MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Take $U=\mathbb{D}_2\setminus \overline{\mathbb{D}_1}$ ($\mathbb{D}_r$ is the open disc centered at 0 with radius $r$) and consider the space $A(U)$ of all functions on $\overline{U}$ which are holomorphic in $U$ and admit a continuous extension to $\overline{U}$ (with obvious operations and the supremum norm).

Is $A(U)$ isomorphic to the disc algebra as a Banach algebra?

EDIT: I tried in fact, to find out, if all unital multiplicative linear functional on $A(U)$ are point-evaluations. Are they?

share|cite|improve this question
I suggest changing your title: $A(U)$ is in no sense a reverse of the disc algebra. Why not "An algebra of holomorphic functions on an annulus"? – Yemon Choi Oct 23 '11 at 23:15
up vote 4 down vote accepted

No. The spectrum of $A(U)$ is $\overline{U}$, while the spectrum of the disc algebra is $\overline{\mathbb D_1}$, and these two spaces aren't homeomorphic.

share|cite|improve this answer
Right. But are the multiplicative linear functionals on $A(U)$ just point-evaluations? – JSanderson Oct 23 '11 at 15:10
Yes, and this is true for more general $U$. You can find a proof in section 2.6 of Kaniuth, A Course in Commutative Banach Algebras (Springer, 2009). – Faisal Oct 23 '11 at 17:15

I'm curious to know where you came across this question - it is the kind of thing people (well, at least one person) like to set as an exercise when giving a graduate course on Banach algebras.

Anyway: an alternative way to see that $A(U)$ is not algebra-isomorphic to the disc algebra is to look at their groups of invertible elements.

Every invertible element of the disc algebra has a logarithm inside the algebra: that is, if $f\in A(\overline{\mathbb D})$ is invertible, then $f=e^g$ for some $g\in A(\overline{\mathbb D})$. (In particular, the group of invertible elements in $A({\mathbb D})$ is path-connected.)

On the other hand, consider the following element of $A(\overline{U})$: let $f(z)=z$. Clearly this is invertible in $A(\overline{U})$. On the other hand, if $g\in A(\overline{U})$ and $f=e^g$ then the restriction of $g$ to the circle of radius $3/2$ would be a continuous single-valued logarithm on that circle, which is impossible for topological reasons. (Pushing this argument further, one finds that the group of invertible elements in $A(\overline{U})$ is not connected.)

For more on this theme, look up the "Arens-Royden theorem". Having said all this, Faisal's answer is by far the simpler and neater way to approach the problem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.