# Ngô Quốc Anh

## October 15, 2021

### Least upper bound axiom, completeness axiom, and Archimedean property and Cantor’s intersection theorem are equivalent

Filed under: Giải Tích 1 — Ngô Quốc Anh @ 17:03

This note concerns the equivalence between the three properties usually taken as an axiom in synthetic constructions of the real numbers. We start with the least upper bound property, call L, which is usually appeared in construction of the real numbers.

(L, least upper bound axiom): If $A$ is a non-empty subset of $\mathbf R$, and if $A$ has an upper bound, then $A$ has a least upper bound $u$, such that for every upper bound $v$ of $A$, there holds $u \leq v$.

The second property, call C, is the completeness of reals.

(C, completeness axiom): If $X$ and $Y$ are non-empty subsets of $\mathbf R$ with the property $x \leq y$ for any $x \in X$ and $y \in Y$, then there is some $c \in \mathbf R$ such that $x \leq c \leq y$ for any $x \in X$ and $y \in Y$.

The third, also last, property, call AC, is the set of two results: the Archimedean property and Cantor’s intersection theorem. These two results often appear as consequences of the construction of reals.

Archimedean property: For any real numbers $x$ and $y$ with $x>0$, there exists some natural number $n$ such that $nx > y$.

Cantor’s intersection theorem: A decreasing nested sequence of non-empty, closed intervals in $\mathbf R$ has a non-empty intersection.

Our aim is to prove that in fact the above three properties (L), (C), and (AC) are equivalent. Our strategy is to show the following direction:

(L) ⟶ (C) ⟶ (AC) ⟶ (L).

(more…)

## April 14, 2019

### Extending functions between metric spaces: Continuity, uniform continuity, and uniform equicontinuity

Filed under: Giải Tích 3, Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:02

This topic concerns a very classical question: extend of a function $f : X \to Y$ between two metric spaces to obtain a new function $\widetilde f : \overline X \to Y$ enjoying certain properties. I am interested in the following three properties:

• Continuity,
• Uniformly continuity,
• Pointwise equi-continuity, and
• Uniformly equi-continuity.

Throughout this topic, by $X$ and $Y$ we mean metric spaces with metrics $d_X$ and $d_Y$ respectively.

CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function $f : X \to Y$ is only assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function $f$ to obtain a new continuous function $\widetilde f : \overline X \to Y$. The following counter-example demonstrates this:

Let $X = [0,\frac 12 ) \cup (\frac 12, 1]$ and let $f$ be any continuous function on $X$ such that there is a positive gap between $f(\frac 12+)$ and $f(\frac12-)$. For example, we can choose

$\displaystyle f(x)=\begin{cases}x^2&\text{ if } x<\frac 12,\\x^3 & \text{ if } x>\frac 12.\end{cases}$

Since $f$ is monotone increasing, we clearly have

$\displaystyle f(\frac12-)-f(\frac 12+)=\frac18.$

Hence any extension $\widetilde f$ of $f$ cannot be continuous because $\widetilde f$ will be discontinuous at $x =\frac 12$. Thus, we have just shown that continuity is not enough. For this reason, we require $f$ to be uniformly continuous.

## November 4, 2014

### Baire properties for open subspaces

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 7:53

This post deals with a classical problem in functional analysis: The Baire space. I am not going to reproduce what we can learn and read from wikipedia; however, to make the post self-contained, following is what the Baire space is.

Loosely speaking, a Baire space $X$ is a topological space in which any one of the following three equivalent conditions is satisfied:

1. Whenever the union of countably many closed subsets of $X$ has an interior point, then one of the closed subsets must have an interior point, i.e. if

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) \ne \emptyset,$

then $\text{int}(C_n) \ne \emptyset$ for some $n$. Here by $C$ we mean a closed subset in $X$.

2. The union of every countable collection of closed sets with empty interior has empty interior, that is to say, i.e if $\text{int}(C_n) = \emptyset$ for all $n$, then

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) =\emptyset.$

3. Every intersection of countably many dense open sets is dense, i.e.

$\displaystyle \overline{\bigcap_{n \geqslant 1} O_n} = X$

provided $\overline{O_n}= X$ for every $n$. Here by $O$ we mean an open subset in $X$.

What I am going to do is to show that every open subset of a Baire space is itself a Baire space, of course, under the subspace topology inherited from $X$. Hence, at the very beginning, we assume throughout this topic that $X$ is a Baire space; hence admits all three equivalent conditions above.

