# 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…)

## 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.

## 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

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

## 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.

## January 8, 2011

### A funny limit involving sine function

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Ngô Quốc Anh @ 2:32

Today, I have been asked to calculate the following limit

$\displaystyle \mathop {\lim }\limits_{n \to + \infty } \sin (\sin \overbrace {(...(}^n\sin x)...))$

for each fixed $x \in [0,2\pi]$. From the mathematical point of view, we can assume $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ as we just replace $x$ by $\sin (\sin x))$ if necessary.

There are three possible cases

Case 1. $x \in (0, \frac{\pi}{2})$. In this case, it is well known that function $\frac{\sin x}{x}$ is monotone decreasing since

$\displaystyle {\left( {\frac{{\sin x}}{x}} \right)^\prime } = \frac{{x\cos x - \sin x}}{{{x^2}}} = \frac{{\cos x}}{{{x^2}}}\left( {x - \tan x} \right) \leqslant 0$

in its domain. Consequently, it holds

## October 3, 2010

### An identity of differentiation involving the Kelvin transform, 2

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 11:36

I found the following interesting identity which is similar to what I have showed recent days [here]. In that entry, we showed that

$\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y$

where $x$ and $y$ are connected by

$\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}$.

For $\lambda>0$ we denote by $y$ the following

$\displaystyle y = \frac{\lambda^2 x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^n}$.

Then we show that

Lemma.

$\displaystyle {\left| x \right|^2}{\Delta _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) - (n - 2)x \cdot {\nabla _x}u\left( {\frac{{{\lambda ^2}x}}{{{{\left| x \right|}^2}}}} \right) = {\left| y \right|^2}{\Delta _y}u\left( y \right) - (n - 2)y \cdot {\nabla _y}u\left( y \right)$.

## September 22, 2010

### An identity of differentiation involving the Kelvin transform

Filed under: Các Bài Tập Nhỏ, Giải Tích 1 — Tags: — Ngô Quốc Anh @ 15:47

This short note is to prove the following

$\displaystyle {\nabla _x}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right) \cdot x = - {\nabla _y}\left( {u\left( y \right)} \right) \cdot y$

where $x$ and $y$ are connected by

$\displaystyle y = \frac{x}{{{{\left| x \right|}^2}}} \in {\mathbb{R}^2}$.

The proof is straightforward as follows.

• Calculation of $\frac{\partial}{\partial x_1}$.

We see that

$\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_1}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_1} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_1}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_1} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_1^2}}{{{{\left| x \right|}^4}}}} \right){x_1} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_1}. \hfill \\ \end{gathered}$

• Calculation of $\frac{\partial}{\partial x_2}$.

Similarly, we get

$\displaystyle\begin{gathered} \frac{\partial }{{\partial {x_2}}}\left( {u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)} \right){x_2} = \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_1}}}{{{{\left| x \right|}^2}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\frac{\partial }{{\partial {x_2}}}\left( {\frac{{{x_2}}}{{{{\left| x \right|}^2}}}} \right){x_2} \hfill \\ \qquad\qquad\qquad= \frac{\partial }{{\partial {y_1}}}\left( {u\left( y \right)} \right)\left( { - \frac{{2{x_1}{x_2}}}{{{{\left| x \right|}^4}}}} \right){x_2} + \frac{\partial }{{\partial {y_2}}}\left( {u\left( y \right)} \right)\left( {\frac{1}{{{{\left| x \right|}^2}}} - \frac{{2x_2^2}}{{{{\left| x \right|}^4}}}} \right){x_2}. \hfill \\ \end{gathered}$

## June 25, 2010

### Some operations on the Hölder continuous functions

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 4:37

This entry devotes the following fundamental question: if $u$ is Hölder continuous, then how about $u^\gamma$ for some constant $\gamma$? Throughout this entry, we work on $\Omega \subset \mathbb R^n$ which is not necessarily bounded.

Firstly, we have an elementary result

Proposition. If $f$ and $g$ are $\alpha$-Hölder continuous and bounded, so is $fg$.

Proof. The proof is simple, we just observe that

$\displaystyle\left| {f(x)g(x) - f(y)g(y)} \right| \leqslant \left| {f(x) - f(y)} \right||g(x)| + \left| {g(x) - g(y)} \right||f(y)|$

which yields

$\displaystyle\frac{{\left| {f(x)g(x) - f(y)g(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}} \leqslant \frac{{\left| {f(x) - f(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega |g(x)|} \right) + \frac{{\left| {g(x) - g(y)} \right|}}{{{{\left| {x - y} \right|}^\alpha }}}\left( {\mathop {\sup }\limits_\Omega |f(y)|} \right)$.

Consequently,

for any positive integer number $n$ and any $\alpha$-Hölder continuous and bounded function $u$, function $u^n$ is also $\alpha$-Hölder continuous and bounded.

