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The following two results are probably the most basic results in the spectral theory:

Let $a$ be an element in a unital Banach algebra $\mathcal{A}$ (over $\mathbb{C}$). The spectrum of $a$, $\sigma(a) = \{\lambda \in \mathbb{C} \mid (\lambda \mathbb{1} - a) \text{ isn't invertible} \}$, is non-empty.

This allows us to define the spectral radius $r(a) = \sup \{ \lvert \lambda \rvert \mid \lambda \in \sigma(a)\}$ for which

$r(a) = \lim_{n \to \infty} \lVert a^n \rVert^{1/n}$

holds. All proofs I've seen of these two theorems are essentially the same. The sketch of the proof of the first proposition goes as follows:

We prove this via contradiction, so suppose $\sigma(a)$ is empty. Define the resolvent of $a$ as the map $$ R:\mathbb{C} \to \mathcal{A}: \lambda \mapsto (\lambda \mathbb{1} - a)^{-1}. $$ It's analytic in the sense that $\phi \circ R$ is analytic for all $\phi$ continuous linear functionals on $\mathcal{A}$. Then it's not so hard to prove $R$ is bounded, so by Liouville's theorem $\phi \circ R$ must be constant for every $\phi$. This then implies $R$ must be constant and zero and thus we arrive at a contradiction.

The proof of the second theorem is very similar: First it's proven that $r(a) \leq \inf_n \lVert a^n \rVert^{1/n}$, which is not so hard. Then all that is left to prove is that $r(a) \geq \limsup_n \lVert a^n \rVert^{1/n}$. Let $\Omega$ be the open disk in $\mathbb{C}$ around $0$ with radius $\frac{1}{r(a)}$ and define $$ f:\Omega \to \mathbb{C}: \lambda \mapsto (\mathbb{1} - \lambda a )^{-1}. $$ We prove that $\phi \circ f$ is analytic for every $\phi$ as before. Then we can argue that $\phi(a^n) \lambda^n$ is a bounded sequence for every $\phi$ and $\lambda \in \Omega$ by using the power series of $\phi \circ f$. By Banach-Steinhaus we then have that $a^n \lambda^n$ is bounded for every $\lvert \lambda \rvert < \frac{1}{r(a)}$. It then follows that $r(a) \geq \limsup_n \lVert a^n \rVert^{1/n}$.

Every text I've encountered so far describe the same proof, maybe in a slightly different fashion but essentially the same. Are there any alternatives to these proofs? Preferably ones that use less complex analysis. I imagine this could be very hard, particularly for the second one. This because that formula is already very reminiscent of the root test for series, so it "screams" power series to me. But I would be very interested if there were any at all.

Edit: I would also like to add my interest in proving only the convergence of $\lVert a^n \rVert^{1/n}$ without relying on the proof described here. So no proof that it's equal to the spectral radius, but simply a proof that it converges.

Demophilus
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    Since nonemptiness of the spectrum isn't a theorem for real Banach algebras, I wouldn't bet money on the existence of a proof avoiding complex analysis. – Daniel Fischer Jun 14 '17 at 21:11
  • @DanielFischer Of course there's that too, of course any other proof would also be welcome even if it uses more complex analysis then these ones. – Demophilus Jun 14 '17 at 21:14
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    Keep in mind that any proof of the general case will imply the Fundamental Theory of Algebra, for which there are no methods nearly as nice as the Complex Analysis proof. It's surprising that the Complex Analysis proof of the Fundamental Theorem of Algebra extends so nicely to this general case. – Disintegrating By Parts Jun 14 '17 at 22:59
  • How would it imply the Fundamental Theorem of Algebra? Is it true that any polynomial is the characteristic polynomial of some matrix? – mathworker21 Jun 26 '17 at 07:20
  • About the Edit: the fact that $|a^n|^(1/n)$ converges is a particular case of something referred to as Fekete's lemma, because $|a^n|$ is submultiplicative : $| a^{n+m} | \leq |a^n| , |a^m|$. – ACL Nov 29 '21 at 19:04

2 Answers2

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You can replace Liouville's theorem by what is used to prove Liouville's theorem, namely Cauchy estimates. For $|\lambda| < 1/\|a^{-1}\|$, $$R(\lambda) = (\lambda 1 - a)^{-1} = \sum_{n=0}^\infty - \lambda^n a^{-n-1}$$ and in particular $$ -a^{-2} = \frac{1}{2\pi i} \oint_{\Gamma(r)} \frac{R(\lambda)}{\lambda^{2}}\; d\lambda $$ where $\Gamma(r)$ is a positively oriented circle centred at $0$ with radius $r < 1/\|a^{-1}\|$. But if the spectrum were empty, this would be true for all $r > 0$. As you take $r \to \infty$, since $\|R(\lambda)\| \le 1/(|\lambda|-\|a\|)$ the norm of the right side goes to $0$, contradiction.

Robert Israel
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  • That's quite nice, thank you! – Demophilus Jun 14 '17 at 23:51
  • Interesting idea! It's more complex analysis, though...it would be interesting to know if there exist proofs circumventing it completely, but I strongly suspect that there are none...can you offer any insights? EDIT: the argument by DisintegratingByParts in the comment under the OP makes it at least extremely implausible – haemi Jun 15 '17 at 15:40
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See theorem 1.2.8 pages 10-13 in A course in commutative Banach Algebras, Eberhard Kaniuth. The first proof is classic, the second one, probably, is what you are looking for.

Norbert
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