## April 25, 2013

### The Cauchy formula for repeated integration

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 23:41

The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress $n$ antidifferentiations of a function into a single integral.

Let $f$ be a continuous function on the real line. Then the $n$-th repeated integral of $f$ based at $a$,

$\displaystyle f^{(-n)}(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1,$

is given by single integration

$\displaystyle f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t.$

A proof is given by induction. Since $f$ is continuous, the base case follows from the Fundamental theorem of calculus

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} f^{(-1)}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\,\mathrm{d}t = f(x);$

where

$\displaystyle f^{(-1)}(a) = \int_a^a f(t)\,\mathrm{d}t = 0.$

## 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

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

## March 10, 2012

### An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$

Note that

$\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$

Therefore,

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$

## November 5, 2011

Filed under: Giải Tích 2, Giải Tích 5, Liên Kết — Tags: — Ngô Quốc Anh @ 0:26

This is not mathematics. I just found an equation so that we can draw a heart in 3D. Indeed, the following equation

$\displaystyle {\left( {{x^2} + \frac{9}{4}{y^2} + {z^2} - 1} \right)^3} - {x^2}{z^3} - \frac{9}{{80}}{y^2}{z^3} = 0$

will generate a heart. I have tried and the following pictures show that fact.

## April 22, 2011

### On Costa-Hardy-Rellich inequalities

This note is to concern a recent result by David G. Costa [here]. Here the statement

Theorem 1.1. For all $a,b\in \mathbb R$ and $u \in C^\infty_0(\mathbb R^N\backslash\{0\})$ one has

$\displaystyle\left| {\frac{{N - 2 - \gamma }}{2}\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} } \right| \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}$

where $\gamma=a+b+1$. In addition, if $\gamma \leqslant N-2$, then

$\displaystyle\widehat C\int_{\mathbb R^N} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} \leqslant {\left( {\int_{\mathbb R^N} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{\mathbb R^N} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right)^{\frac{1}{2}}}$

where the constant $\widehat C=|\frac{N+a+b-1}{2}|$ is sharp.

Here’s the proof.

## April 1, 2011

### Several interesting limits from a paper by Chang-Qing-Yang

Recently, I have learnt from my friend, ZJ, the following result

Assume that $F:\mathbb R \to \mathbb R$ is absolutely integrable. Then

$\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}$

The result seems reasonable by the following observation, for example, we consider the first identity when $t \to +\infty$. Then the factor

$\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}$

decays faster then the exponent function $\exp (2t)$. This may be true, of course we need to prove mathematically, because the integrand contains the term $\exp (-2x)$ which turns out to be a good term since $x \geqslant t$. So here is the trick in order to solve such a problem.

## March 1, 2011

### The implicit function theorem: A PDE example

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 23:29

This entry devotes an existence result for the following semilinear elliptic equation

$-\Delta u + u = u^p+f(x)$

in the whole space $\mathbb R^n$ where $0.

Our aim is to apply the implicit function theorem. It is known in the literature that

Theorem (implicit function theorem). Let $X, Y, Z$ be Banach spaces. Let the mapping $f:X\times Y\to Z$ be continuously Fréchet differentiable.

If

$(x_0,y_0)\in X\times Y, \quad F(x_0,y_0) = 0$,

and

$y\mapsto DF(x_0,y_0)(0,y)$

is a Banach space isomorphism from $Y$ onto $Z$, then there exist neighborhoods $U$ of $x_0$ and $V$ of $y_0$ and a Frechet differentiable function $g:U\to V$ such that

$F(x,g(x)) = 0$

and $F(x,y) = 0$ if and only if $y = g(x)$, for all $(x,y)\in U\times V$.

Let us now consider

$X=L^2(\mathbb R^n), \quad Y=H_+^2(\mathbb R^n), \quad Z=L^2(\mathbb R^n)$.

Let us define

$F(f,u)=-\Delta u + u - u^p-f(x), \quad f \in X, \quad u \in Y, \quad x \in \mathbb R^n$.

It is not hard to see that Fréchet derivative of $F$ at $(f,u)$ with respect to $u$ in the direction $v$ is given by

${D_u}F(f,u)v = - \Delta v + v - p{u^{p - 1}}v$.

Since $-\Delta +I$ defines an isomorphism from $Y$ to $Z$, it is clear to see that our PDE is solvable for $f$ small enough in the $X$-norm.

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