# 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.

## 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 $\int_0^\infty |f (x)|dx<\infty$.

## August 18, 2009

### An other limit supremum of sin function

In the topic we showed that for any irrational $\alpha$ the limit

$\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)$

does not exist. In this topic, we consider the following limit

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right)$.

To be precise, we prove that

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$

for almost every $x \in [0,2\pi]$.

Solution. Let

$\displaystyle A = \left\{ {x \in \left( {0,2\pi } \right): \frac{x} {\pi } \notin \mathbb{Q}} \right\}$.

Then $A$ is a measurable set of measure $2\pi$. Moreover, for any $x \in A$,

$\displaystyle\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$.

Indeed for any $x \in A$, since

$\displaystyle\left\{ {k\frac{x} {\pi } - 2l: l \in \mathbb{Z}} \right\}$

is dense subgroup of $\mathbb R$ there are sequences $\{k_n\}$ and $\{l_n\}$ of $\mathbb Z$ such that

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \left( {{k_n}\frac{x} {\pi } - {l_n}} \right) = \frac{1} {2}$.

Since

$\displaystyle \frac{1} {2} \notin \left\{ {k\frac{x} {\pi } - 2l: k,l \in \mathbb{Z}} \right\}$

$\{k_n\}$ admits a subsequence $\{k'_n\}$ either increasing to $+\infty$ or decreasing to $-\infty$. If $\mathop {\lim }\limits_{n \to \infty } {{k'}_n} = + \infty$ then

Otherwise $\mathop {\lim }\limits_{n \to \infty } \left( { - 3{{k'}_n}} \right) = + \infty$ and

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