Is inversion continuous on normed (not necessarily Banach) algebras?

Question

There is a nice proof here that the map $x\mapsto x^{-1}$ is continuous on the group of invertible elements in a Banach algebra.

I am wondering if the same result holds for normed algebras only? In particular, if $X$ is a normed algebra and $G(X)$ denotes the group of invertible elements in $X$, then for any sequence $(x_n)$ where $G(X)\supset(x_n)_{n\in\mathbb N}\to x\in X$, is it true (1) that $x$ is invertible, and (2) that $({x_n}^{-1})\to x^{-1}?$ Also, are my conditions (1) and (2) equivalent to continuity of $x\mapsto x^{-1}$ on $G(X)$? I think yes, but I'm a bit confused because we are talking about a subspace topology on $G(X)$ induced by a not-necessarily complete norm.

The proof in the linked answer explicitly uses completion in the assumption that a series converges, so it is not sufficient to guarantee continuity in general normed algebras.

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Edit: After more thought, I think my condition (1) is equivalent to $G(X)$ being a topologically closed subset of $X$, and (2) is equivalent to continuity of the inverse map, if we only consider cases where $(x_n)$ converges in $G(X)$. At any rate, I would still like to know whether both (1) and (2) are satisfied.

Answer 1

The question has a trivial answer for (2): yes.

Let $A$ be a unital norm algebra. Then the completion of $A$ becomes a Banach algebra so that $A\subset B$ and $A$ is dense in $B$. Indeed, we already know that the completion of a normed space is a Banach space which contains the initial space as a dense subset, so all one should do is check that multiplication is extended to the completion, which is easy, and then verify the submultiplicativity of the norm on the completion, which again is easy.

Note that since $A$ is unital $B$ is also unital and $1_B=1_A$, because $A$ is dense in $B$. Now let $(x_n)\subset A$ be a sequence of invertible elements in $A$. Then they are invertible in $B$. If $(x_n)$ converges to $x\in A$ and $x$ is invertible, then using the fact that inversion is continuous on Banach algebras, one has that $x_n^{-1}\to x^{-1}$.

Answer 2

As requested by the OP, here is a proof that does not use completeness.

Let $A$ be a (not necessarily complete) unital normed algebra.

Lemma. Given $r$ in $A$, such that $1-r$ is invertible, and $\|r\|<1$, then:

1. The series $ \displaystyle \sum_{k=0}^\infty r^k $ converges to $(1-r)^{-1}$,

2. $\displaystyle \|(1-r)^{-1}-1\| \leq \frac{\|r\|}{1-\|r\|}$.

Proof. The series in (1) is absolutely convergent but, unfortunately, in a normed space this is not enough to guarantee convergence. On the bright side, there are other methods for proving the convergence of a series which do not rely on completeness!

Observe that, for every $n$, one has that $$ (1-r)\sum_{k=0}^n r^k = 1-r^{n+1}, $$ so $$ \lim_{n\to \infty }\sum_{k=0}^n r^k = \lim_{n\to \infty }(1-r)^{-1}(1-r^{n+1}) = (1-r)^{-1}, $$ where we have used that $\|r^{n+1}\| \leq \|r\|^{n+1}$.

To prove (2) we have $$ \|(1-r)^{-1}-1\| = \Big\|\sum_{k=1}^\infty r^k\Big\| \leq \sum_{k=1}^\infty \|r\|^k = \frac{\|r\|}{1-\|r\|}. \tag*{\square} $$

Remark. In a Banach algebra the absolute convergence of $\sum_{k=0}^\infty r^k$ implies convergence, and hence one may prove that $1-r$ is invertible. On the other hand, the hypothesis that $1-r$ is invertible was the key factor making the above proof work, as we are unable to deduce convergence from absolute convergence.

From (2) it is easy to see that inversion is continuous at 1, and hence at every invertible element $a$ in $A$ because if $\{a_n\}_n$ is a sequence of invertible elements converging to $a$, then $$ \|a^{-1}-a_n^{-1}\| \leq \|a^{-1}\|\|1-aa_n^{-1}\| = \|a^{-1}\|\|1-c_n^{-1}\|, $$ where $c_n:=a_na^{-1}$, and clearly $c_n\to 1$.