My Math Forum Routine proof in Functional Analysis

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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 self-adjoint 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 $A : H \rightarrow H$ such that $A(y) \in H$ for all y in H with $A$ being linear and self-adjoint, we will verify that $A : (H, w) \rightarrow (H, w)$ is continuous at zero. With this result we have also that $A : (H, s) \rightarrow (H, s)$ 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 $(x_n)_n$ in H which converges weakly to zero, we have also $(A x_n)_n$ converging weakly to A(0) = 0. Given that $(x_n)_n$ converges weakly to zero, then for any element $T \in H^*= H$, we have that $\langle T , x_n \rangle$ converges to T(0) = 0. In particular, for $Ay \in H^*$, $\langle Ay , x_n \rangle \ \rightarrow \ 0 \ \Rightarrow \\ \langle y, A x_n \rangle \ \rightarrow \ 0 \ \forall y \in H$, using the self-adjoint property of A. This is to say that $(A x_n)_n$ converges to 0 in the $w^*$ topology of $H^*$ which coincides with the weak topology on $H^*$ since H is a Hilbert space. As such, we have the desired result using the weak topology on the domain and co-domain, 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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