Let $A$ be a unital Banach algebra. Denote by $\mathrm{Inv}(A)$ the invertible elements in $A$, and $\mathrm{Inv}_\ell(A)$ the left-invertible elements. That is, $a \in \mathrm{Inv}_\ell(A)$ if and only there exists $b \in A$ such that $ba=1$.

Question: Is $\mathrm{Inv}(A)$ closed in $\mathrm{Inv}_\ell(A)$?

Note: $\mathrm{Inv}(A)$ is obviously open in $\mathrm{Inv}_\ell(A)$ since it is already open in $A$.

Motivation: If $A = B(X)$, the algebra of bounded operators on some Banach space $X$, then this seems to be true [Edit: Perhaps not! see the below comment of Nate Elredge]. Suppose $a_n \in B(X)$ converge (say in operator norm, but the same argument works if $a_n \to a$ in the strong operator norm) to $a \in B(X)$ where each $a_n$ is invertible and $a$ is left-invertible. Obviously $a$ is injective (it has a left-inverse). Since each $a_n$ is surjective and $a_n \to a$, it is quite easy to see that the range of $a$ must be dense in $X$. On the other hand, the range of $a$ is closed (again, this is because $a$ has a bounded left-inverse), so the range of $a$ is all of $X$. Since $a$ is bijective, it has an inverse in $B(X)$ by the bounded inverse theorem. Or, avoiding the latter technology, it follows easily that any (bounded) left-inverse of $a$ is actually the inverse of $a$.

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    $\begingroup$ " Since each $a_n$ is surjective and $a_n \to a$, it is quite easy to see that the range of $a$ must be dense in $X$": Why is that easy? In general the limit of surjective operators need not be surjective. (Consider $a_n = \frac{1}{n} I$ where $I$ is the identity operator; then the limit is 0.) $\endgroup$ – Nate Eldredge Nov 14 '14 at 16:08
  • $\begingroup$ @NateEldredge: You're right, I had temporarily bamboozled myself. The claim of mine which you quoted is not justified. Although, of course, $0$ is not left-invertible in your example... $\endgroup$ – Mike F Nov 14 '14 at 20:42
  • $\begingroup$ Please see my update. $\endgroup$ – Tomek Kania Nov 14 '14 at 21:21
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    $\begingroup$ Indeed, I wasn't claiming it was false (and as @TomekKania shows it is actually true, thanks to some triangle inequality juggling), only that it is not obvious. $\endgroup$ – Nate Eldredge Nov 14 '14 at 21:35
  • $\begingroup$ @NateEldredge: Right. I never read your comment as "the claim is false". Only "the claim needs to be justified". Thanks again for spotting the gap! $\endgroup$ – Mike F Nov 14 '14 at 23:33

Your proof can be generalised easily to arbitrary unital Banach algebras.

Let $A$ be a unital Banach algebra and consider the left regular representation of $A$ on itself, that is, the map $\lambda_a x = ax\;(a,x\in A)$. Since $A$ is unital, $\lambda\colon A\to B(A)$ is isometric.

Now, let $(a_n)_{n=1}^\infty$ be a sequence of invertible elements in $A$ which converges to some left-invertible $a\in A$. In particular, $\lambda_{a_n} \to \lambda_a$ as $n\to \infty$ in the norm topology of $B(A)$. By your argument, $\lambda_a$ is invertible in $B(A)$. Let $T\in B(A)$ be such that $T\lambda_a = \lambda_a T = I_A$. This means that for each $\xi \in A$ we have

$$T (a\cdot \xi ) = a \cdot T\xi = \xi.$$

In particular,

$$a\cdot T1 = 1,$$

so $T1$ is a right-inverse for $a$ in $A$. Since $a$ is both left- and right-invertible, it is invertible.

Edit. Mike, everything's ok. The fact you need is Theorem 2.9 in Aliprantis' Introduction to operator theory.

  • $\begingroup$ This is quite slick, +1. I had thought of embedding $A$ into $B(A)$, but I didn't see how to get from "invertible in $B(A)$" to "invertible in $A$". However, now I don't see why the "proof" I put in the question is valid (see Nate's comment above), so the argument is still incomplete. $\endgroup$ – Mike F Nov 14 '14 at 20:54
  • $\begingroup$ Regarding your edit: Excellent, all is indeed well. Thank you for your help. $\endgroup$ – Mike F Nov 14 '14 at 21:26

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