Spectral radius of an element in a C*-algebra

Question

The following is an proposition of Takesaki's Operator Theory:

For any element $x$ of a Banach algebra ${\cal A}$, we have $$||x||_{sp}=\lim_{n\to \infty}||x^n||^{\frac{1}{n}}$$ Proof:

My question: Why does he use bounded linear functionals of ${\cal A}$ to show that the sequence $\{\frac{x^n}{\lambda^{n+1}}\}$ is bounded? I think because the power series $f(\lambda)=(\lambda - x)^{-1}=\sum_{n\geq 0}\frac{x^n} {\lambda^{n+1}}$ convergences, then $\frac{x^n}{\lambda^{n+1}}\to 0 $ and therefore this sequence is bounded. Am I right? Please help me. Thanks in advance.

Answer 1

Extended discussion...
(Find a draft of the solution below!)

Query

T.A.E. nicely showed Hadamard's criterion saying that the series: $$\sum_{k=0}^\infty A_k$$ converges for $\limsup_{k\to\infty}\|A_k\|^\frac{1}{k}<1$ and certainly diverges for $\limsup_{k\to\infty}\|A_k\|^\frac{1}{k}>1$.

Now, why does one need bounded linear functionals anyway?

Explanation

The point is basically to establish(!) convergence of the series: $$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}\quad(|\lambda|>r_\sigma(A))$$ This is done by drifting into the realm of complex analysis via bounded linear functionals.

Now, that is why one needs bounded linear functionals anyway!

Moreover, the divergence of the series is not(!) sufficient as there might be another expression for the inverse.

As an analog problem consider the real function: $$f(x):=\frac{1}{x^2+1}$$ That one has a power series expansion around zero with radius one - inside it converges and outside it certainly diverges. It is nevertheless analytic on the whole real line just with new power series expansions: $$f(x)=1-x^2+x^4+\ldots\quad(-1<x<1)$$ $$f(x)=\frac12\mp\frac12(x-1)+\ldots\quad(\pm1-\sqrt{2}<x<\pm1+\sqrt{2})$$ So be careful about drawing conclusions from nonexistent a.k.a. divergent expression!!

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Sketchy solution...

Problem

Denote the spectral radius and the 'Hadamard' radius by: $$r_\sigma(A):=\sup\|\sigma(A)\|$$ $$r_H(A):=\limsup_{n\to\infty}\|A^n\|^\frac{1}{n}$$

The goal is to establish: $$r_\sigma(A)=r_H(A)$$

Strategy

On the one hand for all $|\lambda|>r_H(A)$ the inverse is explicitely given by: $$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}$$ by Hadamard's formula as mentioned above.

Therefore $r_H(A)\geq r_\sigma(A)$.

...

On the other hand for all $|\lambda|>r_\sigma(A)$ the inverse is analytic so as well for all bounded functionals: $$\varphi\in\mathcal{A}':\quad\varphi\left((\lambda-A)^{-1}\right)\text{ analytic}$$

So from complex analysis the Laurent series must converge: $$\varphi\left((\lambda-A)^{-1}\right)=\sum_{n=0}^\infty\frac{\varphi(A^n)}{\lambda^{n+1}}$$

But this means that sequence of summands is bounded for every bounded linear functional separately. By the uniform boundedness theorem this applies without them as well and so: $$\limsup_{n\to\infty}\|A^n\|^\frac{1}{n}<|\lambda|$$

Therefore $r_\sigma(A)\geq r_H(A)$.

Especially, again by Hadamard's formula: $$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}$$

Answer 2

You are correct in your observation about proving boundedness without applying linear functionals. In fact, you can take a direct approach using only vector arguments to prove the existence and value of the radius of convergence of a vector power series:

Theorem (Radius of Convergence): Suppose $X$ is a complex Banach space and consider $$ F(\lambda)=\sum_{n=0}^{\infty}\lambda^{n}x_{n},\;\;\; \{ x_{n}\} \subset X,\;\;\lambda\in\mathbb{C}. $$ Then $F$ converges absolutely (and thus converges in $X$) for all $\lambda$ for which $\limsup_{n}|\lambda|\|x_{n}\|^{1/n} < 1$, and does not converge for any $\lambda$ for which $\limsup_{n}|\lambda|\|x_{n}\|^{1/n} > 1$.

