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?

(more…)

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.

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