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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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    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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