Consider the following nonlocal ODEs on $[1,\infty)$.
#1) $$\begin{align} r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\ f(1) &= \alpha \\ \lim_{r\to \infty} f(r) &= 0 \end{align}$$ #2 $$\begin{align} r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\ f'(1) &= \beta \\ \lim_{r\to \infty} f(r) &= 0 \end{align}$$
where $l$ is a positive integer, $\alpha, \beta \in \mathbb{R}$.
Both ODEs are uniquely solvable, and so we can define the Dirichlet to Neumann operator $T: \alpha \mapsto \beta$. I am trying to prove some properties of $T$.
Define the following norm $\lVert \cdot \lVert$: $$\lVert f \lVert^2 := \int_1^{\infty} r^2 f'(r)^2 dr + l(l+1) \int_1^{\infty} f(r)^2 dr$$
It can be shown that $\lVert f \lVert \leq C |f'(1)|$ for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. It then follows that $|f(1)| \leq |\int_1^{\infty} f'(r) dr| \leq C' \sqrt{\int_1^{\infty}r^2 f'^2} \leq C' \lVert f \lVert \leq C'C|f'(1)|$ And so we have for any $f$ solving the above ode, $$|f(1)| \leq \bar C |f'(1)|$$ for some $\bar C$ that is independent of $f$ and $l$.
However, the other way around is not true. In fact, it holds that $$|f'(1)| \leq C \sqrt{l(l+1)}|f(1)|$$ for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. I don't know how to prove this inequality. A weaker inequality is the following, $$\lVert f \lVert \leq C \sqrt{l(l+1)} |f(1)|$$ which I also was not able to prove.
Any help is appreciated.
