Spectrum of Self-adjoint Diagonalizable Operators

Question

Let $L$ be a self-adjoint operator on a Hilbert space $H$. Moreover, assume that $L$ is diagonalizable, that is, $H$ admits an orthonormal basis consisting of eigenvectors of $L$. Is it possible that the spectrum of $L$ contains elements which are not eigenvalues of $L$?

Answer 1

The spectrum contains every accumulation point of the eigenvalues, because the spectrum is closed. If $\{ e_n \}_{n=1}^{\infty}$ is an orthonormal basis of the underlying Hilbert space, and if $\{ r_n \}$ is an enumeration of the rationals in $[0,1]$, then $Lf = \sum_{n=1}^{\infty}r_n \langle f,e_n\rangle e_n$ is bounded, selfadjoint, diagonalizable in your sense, and has spectrum $[0,1]$. In this case $[0,1]\setminus [0,1]\cap\mathbb{Q}$ is an uncountable subset of the spectrum of $L$ that contains no eigenvalue.