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I have a theorem telling me that some property holds for operators that are bounded and weakly sequentially closed. Somehow, I have in mind that boundedness actually implies the weakly sequentially closedness. My reasoning would be this: As boundedness implies closedness and since convex sets (graph is convex) are weakly sequentially closed if they are closed, the second property should be redundant.

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A bounded operator is closed if and only if its domain is, so your implication "boundedness implies closedness" is not true in general. See here for more details.

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    $\begingroup$ In this case I think OP is taking bounded operators defined on the entire space. $\endgroup$ – s.harp Mar 9 '16 at 16:43
  • $\begingroup$ Maybe. If this is the case, I will delete. I'll wait for OP to tell us. $\endgroup$ – Silvia Ghinassi Mar 9 '16 at 16:45

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