# Boundary of invertibles in a normed algebra

A student and I are reading the book Introduction to Banach Spaces and Algebras, by Allan, and we're stuck. Exercise 4.5 says:

Let $A$ be a normed algebra with unit sphere $S$. Let $a\in A$. Then $a$ is a topological divisor of 0 if $$\inf\{\|ab\|+\|ba\|:b\in S \}=0.$$ Prove that every element in the frontier of $G(A)$ is a topological divisor of $0$.

Here $G(A)$ is the collection of invertible elements of $A$. I assume that the question really means to say that $A$ is a unital normed algebra. Then the book already essentially proves this result for Banach algebras (Corollary 4.13).

So if $B$ is the completion of $A$, and if $a$ is still in the frontier of $G(B)$, then we're done (the infimum obviously doesn't change if we replace $S$ by the unit sphere of $B$).

Conversely, if there is an example of $a\in\partial G(A)$ with $a\in G(B)$, then we have a counter-example to the exercise. So my question is:

If $a\in\partial G(A)$ and $B$ is the completion of $A$, then is $a\in\partial G(B)$?

Edit: Embarrassingly, I think I can now answer this!

Let $A$ be the complex polynomials, interpreted as an algebra of continuous functions on the interval $[0,1]$. A little bit of algebra shows that $G(A)$ consists of just the constant polynomials. So $G(A)$ is actually closed (not open, which would be the case if $A$ were Banach). So being careful about what "frontier" means, I guess $G(A)$ is its own frontier. But then the exercise is trivially false, as the frontier of $G(A)$ contains invertibles.

So the exercise seems wrong. But somehow my counter-example seems cheap. So a new question:

Can the frontier of $G(A)$ contain a non-invertible element which is invertible in $B$? Are there examples where $G(A)$ is open?

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It still feels to me that there should be a "simpler" or "more natural" commutative counterexample - some kind of explicitly defined algebra of functions on $[0,1]$, for instance... – user16299 Mar 1 '12 at 23:03

Let $A=\ell^1(F_2)$ sitting inside $B=C_r^*(F_2)$. There exists $a\in A$ which is self-adjoint but whose spectrum in $A$, call it $S$, is not contained in ${\mathbb R}$.
(We can take $a$ to have finite support: I forget the exact formula which works, but it can be found in Palmer Vol. II in one of the sections on "hermitian" groups, the point being that $F_2$ is not hermitian.)
Take a sequence $a_n=a-\lambda_n I$ for scalars $\lambda_n\notin S$ with $\lambda_n\to \lambda$ for some $\lambda \in S \setminus {\mathbb R}$. Then we have a sequence of invertible elements in $A$, which converge in $B$ to an element that is invertible in $B$ but not in $A$. I guess that by taking bicommutants we can get a commutative example from this one, in particular $B=C(X)$ for some $X\subseteq {\mathbb R}$.
Of course this doesn't answer your second question, where (if I understand correctly) you want $G(A)$ to be open in $A$ with respect to the norm from $B$.