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Instead of directing Ackbach to reread the question, can you demonstrate why his counterexample does not address your question?Please read the question again.
$||T||=1$ means that $||Tv|| \le ||v||$ for all $v$.Let T be a bounded linear operator on X and x be in X. Assume ||x||>||T||=1 It follows that ||Tx||>1. Can anyone explain why 0 is not in the closure of T(B), where B is the ball centered at x, with radius 1. Thanks