# 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?

• 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...

EDIT. The weak form of Cauchy's theorem that I mentioned uses only elementary manipulation with absolutely convergent series, no integral is involved, see Liouville's theorem with your bare hands.

• 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

I do not believe that the Hahn-Banach theorem is necessary. At some point I had planned on writing up a blog post verifying this but I lost the motivation...

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.

• Some mathematicians of younger generation tell me that this is ridiculous to think whether Hahn-Banach is used in some proof or not, and to make any efforts to eliminate it. I am of different opinion, and always try to eliminate Hahn-Banach from my proofs if I can :-) And I would like to encourage you to write this blog. – Alexandre Eremenko Dec 19 '12 at 3:37
• 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
• My own motivation was to find a proof of Gelfand-Naimark in ZF (because I had heard that there was a version of Gelfand-Naimark that was true in any topos) but I couldn't figure out how to avoid using Banach-Alaoglu. I think the topos-theoretic version uses locales instead of topological spaces, maybe. – Qiaochu Yuan Dec 19 '12 at 23:32
• @qiaochu: I have a nasty feeling that claim was made shooting from the hip - not that it's false, mind. I think Mulvey and Banaschewski (spelling?) might be the names to check if you want details rather than airily dismissive rhetoric – Yemon Choi Dec 20 '12 at 8:07
• 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

There is a proof without using even complex analysis, and at first glance without using AC. We replace Cauchy integral formula by clever finite sums. Namely, we define the spectral radius as $$r(a):=\lim \|a^n\|^{1/n}$$ (the limit exists by Fekete lemma), by scaling it suffices to consider two cases: $$r(a)=0$$ and $$r(a)=1$$. In the first case $$a$$ is not invertible, since if $$ab=1$$, then $$a^nb^n=1$$ and $$\|a^n\|\cdot \|b\|^n\geqslant 1$$, thus $$r(a)\geqslant 1/\|b\|$$. If $$r=1$$, we prove that there exist $$\lambda$$ on the unit circle such that $$a-\lambda$$ is not invertible. Assume the contrary, then by continuity of the inverse and compactness of the circle we get $$\|(a-\lambda)^{-1}\|\leqslant M$$ for all $$\lambda$$ on the unit circle and certain $$M>0$$. Now we use the rational functions identity $$n^2 a^{n-1}=(1-a^n)^2\sum_{w:w^n=1}w(a-w)^{-2}$$ and take the spectral radius of both right and left hand sides. By using the easy equalities and inequalities $$r(u^n)=(r(u))^n$$, $$r(uv)\leqslant r(u)\cdot \|v\|$$, $$r(1+u)\leqslant 1+r(u)$$ for commuting $$u,v$$ we see that $$r(n^2a^{n-1})=n^2$$ while $$r(RHS)=O(n)$$. A contradiction.

I read this proof in the paper of E. Gorin, who says that it is a simplification of the proof in [Rickart C. E. General Theory of Banach Algebras. Princeton NJ, D. van Nostrand (1960).]