As you probably know, the classical proof of the non-emptiness of the spectrum for an element $x$ in a general Banach algebra over $\mathbb{C}$ can be proven quite easily using Liouville's theorem in complex analysis: every bounded, entire function $\mathbb{C} \to \mathbb{C}$ is constant.

As these two theorems seem closely related and are certainly strong and non-trivial (for instance, both of them easily imply the fundamental theorem of algebra), I wonder if it is also possible to deduce Liouville's theorem from the non-emptiness of spectra for elements in complex Banach algebras. I guess one would like to to apply the Gelfand-Mazur theorem (which is a simple corollary of the above non-emptiness) to the Banach algebra of bounded, entire functions on $\mathbb{C}$ but showing that this is a division algebra is basically the same as showing that it is equal to $\mathbb{C}$ to begin with.

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    $\begingroup$ Interesting question. I can't see an easy way to do it, because in the algebra of bounded analytic functions on a domain $\Omega\subseteq{\mathbb C}$, the spectrum of an element is its range as a function, and this can be shown directly without using the general result for Banach algebras. Hence knowing the spectrum is non-empty is for this algebra somehow not connected to Liouville, as far as I can see $\endgroup$ – user16299 Jan 14 '12 at 21:03
  • $\begingroup$ If I'm not mistaken (and that could very well be :D ), it is enough to find a proof of the following fact: the image of a non constant bounded function on $\mathbb{C}$ is at a positive distance from $0$. This is reasonable, because a bounded function has no poles and it is bounded around $\infty$, therefore it extends to it holomorphically, therefore it has no poles on $\mathbb{CP}^1$, hence it cannot have zeroes. Now one should work a bit to show that the distance from $0$ is actually positive... $\endgroup$ – wisefool Jan 6 '13 at 21:48
  • $\begingroup$ Deduce a theorem from another? When one says that is 'deducing' a theorem from another it is usually meant that one used its statement in the proof. But using a result in a proof can go from using it in a very significant step or in a very trivial way. If there is no restriction to how significant must be the use of a theorem then the answer is going to be easily yes. On the other hand, it might be clear to the eye in many cases, but in general it is hard to define how significant is a part of a proof inside the proof. $\endgroup$ – OR. Jun 26 '13 at 21:19
  • $\begingroup$ If what @user16299 and wisefool are saying is correct, then for any function $f$ in this space, $\varphi(f) \neq 0$ for any multiplicative linear functional $\varphi$. In that case, $f$ must be invertible - am I missing something? $\endgroup$ – Prahlad Vaidyanathan Aug 13 '13 at 16:39
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    $\begingroup$ A proof of nonempty spectra => fundamental theorem of algebra can go as follows, and applies to any normed field $(K,\lVert\cdot\rVert)$. Let $A$ be the algebraic closure of the completion $\hat K$ of $K$. It is standard that there is a unique extension of the norm from $\hat K$ to $A$. Then, for any $a\in A$, let $\lambda$ be in its spectrum (as a $K$-algebra). As $a-\lambda$ is not invertible, we have $a=\lambda$, so $K=A$ is algebraically closed, $\endgroup$ – George Lowther Aug 15 '13 at 22:51

Yes, you can find such a proof in the following article:

Citation: Singh, D. (2006). The spectrum in a Banach algebra. The American Mathematical Monthly, 113(8), 756-758.

The paper is easily located behind a pay-wall (JSTOR) here.

Here is an image of the start:

enter image description here

As promised, the article concludes with a proof of Liouville's Theorem following from Theorem 1 here, that is, using the non-emptiness of $\sigma(a)$ as specified by the OP. Perhaps someone else can find a copy of this article that is freely accessible.


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