[Math] 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: $$(\lambda-A)^{-1}=\sum_{n=0}^\infty\frac{A^n}{\lambda^{n+1}}$$