My Math Forum Routine proof in Functional Analysis

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 August 19th, 2013, 12:37 PM #1 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 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.
 August 26th, 2013, 04:01 AM #2 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: Routine proof in Functional Analysis Indeed, closed graph theorem is a good way. Let $(x_n)$ be a sequence in $H$ converging to  and such that $Sx_n\to y$, $y\in H$. Then for each $n$, $\langle y,Sx_n\rangle=\langle Ty,x_n\rangle$. The LHS converges to $|| y||^2$ while the RHS goes to .
 August 26th, 2013, 11:50 AM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 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.
 August 27th, 2013, 07:27 AM #4 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: Routine proof in Functional Analysis AfroMike, I'm probably misunderstanding your request, but the sketch of proof did use closed graph theorem.
 August 28th, 2013, 10:00 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Routine proof in Functional Analysis We need to show that for any sequence in the graph of $T$ which converges $(x_n , T(x_n)) \rightarrow (x , y)$ where $x_n \in H \ \forall n$, then $y= Tx$ where $T$ is the self adjoint operator. You have established clearly that if $(x_n , T(x_n)) \rightarrow (0 , y)$ then $y= T(0) = 0$. For other $x \in H$ 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?
 August 28th, 2013, 11:33 AM #6 Member   Joined: Nov 2009 From: France Posts: 98 Thanks: 0 Re: Routine proof in Functional Analysis Let $x_n\to x$ and $Tx_n\to y$. Define $x'_n:=x_n-x$. Then $Tx'_n=Tx_n-Tx\to y-Tx=0$ by the case $x=0$. Hence $y=Tx$.
 August 28th, 2013, 12:23 PM #7 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Routine proof in Functional Analysis Ok, I guess its also correct to make the conclusion that since $x_n \rightarrow x$ and $Tx_n \rightarrow y$ then $\langle x_n , y \rangle - \langle Tx_n , x \rangle \rightarrow 0$. Under this assumption, we also have the desired result.

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