# Construct a multiplication operator which has dense point spectrum

By a multiplication operator here we mean an operator

$$Af(t)=m(t)f(t), \qquad f \in D(A)=\{x \in L^2(\mathbb{R} \mid m(t)f(t) \in L^2(\mathbb{R})\}$$

where $m$ is a Borel measurable function on the line. Motivated by Exercise 3.13, pag.104 of Teschl's book (link) I'm trying to construct such an operator having all rationals as eigenvalues.

This amounts to find a Borel measurable function $m$ such that $m^{-1}(\lambda)$ has positive measure for all rational $\lambda$. How to do that?

I find constructing an abstract linear operator on some separable Hilbert space $H$ with $\mathbb{Q}$ as point spectrum not too difficult: take an orthonormal basis $\{e_n\}$ and an enumeration $\{r_n\}$ of $\mathbb{Q}$ then define

$$Af=\sum_{n=0}^\infty r_n(f, e_n)e_n, \qquad f \in D(A)=\left\{g \in H \mid \sum_{n=0}^\infty r_n^2\left\lvert(g, e_n)\right\rvert^2 < \infty\right\}.$$

But how to go from this to a function $m$ as prescribed above?

Thank you.

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It's late here, but if you're just looking for such a Borel function $m$, partition your domain (the reals) into countably many pairwise disjoint Borel sets $A_n$ of positive measure and put $m = \sum_n \lambda_n [A_n]$ where $\{\lambda_n\}_n$ is an enumeration of the rationals and $[A_n]$ is the characteristic function of $A_n$. – t.b. Sep 11 '11 at 1:19
@Theo: Thank you very much! This is a bit embarrassing, though: I should have thought a bit more by myself - it wasn't difficult. For example, choosing $A_n=(n, n+1]$, the function $m$ you construct above is a step function taking all rational values. The corresponding multiplication operator is similar to the operator $A$ of my post if we take $e_n(t)=[(n, n+1]](t)$, which indeed is an orthonormal system of $L^2(\mathbb{R})$ (it is not complete, but that doesn't matter).  Now it is perfectly clear, next time I will try to think twice before posting a question! Thank you again. – Giuseppe Negro Sep 11 '11 at 10:01