When $I_n=\int_0^{\infty} (f_n(x))^2x^2 dx < \infty$ unbounded?

Question

Let $f_n:[0, \infty) \to [0, \infty)$ continuous, and there exists $\{x_n\} \subset [0, \infty)$ such that $|f_n(x_n)| \geq 1$ for all $n\in \mathbb N$ and $x_n \to \infty.$

Assume that $I_n=\int_0^{\infty} (f_n(x))^2x^2 dx < \infty$ for each fixed $n.$

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

Answer 1

You cannot say that $\{I_n\}$ is bounded. Let's build some trouble...

Start with $f_n=e^{-x}$ so that we know $I_n<\infty$. Now, this is not the $f_n$ we'll use because it fails the bump test, $|f_n(x_n)|\geq 1$. So, we need to add some bumps. To each $f_n$ add something like $$ h_n(x) = e^{\frac{-1}{1-|x-n|^2}} \text{ for $x\in [n-1,n+1]$, else 0} $$

i.e. in programming speak $f_n^{new} = f_n^{old}+h_n$. Now everything is good. To make $I_n\to \infty$ just update $f_n$ one more time with $f_n^{new} = n\cdot f_n^{old}$

You also cannot say $I_n$ will be unbounded. Just use the traveling bump $h_n$, you may need to rescale it but it'll work.