I tried to prove the following:

If $A$ is a unital Banach algebra and $r(a)$ denotes the spectral radius then $r(a^n) = (r(a))^n$.

Could somebody please tell me if I got this proof right? Thanks. Proof:

It follows from the submultiplicativity of the norm that $$ r(a^n) = \lim_{k \to \infty} \|(a^n)^k\|^{{1\over k}} \le \lim_{k \to \infty } \|a^k\|^{{n \over k}} = (r(a))^n$$

Now by contradiction assume that $r(a^n)< r(a)^n$. Then there exists a sequence $\lambda_k$ in $\sigma (a)$ such that $$ \lim_{k \to \infty} |\lambda_k|^n > \sup_{\lambda \in \sigma (a) } |\lambda|^n$$

which is a contradiction. (Since the spectrum is compact so the $\sup$ on the RHS is an element of the spectrum. Also we know that $p(\sigma(a))= \sigma (p(a))$ for polynomials $p$)

I would still appreciate feedback. Thank you.

  • $\begingroup$ The first part is entirely correct. The second part I would like you to tell me why it is a contradiction, and what you really mean. $\endgroup$ – Ukhrir Jun 27 '14 at 12:08
  • $\begingroup$ @Ukhrir Sorry that was a typo. I corrected it now. I also added the explanation you request. $\endgroup$ – Student Jun 27 '14 at 12:12
  • $\begingroup$ You're still using a result relating the spectra of $a^n$ and $a$ without explicitly stating it, that was just what I wanted you to do :). $\endgroup$ – Ukhrir Jun 27 '14 at 12:21
  • $\begingroup$ @Ukhrir I edited it again. Is it this that you meant? $\endgroup$ – Student Jun 27 '14 at 12:42
  • $\begingroup$ @Student: The proof above is kind of a merge of actually three proofs: $p(\sigma(a))=\sigma(p(a))$ and $r(a)=\liminf\|a^k\|^{\frac{1}{k}}$ and $\liminf\|a^k\|^{\frac{1}{k}}=\lim\|a^k\|^{\frac{1}{k}}$ $\endgroup$ – C-Star-W-Star Jun 27 '14 at 15:30

This is not a comment on your proof, but a different (simpler, I think) way to prove the statement.

$$r(a^n) = \lim_{k\rightarrow \infty}||(a^n)^k||^{\frac 1k} = \lim_{k\rightarrow \infty}||a^{nk}||^{\frac 1k} = \lim_{k\rightarrow \infty}(||a^{nk}||^{\frac 1{nk}})^n = (\lim_{k\rightarrow \infty}||a^{nk}||^{\frac 1{nk}})^n = r(a)^n $$

  • $\begingroup$ Thank you. This is very nice, I like it. Though, I am also still interested if my own, longer proof is correct so I will hope for another answer. $\endgroup$ – Student Jun 27 '14 at 13:59
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    $\begingroup$ Sure thing. For what it's worth, your version seems to be correct to me, though the jump from the radii to the spectra is nuanced enough that I will reserve judgement until I check it with a pen and paper in hand (which I have not done yet). $\endgroup$ – Aahz Jun 27 '14 at 14:34

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