Proof: It is straightforward to show that $F$ converges absolutely if $\limsup_{n}|\lambda|\|x_{n}\|^{1/n} < 1$ because, in such a case, there exists a positive integer $N$ such that $$ \sup_{n \ge N}|\lambda|\|x_{n}\|^{1/n} \le r < 1, $$ which gives $$ \sum_{n=N}^{\infty}|\lambda|^{n}\|x_{n}\| < r^{N}/(1-r) < \infty. $$ Absolute convergence implies convergence because $X$ is complete, which proves that the vector power series converges absolutely for all $\lambda$ for which $\limsup_{n}|\lambda|\|x_{n}\|^{1/n} < 1$.

On the other hand, if the $\limsup$ is greater than $1$, then there exists a subsequence $\{ n_{k}\}_{k=1}^{\infty}$ of the integers for which $|\lambda|\|x_{n_{k}}\|^{1/n_{k}} \ge r > 1$ which prevents the general term of $\sum_{n=0}^{\infty}\lambda^{n}x_{n}$ from converging to $0$ and, hence, also prevents the sum from converging. $\Box$

Once you have this result, then you can consider the Laurent series for $$ f(\lambda) = (x-\lambda 1)^{-1}, $$ where $x$ is an element of a complex unital Banach algebra with unit $1$. The author shows that $f$ is holomorphic for $|\lambda| > r_{\sigma}(x)$, where $r_{\sigma}(x)$ is the spectral radius; and $f$ has the Laurent series expansion $$ f(\lambda) = \sum_{n=0}^{\infty}\frac{1}{\lambda^{n+1}}x^{n},\;\;\;\; |\lambda| > r_{\sigma}(x), $$ and the series cannot converge absolutely for any $\lambda_{0}$ such that $|\lambda_{0}| < r_{\sigma}(x)$ because, if it does, then it is easily verified that the series must converge to $y$ such that $y(x-\lambda 1)=(x-\lambda 1)y = 1$ for all $\lambda$ for which $|\lambda| \ge |\lambda_{0}|$. Hence, $$ r_{\sigma}(x) = \limsup_{n}\|x^{n}\|^{1/n}. $$

Answer 3

Takesaki (and most authors) do this because it is technically simpler. It reduces directly to standard complex analysis. Otherwise, it would be necessary to develop a little of the theory of complex analysis with values in a Banach space and a little integration theory so as to express Cauchy's formula, or use a special method. The extension of complex analysis is very easy by copying standard complex analysis. The integration theory is only easy. The simplest way is to define an integral on step maps from an interval to the Banach space and extend by continuity (the regulated integral). The shortest way is to use a weak integral (the Pettis integral), but that uses exactly the same linear functional reduction that we are trying to avoid. The most complete way is full measure theory going directly to a strong integral (the Bochner integral), (make it more technical to be the most complete).

Rudin's books use both methods. Rudin's books have unusual mixtures of complex and real and functional analysis, so he is ideally placed to use the most natural method without having to take a detour to prove theorems to support it. However, in Real and Complex Analysis, he only develops standard complex analysis and uses the weak method for the spectral radius formula. In Functional Analysis he doesn't develop standard complex analysis, but proves in an early chapter the nontrivial theorem that weakly analytic functions to a Frechet space are strongly analytic (the converse is obvious). Before that he takes 5 pages to develop weak integration for more general spaces. That Cauchy's formula holds in the weak sense is an easy part of the proof. Even the weak integration needs much more than the Hahn-Banach theorem due to its generality.

Then in the proof of the spectral radius formula, Rudin could reasonably use extended complex analysis after saying that it follows in the usual way from the integral formula (his integral is only weak, but estimates made using it are strong). However, Rudin's style is to not motivate any of this, and to give full details of minimal proofs. So instead of proving and/or using the Laurent expansion that follows from Cauchy's formula, he gives minimal details that repeat a little of the proof of the Laurent expansion. Mixing these details makes the proof unnatural and almost twice as long as the proof in Real and Complex Analysis although the latter has fewer prerequisites.

I don't know of any more direct proofs. The magic of analyicity seems to be required to give a radius of convergence. Strong analyticity is obvious for the resolvent function. Standard complex analysis only obviously implies that the the Laurent expansion holds weakly down to the radius. We need to prove that it holds strongly, and it is inelegant to use the harder (?) theorem that this is automatic.