# Is Every Symmetric Operator on the Schwartz Space Essentially Self-Adjoint?

More generally, suppose $S$ is a subspace of a Hilbert space $H$ that contains an orthonormal basis of $H$ (For example- the Schwartz space inside $L^2(\mathbb{R}^n)$). If $A:S \rightarrow S$ is symmetric, is $A$ necessarily essentially self-adjoint? That is, does $A$ have a unique self-adjoint extension?

This seems like it would be a standard theorem if it were true, and my inability to find such a statement on, say, Wikipedia, suggests that it is probably false.

This seems elementary- but my first attempts at a proof have not been successful. Maybe somebody knows of a good counterexample?

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Note that by the Gram-Schmidt, every dense subspace $S$ contains an orthonormal basis. Thus here the operator $A$ is just symmetric, densely defined, with $A(S)\subset A$, and the standard counterexamples work (e.g. $i d/dx$ on the half-line). Or am I missing something? – Pietro Majer Nov 5 '12 at 7:50

One needs additional assumptions to obtain essential self-adjointness. In particular, it is sufficient that S contains a dense set of analytic vectors. The whole subject is explained very well in Vol.2 of "Methods of Modern Mathematical Physics" by Reed and Simon.

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Thanks! I'll check it out. – Alex Zorn Nov 5 '12 at 15:44