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August 19th, 2013, 12:41 PM  #1 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Routine proof in Functional Analysis
I have a rather routine problem, but cannot quite get a finger on it. I need to prove that any selfadjoint linear map defined everywhere on a Hilbert space is bounded. I understand that either the closed graph theorem, or the uniform boundedness principle may be adequately applied; but I have blanked out. Please any clearly stated proof is more than welcome.

August 21st, 2013, 10:57 AM  #2 
Senior Member Joined: Dec 2012 Posts: 372 Thanks: 2  Re: Routine proof in Functional Analysis
Ok, so I have deviced a proof using neither of the two theorems. Given such that for all y in H with being linear and selfadjoint, we will verify that is continuous at zero. With this result we have also that is continuous at zero, giving us continuity of A on H; as a consequence of A being linear. It suffices to check that for any sequence in H which converges weakly to zero, we have also converging weakly to A(0) = 0. Given that converges weakly to zero, then for any element , we have that converges to T(0) = 0. In particular, for , , using the selfadjoint property of A. This is to say that converges to 0 in the topology of which coincides with the weak topology on since H is a Hilbert space. As such, we have the desired result using the weak topology on the domain and codomain, which in turn gives us continuity in the normed sense as initially proposed. I am not satisfied though. I still hope for some analysis gurus to someday give proofs using the closed graph theorem and uniform boundedness principle. 

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analysis, functional, proof, routine 
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