# Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem.

This problem was posed on Math StackExchange some time ago, but it did not garner any solutions there. I think that it is interesting enough to be posed here on Math Overflow, so here it goes.

Let $\mathcal{A}$ be a unital Banach algebra over $\mathbb{C}$, with $\mathbf{1}_{\mathcal{A}}$ denoting the identity of $\mathcal{A}$. For each $a \in \mathcal{A}$, define the spectrum of $a$ to be the following subset of $\mathbb{C}$:

$${\sigma_{\mathcal{A}}}(a) \stackrel{\text{def}}{=} \lbrace \lambda \in \mathbb{C} ~|~ \text{ a - \lambda \cdot \mathbf{1}_{\mathcal{A}}  is not invertible} \rbrace.$$

With the aid of the Hahn-Banach Theorem and Liouville's Theorem from complex analysis, one can prove the well-known result that ${\sigma_{\mathcal{A}}}(a) \neq \varnothing$ for every $a \in \mathcal{A}$. All proofs that I have seen of this result use the Hahn-Banach Theorem in one way or another (a typical proof may be found in Walter Rudin's Real and Complex Analysis). Hence, a natural question to ask would be: Can we remove the dependence of this result on the Hahn-Banach Theorem? Is it a consequence of ZF only? Otherwise, if it is equivalent to some weak variant of the Axiom of Choice (possibly weaker than the Hahn-Banach Theorem itself), has anyone managed to construct a model of ZF containing a Banach algebra that has an element with empty spectrum?

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Wasn't this question raised by Qiaochu Yuan? math.stackexchange.com/questions/157217/… –  Yemon Choi Dec 19 '12 at 4:05
I wasn't referring to Qiaochu's question, but it sure is a surprise to see that his asks almost the same thing. –  Leonard Dec 19 '12 at 5:03

I think Hahn-Banach can be eliminated from the usual proof, but being a non-expert in set theory, I cannot guarantee that the proof is completely independent of the axiom of choice.

Here is a sketch of a basic calculus proof. A function $U\to B$ from a region $U\subset C$ to a Banach space $B$ is called analytic if every point has a neighborhood where it is represented by a convergent Taylor series. You can prove a weak form of Cauchy theorem which says that if a function is analytic in $| z | < R \leq \infty$ then its Taylor series has radius of convergence at least $R$. It seems that this does not use the axiom of choice. Then you prove that Cauchy inequalities hold (there is a simple algebraic proof of this, see my answer to Liouville's theorem with your bare hands), and derive the Liouville theorem for Banach-space-valued functions.

Then again it is an elementary fact that if $a-\lambda_0 1$ has has a bounded inverse then the resolvent is an analytic function (in the sense I defined above) in a neighborhood of $\lambda_0$. Then you show that if the resolvent exists everywhere then it tends to $0$ as $\lambda\to\infty$. Then it seems to me that you obtain a proof without Hahn-Banach by applying the Liouville theorem to the resolvent.

Sorry if I missed something...

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It appears you have a mismatched dollar sign or something at the end of the second paragraph. –  Alex Becker Dec 19 '12 at 3:29
MathJack sometimes works funny. I had no misprint but struggled for several minutes to make MathJack understand my text:-) –  Alexandre Eremenko Dec 19 '12 at 3:39
+1/ I seem to remember trying exactly this approach some years ago when trying to teach a Banach algebras course - the ideas were stolen from some of the integral-free proofs of results like the maximum modulus theorem in 1CV. I haven't looked at the notes for a long time, though... –  Yemon Choi Dec 19 '12 at 4:08
Leonard, unfortunately I have no tips. I just alter the text in all ways delete and type again, until it works. Sometimes, when I type a longer answer, it just gets stuck, and no way out of it. So I prefer to type in a separate window, then paste. This usually works. –  Alexandre Eremenko Dec 19 '12 at 5:14
Some things are problematic, such as *, <, \{, \\ in math formulae because they sometimes are interpreted as markdown, html or escaped characters and the parser gets confused. If you avoid those, then things work reasonably well: use \ast, \lt, \lbrace to get *, <, \{ and use \cr instead of \\. –  Theo Buehler Dec 19 '12 at 10:15
The idea is that you can prove Liouville's theorem in the Banach space setting directly without using Hahn-Banach to reduce to the case of $\mathbb{C}$ (I asked whether this was possible in this math.SE question). Most of the steps in the proof are exactly the same; the only one that isn't, as far as I can tell, is the fundamental theorem of calculus, which is usually proven using the mean value theorem but which can instead be proven following the answers to this math.SE question.
One reason to eliminate the Hahn-Banach theorem in arguments about Banach spaces is to find a proof that would work for $p$-normed spaces, such as $L_p$ with $p<1$. –  Bill Johnson Dec 19 '12 at 15:18
I think the topos version must use locales or other "pointless" approaches, since otherwise you will have problems producing a character on the unital commutative C-star algebra $\ell^\infty/c_0$ ... –  Yemon Choi Dec 20 '12 at 8:09