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Given a (separable) Hilbert space H and an unbounded densely defined linear operator $T:{\cal D}(T) \to $H such that ${\cal D}$ is diagonalizable (it means $\exists$ an O.N.B. of H such that all basis elements are eigenvectors of $T$). Is it possible for $T$ to have non-point spectrum, I mean, can exist a $\lambda \in \sigma(T)$ such that $\lambda$ is not an eigenvalue of $T$?

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Take $T$ to be the inverse of a bounded/continuous, self-adjoint operator with eigenvalues (an orthonormal basis) all rationals between $0$ and $1$. Then $T$ has an orthonormal basis of eigenvectors, with eigenvalues all rationals above $1$. Spectra are closed...

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