There are a handful of results establishing conditions on the measurable real-valued function $V(x)$ under which the operator:
$$-\Delta + V(x)$$
Is (essentially) self-adjoint on $L^2(\mathbb{R}^n)$. My question is: Is the operator $V(x)$ by itself always self-adjoint? My instinct is no: Something sufficiently ugly like:
$$V(x) = \frac{1}{\sin(1/x)}$$
With infinitely many singularities near the origin ought to do it. And, if it is not true that multiplication by all measurable functions is self-adjoint, is there a relatively simple necessary and sufficient condition? Being integrable on every compact set seems sufficient but perhaps too strong.
Thoughts?
Weidmann: Linear Operators in Hilbert spaces
Example 1. (page 51) is where the multiplication operator is defined. Example 2 on page 91 shows that for an arbitrary measurable $V$ you always get a (densely defined) selfadjoint (but not necessarily bounded) operator. If you want it a bounded operator, you need it essentially bounded.
Summarizing, there is no restriction at all.