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August 19th, 2013, 12:37 PM   #1
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Routine proof in Functional Analysis

This problem is quite routine, but I am not quite getting a finger on it. I need to prove that any self-adjoint linear map defined everywhere on a Hilbert space is bounded. I understand that either the closed graph theorem or uniform boundedness principle can be adequately applied, but I have blanked out. Please, any clearly stated proof is most welcome.
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August 26th, 2013, 04:01 AM   #2
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Re: Routine proof in Functional Analysis

Indeed, closed graph theorem is a good way. Let be a sequence in converging to and such that , . Then for each , . The LHS converges to while the RHS goes to .
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August 26th, 2013, 11:50 AM   #3
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Re: Routine proof in Functional Analysis

Yes, Girdav, I see you have indicated continuity at 0 which in turn gives us continuity everywhere. If you can obtain and post the proof using either the closed graph or uniform boundedness principle, I'll be more than grateful.
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August 27th, 2013, 07:27 AM   #4
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Re: Routine proof in Functional Analysis

AfroMike, I'm probably misunderstanding your request, but the sketch of proof did use closed graph theorem.
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August 28th, 2013, 10:00 AM   #5
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Re: Routine proof in Functional Analysis

We need to show that for any sequence in the graph of which converges where , then where is the self adjoint operator. You have established clearly that if then . For other the result is not so clear. This is the reasoning I have to prove using the closed graph theorem; perhaps you have a more concise approach?
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August 28th, 2013, 11:33 AM   #6
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Re: Routine proof in Functional Analysis

Let and . Define . Then by the case . Hence .
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August 28th, 2013, 12:23 PM   #7
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Re: Routine proof in Functional Analysis

Ok, I guess its also correct to make the conclusion that since and then .
Under this assumption, we also have the desired result.
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