TITLE: Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem.
QUESTION [16 upvotes]: 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?

REPLY [4 votes]: 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).]