$I_n=\int_0^{\infty} |f_n(t)|^2 t^2 dt < \infty \implies I_n \to \infty $

Question

Let $g_n:[0, \infty)\to \mathbb R$ be a differentiable functions, and $|g_n(n)| \geq 1$ for $n\in \mathbb N.$ Assume that $g_n \neq g_m $ for $n\neq m.$

Put $f_n(t)=g'_n(t)$ for $t\in [0, \infty).$ Assume that $A_n= \{t\in [0, \infty): f_n(t)=0 \} \subset \mathbb N$ for all $n\in \mathbb N,$ and $I_n=\int_0^{\infty} |f_n(t)|^2 t^2 dt < \infty$ for all $n\in \mathbb N.$

Question: Can we say $\{I_n \}_{n\in \mathbb N}$ is unbounded in $\mathbb R$?

Note: For instance, $g_n(x)=e^{-(x-n)^2}$ satisfies the above situation.

Answer 1

The answer is NO.

Let $f(x) = e^{-x^2/2} \sin \pi x$ and $g(x) = 1 + \int_0^x f(s) \, ds$. Set $g_n(x) = g(x - n)$. Then the set of $I_n$ is clearly bounded by $$ \int_{-\infty}^\infty e^{-x^2}(\sin \pi x)^2 x^2 dx = \frac{1}{4} e^{-\pi ^2} \sqrt{\pi } \left(-1+e^{\pi ^2}+2 \pi ^2\right) $$

Correction

For the functions chosen above, the $I_n$ are in fact not bounded. Here is a modification.

Let $f$ be defined as before. For $n \in \mathbb{N}$ let $J_n = \int_0^\infty |f(x-n)|^2 x^2 dx$. Now set $$ f_n(x) = \frac{f(x-n)}{\sqrt{J_n}} \, . $$ Then $\int_0^\infty |f_n(x)|^2 x^2 dx = 1$. And then define $$ g_n(x) = 1 + \int_n^x f_n(s) \, ds \, . $$ That should do it.