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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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).]