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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?

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I'm fairly sure that multiplication by $V$ will be self-adjoint if $V\in L^2_\text{loc}$. EDIT: This source seems to corroborate: books.google.com/… –  Alex Youcis Feb 21 '13 at 8:51
    
It says "In conclusion, we note that the condition of local square-integrability can be weakened to the condition that V(x) be finite almost everywhere and measurable".. that seems like basically not a restriction at all. –  Alex Zorn Feb 21 '13 at 17:19

1 Answer 1

up vote 2 down vote accepted

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.

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