Justification of discrete spectrum for V(x) unbounded at $\pm \infty$ in Pauling and Wilson In Pauling and Wilson, Introduction to Quantum Mechanics, they offer the following intuitive reason for the discrete spectrum of a potential which is unbounded at $\pm \infty$:



This is interesting but I'm not sure I "buy" it in it's current form. Is there a way to make this argument mathematically precise?
 A: You can think about "solving X equations in Y unknowns". When $X>Y$, you generally expect infinite solutions, when $X=Y$ you generally expect a unique solution, when $X<Y$ you generally expect no solutions. This kind of statement is not always mathematically rigorous but you can usually argue it rigorously in specific circumstances.
Pick an energy $W$ and an arbitrary point $x$. You can pick any two real values for $\psi(x)$ and $\psi'(x)$: You have two continuous adjustable parameters here. Oops, not really. Because the differential equation is homogeneous, you can assume (without loss of generality) that $\psi(x)$ is either 1 or 0. So you actually have only one continuous adjustable parameter. Let's assume from now on that $\psi(x)=1$, because the $\psi(x)=0$ case is actually equivalent to a special case of $\psi(x)=1$ with the slope going to $\pm \infty$.
"Going to zero on the right" is a constraint, so we have one constraint and one adjustable parameter. We generally expect a unique solution. In fact, I think you can rigorously prove that there is a unique solution here, because the asymptotic behavior is related monotonically to $\psi'(x)$.
Similarly, "going to zero on the left" is a constraint, so there is a unique value of $\psi'(x)$ that satisfies this constraint.
Now, we define a function $R(W)$ which is the value of $\psi'(x)$ that makes the solution go to zero on the right at energy $W$; and similarly $L(W)$ on the left. Solutions occur at crossings where $L(W) = R(W)$.
Usually when you draw two 1D curves, they cross each other only at discrete points, rather than, say, perfectly overlapping over a continuous interval.
Can we rigorously prove that you don't have the weird circumstance where $L(W) = R(W)$ over a whole continuous interval of different $W$? Presumably you can prove it somehow, but I'm not sure of the details...
The first thing I would try is to find an expression for the derivatives $L'(W)$ and $R'(W)$, in terms of $V$, and hopefully I would be able to prove that they cannot be equal over a whole interval. Something like that...
