Added citation.

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Bokuan Li
2026-05-31 19:30:12 -04:00
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year={1993},
publisher={Taylor \& Francis}
}
@book{FollandHarmonic,
title={A Course in Abstract Harmonic Analysis},
author={Folland, G.B.},
isbn={9781498727150},
series={Textbooks in Mathematics},
url={https://books.google.ca/books?id=z-GYCgAAQBAJ},
year={2016},
publisher={CRC Press}
}

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\label{proposition:spectral-radius-hadamard}
Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$.
\end{proposition}
\begin{proof}
\begin{proof}[Proof, {{\cite[Theorem 1.8]{FollandHarmonic}}}. ]
Let $r = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$.
Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ is the expansion of $R_x$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^*$, then by the preceding discussion, the series $\sum_{n = 0}^\infty \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$,