Limit-circle and limit-point at endpoints

I was wondering if the following holds:

If you have an ODE $$-y''(x) + q(x) y(x) = \lambda y(x)$$ on a finite interval $(a,b)$ and you know that this equation is limit-circle or limit-point at the end-points.

If you now add a nice smooth + bounded -potential $V \in C^{\infty}(\mathbb{R})$ to your current potential, so that you end up with the ODE

$$-y''(x) + (q(x)+V(x)) y(x) = \lambda y(x),$$

is it still clear that your differential equation is limit-circle or limit-point at the endpoints?

I mean, somehow I feel that this statement should hold, as it is somehow natural to assume that a nice potential should keep the nice properties of the operator, but I could not find a reference for this.

• @TobiasHurth: $V$ bounded is definitely sufficient; I had $V\in C^{\infty}(a,b)$ (open!) in mind. I in fact wasn't sure about the exact situation the OP had in mind as both $(a,b)$ and $\mathbb R$ are mentioned as intervals. – Christian Remling Oct 21 '14 at 18:02
• @TobiasHurth: Yes, I think such an argument should also work. (One inconvenience is that one doesn't really have precise information on what the solutions look like.) – Christian Remling Oct 21 '14 at 18:12

The LP/LC classification is certainly not affected by bounded perturbations. One easy argument is to observe that a bounded $V$ is also bounded as an operator, and the deficiency indices are stable under bounded perturbations.
• @Fabiano: Yes, this is right; also, LP/LC is a local issue, so I can cut the interval into two halves and then apply the argument (this shows that there is no swapping of LP/LC between endpoints in the $(1,1)$ case). – Christian Remling Oct 22 '14 at 17:58