# Spectral radius for linear compact maps

Prove or disprove the following assertions for a linear map $C$ from a Banach space $X$ into itself:

a) If C is compact then its spectral radius equals the maximum of the absolute value of $C$

I'm not sure what the definition of absolute value of an operator is. But the spectral value is defined as $|\sigma(M)| = max_{\lambda \in \sigma(M)} |\lambda|$

edit: I split it up to two questions

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I can't make sense of the question, and you also don't know what it means. Where did it come from? – Jonas Meyer Dec 12 '12 at 15:23
An old exam, maybe there was a miss print. – Johan Dec 12 '12 at 16:09
If the absolute value of $C$ is the norm of $C$, then the answer is no. It becomes YES though, assuming further that $C$ is self-adjoint. – Yiorgos S. Smyrlis Dec 30 '15 at 10:40

This is false, consider Volterra operator on $C[0,1]$. Radius $0$ but $\mid Cf \mid \ge 1$, just take $\chi_{[0,1]}$

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If your Banach space is $\mathbb R^2$ and $C=\left(\begin{matrix}0&1\\ 0&0 \end{matrix}\right)$, then $\lvert C\rvert=1$, while its spectral radius is $0$.

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