2
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ 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"? $\endgroup$
    – Yemon Choi
    Commented Oct 23, 2011 at 23:15

2 Answers 2

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Right. But are the multiplicative linear functionals on $A(U)$ just point-evaluations? $\endgroup$
    – JSanderson
    Commented Oct 23, 2011 at 15:10
  • $\begingroup$ 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). $\endgroup$
    – Faisal
    Commented Oct 23, 2011 at 17:15
3
$\begingroup$

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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .