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Suppose $T$ is an everywhere defined linear map from a Hilbert space $\mathcal{H}$ to itself. Suppose $T$ is also symmetric so that $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y\in\mathcal{H}$. Prove that $T$ is a bounded directly from the uniform boundedness principle and not the closed graph theorem.

This is problem III.13 in the Reed-Simon volume 1. Hints are welcome.

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For a solution that does use the closed graph theorem, see… – Nate Eldredge May 19 '13 at 23:59
up vote 7 down vote accepted

Hint: Consider the family of linear functionals $f_x$ defined by $f_x(y) = \langle Tx,y \rangle$, as $x$ ranges over the unit ball of $\mathcal{H}$.

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Both proofs are valid for a linear operator $T : X \to X'$ on a Banach space $X$, that is symmetric in the sense that $\langle T x, y \rangle = \langle T y, x \rangle$, $x, y \in X$, where $\langle \cdot, \cdot \rangle$ is the duality pairing.

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