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I'm reading a book functional analysis, and reading and have seen an example of somebody please help me if you can. The example that I've seen is the following:

If $A$ is a normed algebra and $a,b\in A$ so that $ab=ba$, then $$v(a+b)\leq v(a)+v(b),$$ and $$v(ab)\leq v(a)v(b).$$

Remarks: Let $A$ is a normed algebra whith unit. For $x\in A$ number $$v(x)=\inf{\{\vert x^n \vert}^{1/n}: n\in\mathbb{N}\}$$ is spectral radius of element $x$.

Please help me. Thanks for your attention.

Madrit Zhaku
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  • Shouldn't it be $v(ab)\geq v(a)v(b)$? – Sungjin Kim Oct 23 '13 at 17:43
  • no sir should not be so, at least as I wrote the book I wrote example $v(ab)≤v(a)v(b)$ – Madrit Zhaku Oct 23 '13 at 17:45
  • I see, I was confused with other thing. – Sungjin Kim Oct 23 '13 at 17:52
  • First show that $\nu(x) = \lim\limits_{n\to\infty} \lvert x^n\rvert^{1/n}$. Then $\nu(ab) \leqslant \nu(a)\nu(b)$ for commuting $a,b$ is straightforward. For $\nu(a+b) \leqslant \nu(a)+\nu(b)$, I don't see how to prove it without a bit of effort from the limit property, but it is doable. – Daniel Fischer Oct 23 '13 at 17:52
  • Mr. seemed typical example, maybe I was wrong and so I will pray that if you have the opportunity to help me complete solution to the example given, much will thank you – Madrit Zhaku Oct 23 '13 at 17:56

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Denote $v_n(a)=|a^n|^{1/n}$ and $w_n(a)=\ln |a^n|$. It is easy to check that $w_{k+l}(a)\leq w_k(a)+w_l(a)$. By Fekete's lemma we get there exist the limit $$ \lim\limits_{n\to\infty}\frac{w_n(a)}{n}=\inf\limits_{n\in\mathbb{N}}\frac{w_n(a)}{n} $$ Since $\ln v_n(a)=w_n(a)/n$, then from previous equality we get $$ v(a) =\inf_{n\in\mathbb{N}} v_n(a) =\lim\limits_{n\to\infty} v_n(a) $$

Note that $$ v_n(ab) =|(ab)^n|^{1/n} =|a^nb^n|^{1/n} \leq(|a^n||b^n|)^{1/n} =|a^n|^{1/n}|b^n|^{1/n} =v_n(a)v_n(b) $$ then we take limits as $n\to\infty$ and get $$ v(ab)\leq v(a)v(b)\tag{1} $$

We will need one more property of $v$ function. For any $\lambda\in\mathbb{C}$ we have $$ v_n(\lambda a) =|(\lambda a)^n|^{1/n} =|\lambda ^n a^n|^{1/n} =(|\lambda|^n |a^n|)^{1/n} =|\lambda| |a^n|^{1/n} =|\lambda|v_n(a) $$ hence after taking the limits we see that $$ v(\lambda a)=|\lambda| v(a)\tag{2} $$

To prove $(2)$ consider arbitrary $s>v(a)$ and $t>v(b)$. Then define $c=s^{-1}a$ and $d=t^{-1}b$. From $(2)$ we know that $v(c)<1$ and $v(d)<1$. From definition of $v$ we derive that there exists $M\in\mathbb{N}$ such that $|c^n|^{1/n}<1$ and $|d^n|^{1/n}<1$ for all $n>M$. Hence there exist $\alpha>0$ such that $|a^n|<\alpha$, $|b^n|<\alpha$ for all $n\in\mathbb{N}$. Now we have estimation $$ |(a+b)^n| =\left|\sum\limits_{k=0}^n{n \choose k}a^k b^{n-k}\right| \leq\sum\limits_{k=0}^n{n \choose k}|a^k| |b^{n-k}| =\sum\limits_{k=0}^n{n \choose k}s^k t^{n-k}|c^k| |c^{n-k}|\\ \leq \alpha^2\sum\limits_{k=0}^n{n \choose k}s^k t^{n-k} =\alpha^2(s+t)^n $$ Note that binomial formula is valid because $ab=ba$. Therefore $$ v(a+b) =\lim\limits_{n\to\infty}|(a+b)^n|^{1/n} \leq\lim\limits_{n\to\infty}\alpha^{2/n}(s+t) =s+t $$ Since $s>v(a)$ and $t>v(b)$ are arbitrary, then $$ v(a+b)\leq v(a)+v(b)\tag{3} $$

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Norbert
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