If f is a closed linear functional defined on a dense subspace of a Banach space X, and, consider f1 which is an extension of f to X, is there a way to show that f1 is also closed without invoking the continuity of f?
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1$\begingroup$ If closed means "with a closed graph", in general $f1$ needs not to be closed: if $f$ is defined on a dense subspace of infinite co-dimension (e.g.$f=0$ there), its extension $f1$ can be whatever, thus also discontinuous, on an algebraic complement of it. $\endgroup$– Pietro MajerCommented Feb 15, 2012 at 11:27
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$\begingroup$ Well "closed" here means a closed map i.e. if xn-->x and f(xn)-->y then if f is closed, y = f(x). I believe, in such a case f1 is necessarily closed. This is because, it can be easily shown that for linear functionals f closed <--> f continuous. So by uniqueness of the hahn-banach extension of f from a dense subspace of X to X itself, f1 is guaranteed to be continuous and as a result , it is closed. But, I don't want to use continuity arguments and want to prove it independently. $\endgroup$– Abhi. ACommented Feb 16, 2012 at 4:44
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