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Let $A$ be an operator on a Banach space, possibly unbounded, such that its resolvent $(\lambda - A)^{-1}$ is compact. Is $A$ then a Fredholm operator of index 0? My feeling is yes but I cannot prove it.

Thank you for the help.

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    Why do you feel this should be true in the case where $A$ is bounded? My *feeling* is that your question would make more sense if $A$ is densely-defined, closed and unbounded -- is that the particular case you had in mind2011-10-30
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    Also, I don't know what it means for an unbounded densely-defined operator to be Fredholm. What was the definition you had in mind?2011-10-30

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