Let us assume , $u$ is $\alpha$-Hölder continuous and bounded, $\gamma>0$ is a constant. Let $n =\left\lfloor \gamma \right\rfloor$. Since $\gamma \in \mathbb R$, we may assume $u$ is also bounded away from zero, that means there exist two constants $0 such that

$m.

We now study the $\alpha$-Hölder continuity of $u^\gamma$. Observe that function

$\displaystyle f(t) = {t^{\left\lceil \gamma \right\rceil }}, \quad t>0$

is sub-additive in the sense that

$\displaystyle f(t_1+t_2) \leqslant f(t_1)+f(t_2), \quad \forall t_1,t_2>0$.

## February 25, 2010

### Double Kelvin transform being the identity map

In this topic, we proved a very interesting property involving the Laplacian of the Kelvin transform of a function. Recall that, for a given function $u$, its Kelvin transform is defined to be

$\displaystyle {u^\sharp }(x) = \frac{1}{{{{\left| x \right|}^{n - 2}}}}u\left( {\frac{x}{{{{\left| x \right|}^2}}}} \right)$.

We then have

$\displaystyle\Delta {u^\sharp }(x) = |{x^\sharp }{|^{n + 2}}\Delta u\left( {{x^\sharp }} \right)$

where the inversion point $x^\sharp$ of $x$ is defined to be

$\displaystyle {x^\sharp } = \frac{x}{{\left| x \right|^2}}$.

Therefore, the Kelvin transform can be defined to be

$\displaystyle {u^\sharp }({x^\sharp }) = \frac{1}{{{{\left| {{x^\sharp }} \right|}^{n - 2}}}}u\left( x \right)$.

We also have another formula

$\displaystyle\Delta {u^\sharp }({x^\sharp }) = |{({x^\sharp })^\sharp }{|^{n + 2}}\Delta u\left( {{{({x^\sharp })}^\sharp }} \right) = {\left| x \right|^{n + 2}}\Delta u(x)$.

The right hand side of the above identity involves $\Delta u(x)$, we can rewrite this one in terms of $(\Delta u)^\sharp$. Actually, we have the following

$\displaystyle {(\Delta u)^\sharp }({x^\sharp }) = \frac{1}{{{{\left| {{x^\sharp }} \right|}^{n -2}}}}\Delta u(x)$

which gives

$\displaystyle\Delta {u^\sharp }({x^\sharp }) = {\left| x \right|^{n + 2}}\Delta u(x) = {\left| x \right|^{n + 2}}{\left| {{x^\sharp }} \right|^{n - 2}}{(\Delta u)^\sharp }({x^\sharp }) =|x|^4 {(\Delta u)^\sharp }({x^\sharp })$.

Today, we study the a more general form of the Kelvin transform. The above definition of the Kelvin transform is with respect to the origin and unit ball in $\mathbb R^n$. We are now interested in the case when the Kelvin transform is defined with respect to a fixed ball. The following result is adapted from a paper due to M.C. Leung published in Math. Ann. in 2003.

Define the reflection on the sphere with center at $\xi$ and radius $a>0$ by

$\displaystyle {R_{\xi ,a}}(x) = \xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}, \quad x \ne \xi$.

It is direct to check that

$\displaystyle {R_{\xi ,a}}({R_{\xi ,a}}(x)) = x,\quad \forall x \ne \xi$.

We observe that

$\displaystyle \xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}} \ne \xi , \quad \forall x \ne \xi$.

The  Kelvin transform with center at $\xi$ and radius $a>0$ of a function $u$ is given by

$\displaystyle u_{\xi ,a}^\sharp (x) = \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}u\left( {\xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}} \right), \quad \forall x \ne \xi$.

We remark that if $\xi$ and $a$ are fixed, and if $u(y)\gg 1$, then $u_{\xi ,a}^\sharp (x) \gg 1$ as well, where $y=R_{\xi, a}(x)$. In addition, $R_{\xi, a}$ sends a set of small diameter not too close to $\xi$ to a set of small diameter. We verify that double Kelvin transform is the identity map. We have

$\displaystyle\begin{gathered} (u_{\xi ,a}^\sharp )_{\xi ,a}^\sharp (x) = \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}u_{\xi ,a}^\sharp \left( {\xi + {a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}} \right) \hfill \\ \qquad\qquad= \frac{{{a^{n - 2}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}\frac{{{a^{n - 2}}}}{{\frac{{{a^{2(n - 2)}}}}{{{{\left| {x - \xi } \right|}^{n - 2}}}}}}u\left( {\xi + {a^2}\frac{{{a^2}\frac{{x - \xi }}{{{{\left| {x - \xi } \right|}^2}}}}}{{\frac{{{a^4}}}{{{{\left| {x - \xi } \right|}^2}}}}}} \right) \hfill \\ \qquad\qquad= u(x), \quad\forall x \ne \xi . \hfill \\ \end{gathered}$

## 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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