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?
Best Answer
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.