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X is a Banach space and $T \in K(X)/\{0\}$ and following is given (Definitions of point spectrum etc. together with $\{0\}=\sigma(T)$) \begin{align} \{0\}=\sigma(T):=& \mathbb{K}/\{\lambda\in \mathbb{K}|ker(\lambda*Id-T)=\{0\}\cap Im(\lambda*Id-T)=X\}\\ \sigma_p(T):=&\{\lambda\in\sigma(T)|ker(\lambda*Id-T)\neq\{0\}\}\\ \sigma_c(T):=& \mathbb{K}/\{\lambda\in \mathbb{R}|ker(\lambda*Id-T)=\{0\}\cap Im(\lambda*Id-T)\neq X \cap \overline{Im(\lambda*Id-T)} = X\}\\ \sigma_r(T):=& \mathbb{K}/\{\lambda\in \mathbb{R}|ker(\lambda*Id-T)=\{0\}\cap \overline{Im(\lambda*Id-T)}\neq X\} \end{align}

The task now is to find out if there are Banach spaces X and Operators $T \in K(X)/\{0\}$ with

$$ i) \sigma_p(T)= \{0\} $$ $$ii) \sigma_c(T)= \{0\} $$ $$iii) \sigma_r(T)= \{0\} $$

For i) I guess its wrong because $ker(-T) =\{0\}$

I dont know about the other cases though.

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    $\begingroup$ It looks like cases (i)-(iii) are the same! $\endgroup$ – Robert Lewis Jan 6 '18 at 15:45
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    $\begingroup$ Thanks for correction. Its the other spectra of course. $\endgroup$ – SquareJoe Jan 6 '18 at 17:19

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