Convergence of the sequence $nq^n$ for a topologically nilpotent element $q$ of a topological ring $R$.

Question

Let $R$ be a topological ring and $q$ an element of $R$ such that \begin{align*} \lim_{n\to \infty}q^n = 0, \end{align*} then we call $q$ topologically nilpotent. Do we also have \begin{align*}\lim_{n\to\infty}nq^n=0?\end{align*} If $q$ is nilpotent, this is clear as $nq^n$ is eventually constantly $0$. We will therefore assume that $q$ is topologically nilpotent, but not nilpotent.

The result is also clear when the topology is induced by an absolute value on $R$, that is a norm $||-||\colon R\to \mathbb{R}_{\geq 0}$ such that $||a\cdot b|| = ||a||\cdot ||b||$. In this case, $0=\lim_{n\to\infty}||q^n|| = \lim_{n\to\infty}||q||^n$ implies $||q||<1$. We then have \begin{equation} \lim_{n\to \infty} n||q^n|| = \lim_{n\to\infty}n||q||^n = 0. \end{equation} I would like to know whether this holds in more general topological rings. If not all topological rings, on what class of rings does it hold?

Answer 1

I don't know yet if this is true in general, but can rule out most familiar examples.

This holds for the following two classes of examples that cover most practical examples from algebra/number theory/algebraic geometry/nonarchimedean analsyis, and (archimedean) functional analysis:

• Topological ring for which a neighborhood basis of $0$ consists of abelian subgroups (aka non-archimedean rings, including Huber rings, Tate algebras, $I$-adic topology on a commutative ring $R$, and Adele rings of global fields.

• Topological rings with submultiplicative norm (which includes Banach algebras).

It's trivial to establish the first case. Let's show the second one.

Since $x^n\rightarrow 0\Leftrightarrow \|x^n\|\rightarrow 0$, there exists $m$ such that $\|x^m\|\le\frac{1}{2}$.

$$\|x^n\|=\|x^{mq+r}\|\le \|x\|^r\|x^m\|^q\le \max(1, \|x\|^m)(\frac{1}{2})^{\frac{n}{m}-1}=O(2^{-\frac{n}{m}})$$

By the triangle inequality,

$$\|nx^n\| = \|x^n + \cdots + x^n\|\le n \|x^n\|=O(\frac{n}{2^{n/m}})\xrightarrow{n\rightarrow \infty} 0$$

Answer 2

If $R$ is a topological ring whose topology comes from a family of submultiplicative semi-norms, then $q^n\to 0$ implies $nq^n\to 0$. In particular any $m$-convex locally convex Frechet algebra has this property.

The Aren's algebra $R = L^\omega[0, 1] = \bigcap_{p\geq 1} L^p[0, 1]$ with semi-norms $\|f\|_p = \left(\int_0^1 |f|^p\right)^{1/p}$ and multiplication $(fg)(x) = f(x)g(x)$ is a locally convex Frechet algebra which is not $m$-convex.

If $q(x) = x$, then $\|q^n\|_p = \frac{1}{(np+1)^{1/p}} \to 0$ but $\|nq^n\|_p = \frac{n}{(np+1)^{1/p}}\not\to 0$.

• The space $L^\omega$ and convex topological rings by Arens