For $p$, $f$ sufficiently differentiable functions: \begin{equation} \int_{-\infty}^{\infty} f^2 p = 1 \qquad \text{(i.e. $f \in L^2_p$)} \end{equation}
I want to conclude $\lim_{x\to\pm\infty} f^2 p = 0$. Is it possible to conclude this from the following properties:
- $p > 0$, $\int_{-\infty}^{\infty} p = 1$, i.e. $p \in L^1$ is a probability density function
- $\int_{-\infty}^{\infty} fp = 0$
- $\lim_{x\to\pm\infty} p = 0$
- $\lim_{x\to\pm\infty} p' = 0$ but $\lim_{x\to\pm\infty} \frac{p'}{p}$ possibly unbounded
- $\lim_{x\to\pm\infty} f$ exists, possibly infinite
- $f$ strictly monotonic near the boundary, i.e. for $x < x_0$ or $x_1 < x$ for some finite $x_0 < x_1$
- $f$ and $pf'$ are locally absolutely continuous
If not, is there an additional condition on $p$ to allow this conclusion? It should be okay to assume $p$ is strictly monotonic in a neighborhood of the boundary, like it is noted for $f$. Would it help if this would hold for $f'$ and/or $p'$? Then $f$ (and $p$) would be convex (or concave) in a neighborhood of the boundary.
I tried to apply Barbalat's lemma but I don't see how to conclude that $f^2 p$ would be uniformly continuous, because the domain is not compact. Given the limit exists, $p$ is uniformly continuous but this does not imply uniform continuity for the product on a non-compact domain. Looking at $(pf^2)' = p' f^2 + 2pff'$ does not seem to allow a conclusion as not enough is known about each summand's behaviour. I wonder if it's possible to conclude something from $f$ being locally absolutely continuous.