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This question arose from the proof of proposition $1.11(e)$ in chapter $8$ of John B. Conway's A Course in Functional Analysis. This portion of the proposition can be stated:

Let $\mathscr{A}$ be a $C^*$-algebra, and let $a\in\mathscr{A}$ be given. If $a=a^*$, then $\|a\|=r(a)$.

(Here, $r(a)$ denotes the spectral radius of $a$.)

The proof, as stated in the book, proceeds as follows:

Since $a^*=a$, $\|a^2\|=\|a^*a\|=\|a\|^2$; by induction, $\|a^{2n}\|=\|a\|^{2n}$ for $n\geq1$ That is, $\|a^{2n}\|^{1/2n}=\|a\|$ for $n\geq1$. Hence $r(a)=\lim\|a^{2n}\|^{1/2n}=\|a\|$.

Now I was able to show by induction that $$ \|a^{2^n}\|=\|a\|^{2^n} \qquad (n\geq1),$$ from which the result follows, but I could not prove it as it is stated in the book.

So my question is: How can we prove (presumably by induction) that $\|a^{2n}\|^{1/2n}=\|a\|$ for $n\geq1$? Is this simply an error in the book, or can it be done?

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  • $\begingroup$ I suppose if someone can point me to an errata for the book with this being mentioned, that would be an acceptable answer. $\endgroup$ – Aweygan Aug 21 '16 at 2:25
  • $\begingroup$ Seems likely to me this is just an error. $\endgroup$ – Eric Wofsey Aug 21 '16 at 2:27
  • $\begingroup$ @EricWofsey That's what I was thinking, as it would be a small error in typesetting. But still I am curious. $\endgroup$ – Aweygan Aug 21 '16 at 2:31
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This was corrected in a later edition ...

enter image description here..

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  • $\begingroup$ Thanks, that's what I get for being cheap and buying used and previous editions of books. $\endgroup$ – Aweygan Aug 21 '16 at 3:34
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It is definitely a typo. The proof of $\|a^{2n}\|=\|a\|^{2n}$ cannot be trivial, since for instance it implies that $\|a^3\|=\|a\|^3$ (which I don't think can be easily obtained from the axioms): $$ \|a^3\|^2=\|a^{2\times3}\|=\|a\|^{2\times3}=(\|a\|^3)^2. $$ So, even if one can find an argument for the formula for $2n$, it is not worth it for Conway's proof, as any subsequence is good enough.

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  • $\begingroup$ Thanks, I do believe I remember seeing that before as well. $\endgroup$ – Aweygan Aug 21 '16 at 3:33

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