f)=f(1)$. We may endow it with the $I$-adic topology, where $I=\ker(\varepsilon_*)$. In particular, a filter of neighborhoods of $0$ is given by $I^n$ for $n\in\mathbb{N}$.

We may also consider its finite (or Sweelder) dual $$\Bbbk[X]^\circ=\left\{f\in\Bbbk[X]^*\mid\ker(f)\supseteq I,\textrm{ for }I\textrm{ a finite-codimensional ideal in }\Bbbk[X]\right\}$$ which is a subalgebra of $\Bbbk[X]^*$ and it is endowed with the augmentation $\varepsilon_\circ:\Bbbk[X]^\circ\to\Bbbk$ sending $g$ to $g(1)$ as well. It can be endowed with the $J$-adic topology, where $J=\ker(\varepsilon_\circ)$, or with the linear topology induced by the filter of neighborhoods $\Bbbk[X]^\circ\cap I^n$ for $n\in\mathbb{N}$.

I'm asking myself: are these two topologies equivalent?

Of course, $J=\Bbbk[X]^\circ\cap I$ and $J^n\subseteq \Bbbk[X]^\circ \cap I^n$ for all $n\geq 2$, whence the $J$-adic topology is finer than the induced one, but can we claim the converse?

Any suggestion is really welcome.

[Edit 12.03.2017]

An equivalent formulation of the problem could be the following: consider the algebra $B$ of all infinite sequences $f=\left(f_n\right)_{n=0}^{\infty}$ with the Hurwitz product $$\left(f_n\right)_{n=0}^{\infty} \cdot \left(g_n\right)_{n=0}^{\infty}=\left(\sum_{h+k=n}\binom{n}{k}f_hg_k\right)_{n=0}^{\infty}$$ and let $\mathfrak{m}:=\left\{f\in B\mid f_0=0\right\}$ be its unique maximal ideal. Let $A\subseteq B$ be the subalgebra of all linearly recursive sequences, i.e. those sequences $f$ that satisfies a relation of the form $$f_n=c_1f_{n-1}+c_2f_{n-2}+\cdots+c_rf_{n-r} \quad \text{for} \quad n\geq r,$$ where $c_1,\ldots,c_r\in\Bbbk$. Set $\mathfrak{n}:=A\cap \mathfrak{m}$.

Are the $\mathfrak{n}$-adic topology and the topology induced by the $\mathfrak{m}$-adic topology on $A$ equivalent?