# In unital Banach algebra $r(a^n) = (r(a))^n$

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.

• 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. – Ukhrir Jun 27 '14 at 12:08
• @Ukhrir Sorry that was a typo. I corrected it now. I also added the explanation you request. – Student Jun 27 '14 at 12:12
• 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 :). – Ukhrir Jun 27 '14 at 12:21
• @Ukhrir I edited it again. Is it this that you meant? – Student Jun 27 '14 at 12:42
• @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 $\liminf\|a^k\|^{\frac{1}{k}}=\lim\|a^k\|^{\frac{1}{k}}$ – C-Star-W-Star Jun 27 '14 at 15:30

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