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 $T$?

## 1 Answer

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...