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Every proof of the spectral radius formula $r(\xi) = \limsup \|\xi^n\|^{1/n}$ for Banach algebras (that I have seen) uses the fact, that if $\zeta \in \mathbb{C}$ with $|\zeta| \geq s > r(\xi)$ (for $s>0$), then the series $$\sum \xi^n\zeta^{-n}$$ converges uniformly and absolutely on $\{\zeta \colon |\zeta| \geq s\}$. But isn't this already using the fact, that $r(\xi) \geq \limsup \|\xi^n\|^{1/n}$, which we essentially wanted to prove in the first place?

I guess the question basically boils down to why holomorphicity of the function $\zeta \mapsto (\zeta-\xi)^{-1}$ on the resolvent $\rho(\xi)$ of $\xi$ implies, that $(\zeta-\xi)^{-1} = \sum \xi^n\zeta^{-n-1}$ for all $|\zeta| > r(\xi)$.

h3fr43nd
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    $A=-aI$ shows that $I-a^{-1}A$ can be invertible. – Kavi Rama Murthy Apr 11 '20 at 07:29
  • @KaviRamaMurthy Do you understand the explanation given here: https://math.stackexchange.com/a/2651815/485385 ? – h3fr43nd Apr 12 '20 at 17:08
  • @Cameron Williams I have seen that one, my comment above yours lists exactly the same link. Yet I fail to understand the explanation given there. – h3fr43nd Apr 16 '20 at 11:53
  • Oh blerp haha I went on a deep dive here on MSE and forgot you linked something by the time I got back. I think this approach is pretty clean: https://chiasme.wordpress.com/2013/05/17/gelfands-spectral-radius-formula/ . – Cameron Williams Apr 16 '20 at 12:07
  • Yes I have seen that one too haha. Yet I think this also uses the statement I couldn't figure out how to deduce, since in Claim 3 the author tells us, that $f(\lambda) = \frac{-1}{\lambda} \sum \varphi(A^n)$ for every $|\lambda| > s_A$. Why does this follow from the uniqueness of power series expansion if we do not even know if the said series converges for $|\lambda| > s_A$? I mean by holomorphicity we know that there is a series expansion yes, but how can we be sure it is that one? – h3fr43nd Apr 16 '20 at 12:15

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I will use the symbol $a$ for a nonzero element in a Banach algebra and $\lambda$ for a complex number.

Suppose $|\lambda|\gt \|a\|\gt0$. Then $\frac{\|a\|}{|\lambda|}\lt 1$ and hence $(1-\lambda^{-1}a)^{-1}$ exists. Also, $$(1-\lambda^{-1}a)^{-1}=\sum_{n\geq 0}\lambda^{-n}a^n.$$ Now, $(1-\lambda^{-1}a)^{-1}=\lambda(\lambda-a)^{-1}$. Thus, $$(\lambda-a)^{-1}=\sum_{n\geq 0}\lambda^{-n-1}a^n\tag1\label1.$$

In other words, if $\lambda^{-1}\in B(0,1/{\|a\|})\setminus\{0\}$(the deleted open ball of radius $\frac1{\|a\|}$ centred at $0$), then $(\lambda-a)^{-1}$ exists and the holomorphic function $\lambda^{-1}\mapsto (\lambda-a)^{-1}$ has a power series representation given as in $\eqref1.$

Notice that $r(a)\leq \|a\|$, so $B(0,1/{\|a\|})\subseteq B(0,1/{r(a)}).$ Now, if $0\lt|\lambda^{-1}| \lt \frac1{r(a)}$(i.e. $|\lambda|\gt r(a)\gt0$), then $(\lambda-a)^{-1}$ exists and the holomorphic function $\lambda^{-1}\mapsto (\lambda-a)^{-1}$ has a power series expansion on $B(0,1/{r(a)})\setminus\{0\}.$ Since we already have a power series on a smaller neighbourhood, the uniqueness implies that the same should hold on $B(0,1/{r(a)})\setminus\{0\}$ as well.

cqfd
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  • That's very well written! I guess the uniqueness part is due to the identity theorem (https://en.wikipedia.org/wiki/Identity_theorem) and the proof is analogous for Banach algebra-valued analytic functions right? – h3fr43nd Apr 17 '20 at 08:53
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    @h3fr43nd Thanks for the bounty. I wrote this answer based on the proof of the spectral radius formula given in this text. Their proof is slightly different than the one given above. You might want to have a look at it. – cqfd Apr 17 '20 at 09:32
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    I will do that! The proof I had in mind was the proof given in the functional analysis book of Lax. He uses the fact that $$\xi^n = \oint_C \frac{\zeta^n}{\zeta-\xi} d\zeta$$ if C is a circle with radius greater than $r(\xi)$. He didn't explain however why we could argue this identity would hold (since we need uniform convergence, which he only showed for radii greater than $|\xi|$), since I guess he assumed this to be a trivial fact. If you are curious for the proof, it is on page 196 and the book is freely available online. – h3fr43nd Apr 17 '20 at 09:48