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.

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