maximum between two successive zeros of Sturm-Liouville problem 
Let $x_1,x_2,\cdots $ be zeros of $y"+q(x)y=0$ arranging in increasing order, where $q(x)>0$ and continuously increasing in $[x_1,\infty)$.
Let $b_n=\max_{x_n\le x\le x_{n+1}} |y(x)|$. Prove that $b_n$ is strictly decreasing and if $q(x)\to C$ for some constant $C$ when $x\to \infty$, then there exists a constant $B$ such that $b_n\to B>0$ when $x\to \infty$.

I have used Sturm comparison theorem to show that $x_{n+1}-x_n$ is decreasing with $n$ and tends to $\pi/\sqrt{C}$, but I don't know how to deal with function between those zeros. Thanks for any help.
 A: Denote $\xi_k$ the root of $y'$ in $(x_k,x_{k+1})$, so that $b_k=|y(\xi_k)$ and $q_k=q(x_k)\le q(x)$ for $x>x_k$. Then consider
$$
\frac12\frac{d}{dt}(y'^2+q_ky^2)=y'(y''+q_ky)=-yy'(q-q_k)
$$
Assuming $\xi_{k-1}$ is a maximum and $\xi_k$ a minimum, the right side is non-positive for $x\in[\xi_{k-1},\xi_k]$ since
\begin{array}{l|ccc|c}
x&y(x)&y'(x)&(q(x)-q_k)&-yy'(q-q_k)\\\hline
\in[\xi_{k-1},   x_k) &>0 &<0 &\le 0 &  \le 0  \\
\in[      x_k, \xi_k] &<0 &<0 &\ge 0 &  \le 0
\end{array}
A similar analysis holds in the other case.
Thus $(y'^2+q_ky^2)$ is falling on the interval $[\xi_{k-1}, \xi_k]$ and thus
$$
0^2+q_ky(\xi_{k-1})^2\ge 0^2+q_ky(\xi_{k})^2.
$$
Using $q_{k-1}\le q(x)\le q_k$ on $[\xi_{k-1},   x_k)$ and $q_{k}\le q(x)\le q_{k+1}$ on $[      x_k, \xi_k]$ allows to give lower bounds for the right side and thus upper bounds for the decrease in the amplitudes,
$$
0\ge q_k(b_k^2-b_{k-1}^2)\ge -2\int_{\xi_{k-1}}^{x_k}(q_{k-1}-q_k)yy'dx-2\int_{x_k}^{\xi_{k}}(q_{k+1}-q_k)yy'dx
\\
=-(q_k-q_{k-1})b_{k-1}^2-(q_{k+1}-q_k)b_k^2
$$
so that finally
$$
b_{k-1}^2\ge b_k^2\ge\frac{q_{k-1}}{q_{k+1}}b_{k-1}^2
$$
