# Ngô Quốc Anh

## September 8, 2012

### CE: Liminf of products

Filed under: Counter-examples, Giải Tích 1 — Ngô Quốc Anh @ 4:46

Today we discuss the inferior limit of the product of two functions. Let us take the following simple question:

Question. Given a function $f:\mathbb R^n \to \mathbb R$ with the following property

$\displaystyle \liminf_{|x| \to +\infty} f(x) \geqslant 0,$

do we always have the following

$\displaystyle \liminf_{|x| \to +\infty} |x|^5f(x) \geqslant 0.$

It turns out that the statement should be hold since $|x|>0$. Unfortunately, since the function $|x|$ blows up of order $5$, the behavior of the product $|x|^5f(x)$ depends on the order of decay of the function $f$. Let take the following counter-example.

We consider the function

$\displaystyle f(x)=-\frac{1}{1+|x|}, \quad x \in \mathbb R^n.$

Although the function $f$ is everywhere negative, there holds

$\displaystyle \liminf_{|x| \to +\infty} f(x) =0.$

Now it is clear that

$\displaystyle \liminf_{|x| \to +\infty} |x|^5f(x)=\liminf_{|x| \to +\infty} \frac{-|x|^5}{1+|x|}=-\infty.$

By definition, one can easily prove that the statement of the question holds if we have

$\displaystyle \liminf_{|x| \to +\infty} f(x)> 0.$