Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:24:27Z http://mathoverflow.net/feeds/question/116749 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116749/spectra-of-elements-of-a-banach-algebra-and-the-role-played-by-the-hahn-banach-th Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem. Leonard 2012-12-19T02:50:06Z 2012-12-19T03:42:04Z <p>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.</p> <p>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 $\mathbb{C}$:</p> <p>$${\sigma_{\mathcal{A}}}(a) \stackrel{\text{def}}{=} \lbrace \lambda \in \mathbb{C} ~|~ \text{ a - \lambda \cdot \mathbf{1}_{\mathcal{A}}  is not invertible} \rbrace.$$</p> <p>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 <em>Real and Complex Analysis</em>). 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?</p> http://mathoverflow.net/questions/116749/spectra-of-elements-of-a-banach-algebra-and-the-role-played-by-the-hahn-banach-th/116750#116750 Answer by Qiaochu Yuan for Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem. Qiaochu Yuan 2012-12-19T03:15:20Z 2012-12-19T03:15:20Z <p>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... </p> <p>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 <a href="http://math.stackexchange.com/questions/157217/liouvilles-theorem-for-banach-spaces-without-the-hahn-banach-theorem" rel="nofollow">this math.SE question</a>). 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 <a href="http://math.stackexchange.com/questions/185305/is-the-fundamental-theorem-of-calculus-independent-of-zf" rel="nofollow">this math.SE question</a>. </p> http://mathoverflow.net/questions/116749/spectra-of-elements-of-a-banach-algebra-and-the-role-played-by-the-hahn-banach-th/116751#116751 Answer by Alexandre Eremenko for Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem. Alexandre Eremenko 2012-12-19T03:26:21Z 2012-12-19T03:42:04Z <p>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.</p> <p>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 $| z | &lt; R \leq \infty$ 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), and derive the Liouville theorem for Banach-space-valued functions.</p> <p>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 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.</p> <p>Sorry if I missed something...</p>