If $H$ is a Hilbert space and $A:H \rightarrow H$ is a linear operator such that $\langle Ax, y\rangle = \langle x, Ay\rangle, \forall x, y \in H$ then how to show $A$ is a bounded linear operator?
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3Use Banach Steinhaus. – copper.hat Aug 07 '16 at 06:14
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4Does this answer your question? Hellinger-Toeplitz theorem use principle of uniform boundedness – ViktorStein Mar 07 '20 at 00:34
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Solution 1: Use Closed Graph Theorem like Robert Israel suggested.
Solution 2: Note that if $\|x\| = 1$ then for all $y \in H$ we have that $|\langle Ax, y \rangle| = |\langle x, Ay \rangle| $ $\leq \|x\| \cdot \|Ay\| = \|Ay\|$. Hence by the uniform boundedness principle we see that $\sup_{\|x\|=1} \| \langle Ax, \cdot \rangle\| < \infty$. Use this as a hint to proceed.
shalin
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Hints:
- Adjoint operators are closed.
- Closed graph theorem.
EDIT: BTW, this is called the Hellinger-Toeplitz theorem.
Robert Israel
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A proof using the closed graph theorem can be found e.g. here. Why do we need the closeness of the adjoint? Or are those two ideas for two proofs? – ViktorStein Mar 07 '20 at 00:36