This is somewhat unrelated to what I normally do in mathematics, which is why it may be obvious to some of you, but I was puzzled by this:

If we look for classical solutions on $[0,1]$ to

$$-y''(x) = \lambda y(x)$$ with initial conditions $y(0)=1, y'(0)=0$ then the solution is $y_{\lambda}(x)=\cos(\sqrt{\lambda}x).$ Now consider $f(\lambda)=y_{\lambda}(1)$ Now, if we allow $\lambda$ to be complex (which is why I call it $z$ in the following) and always pick the correct branch of the square root, then a plot showed me that we have in fact $ \frac{1}{\left\lvert f(z) \right\rvert} =O\left({\left\lvert \textbf{Im}(z) \right\rvert^{-1}}\right)$ by which I mean that there is a constant $C_R>0$ such that

$\frac{{\left\lvert \textbf{Im}(z) \right\rvert}}{\left\lvert f(z) \right\rvert} \le C_R$ for $z \in \mathbb{C}\backslash \mathbb{R}\cap B(0,R) $ and $B(0,R)$ is an arbitrary ball in the complex numbers.

Numerically, I found some evidence that if $V$ is an even and let's say smooth function with respect to $\frac{1}{2}$ then the same holds for solutions to $-y''(x)+V(x)y(x)=\lambda y(x)$ with the same initial conditions. I am not really sure how to show this theoretically, if it is even true?

Best

Kinzlin