is bounded in for each . Hence Banach Steinhauss implies is bounded in .
Hi,
The question: Let X, Y be Banach spaces, be a linear map. Prove that if is a continuous linear functional for every continuous linear functional , then is continuous.
The hypothesis of the problem (that X and Y both be Banach) suggest to me the Open Mapping Theorem, but we don't have T mapping onto Y, and T(X) may not be closed (which we need for it to be complete as a subspace of the complete space Y).
As always your help is much appreciated.
Thanks for your response. I take it that B is the closed unit ball (to have completeness). However, the Banach Steinhaus theorem requires that, for each point, a family of operators all be bounded by some number. Although is bounded by some for each , the supremum of all such may be infinite!
Are there any other suggestions?
Thanks again.
EDIT: Nevermind: it turns out to be a straightforward application of the CGT.
You are right, the result is true even with no completeness. Being weakly bounded is equivalent to be norm bounded, it is a fundamental result in Functional Analysis, valid for locally convex spaces even non barrelled (spaces satisfying Banach Steinhauss). Your hypothesis clearly implies that T(B) is weakly bounded.