if $f$ is such that : $\int_{-\infty}^{\infty} x^2f(x) \mathrm{d}x \leq M$ then $f = O(x^3)$

Question

I am wondering if the following is true :

Let $f$ be a positive piecewise continuous function such that : $$\int_{-\infty}^{\infty} x^2f(x) \mathrm{d}x \leq M$$ Then can we conclude that $f = O(x^{-3})$ at $\pm \infty$ ?

I tried to see if a proof by contradiction works here yet what I am doing doesn't lead anywhere :

Suppose there is a stricly increasing sequence $(x_n)$ such that : $f(x_n) \geq \frac{1}{x_n^2}$. But then what to do ? Since we can just say this on single point it's impossible to integrate (since it's discrete)..

Thank you !

Answer 1

The answer is no. Let $f(x)=1/n^2$ in $[n,n+1/n^a]$ for $n\in\mathbb{N}^+$ and zero otherwise then $$\int_{-\infty}^{\infty} x^2f(x)\,dx =\sum_{n=1}^{\infty} \frac{1}{n^2}\int_n^{n+1/n^a}x^2dx=\frac{1}{3}\sum_{n=1}^{\infty} \frac{1}{n^2} ((n+1/n^a)^3-n^3)\\ =\frac{1}{3}\sum_{n=1}^{\infty} \frac{1}{n^2}\cdot \left(\frac{3n^2}{n^a}+\frac{3n}{n^{2a}}+\frac{1}{n^{3a}}\right)$$ It follows that for $a>1$ the RHS is convergent.

How can you modify $f$ in order to have a positive piecewise continuous function?

Answer 2

Define $f$ to be $0$ everywhere except that on $[n^{10},n^{10}+n^{-100}]$ it takes value $n^{40}$ (where $n\in \mathbb{N}$). This is piecewise constant, so in particular piecewise continuous, and we have $$ \int_0^\infty x^2 f(x)\, dx < \sum_{n=1}^\infty (n^{10}+n^{-100})^2 n^{-60} < 1. $$ However if $x=n^{10}$ then $f(x) = n^{40}$, which is not bounded by a constant times $x^3 = n^{30}$.

Answer 3

If you assume that $f\in C^1$ and that $$\int_{-\infty}^\infty |f'(y)|\, dy<\infty, $$ then it is almost true. Indeed, with these assumptions the limit $$ \lim_{|x|\to \infty} f(x) $$ must exist, and so the limit of $x^2f(x)$ exists also, perhaps infinite. However, for the integral $$ \int_{-\infty}^\infty x^2 f(x)\, dx $$ to be finite, it is necessary that $$ \lim_{|x|\to \infty} x^2 f(x)=0, $$ that is, $f(x)=o(|x|^{-2})$. (You wanted $f(x)=O(|x|^{-3})$ which is slightly more than this).