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

## September 4, 2010

### CE: Convergence of improper integrals does not imply the integrands go to zero

Filed under: Counter-examples, Giải Tích 3 — Ngô Quốc Anh @ 12:31

This entry devotes a similar question that raises during a course of series. We all know that for a convergent series of (positive) real number $\displaystyle \sum_{n=1}^\infty a_n$

it is necessary to have $\displaystyle\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$.

This is the so-called $n$-th term test. A natural extension is the following question

Question. Suppose $f(x)$ is positive on $[0,\infty)$ and $\displaystyle\int_0^{ + \infty } {f(x)dx}$

exists. Must $f(x)$ tend to zero as $x \to +\infty$?

## February 9, 2010

### CE: Integral calculus

Question: If the integral $\displaystyle \int_0^\infty f (x)dx$

converges and a function $y = g(x)$ is bounded then the integral $\displaystyle \int_0^\infty f (x)g(x)dx$

converges.

Observation: It seems since $g(x)$ is bounded by a constant called $M$, the integral $\int_0^\infty f (x)g(x)dx$ is then dominated by $M$ times $\int_0^\infty f (x)dx$ which becomes a finite number.

Counter-example: The integral $\displaystyle \int_0^\infty \frac{\sin x}{x}dx$

converges and the function $g(x)=\sin x$ is bounded but the integral $\displaystyle \int_0^\infty \frac{\sin^2 x}{x}dx$

diverges.

Explanation: What we thought is the following estimate $\displaystyle\left| {\int_0^\infty {f(x)g(x)dx} } \right| \leqslant \int_0^\infty {\left| {f(x)g(x)} \right|dx} \leqslant M\int_0^\infty {|f(x)|dx}$.

However, the convergence of $\int_0^\infty f (x)dx$ is not sufficient to imply that $\int_0^\infty |f (x)|dx<\infty$.