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?