# Ngô Quốc Anh

## 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.

## 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.

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

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