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


\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).


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<m<M<\infty such that

m<u(x)<M \quad \forall x \in \Omega.

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


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


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}.


\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

\mathop {\lim }\limits_{n \to \infty } \sin \left( {{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( {...

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

\mathop {\lim }\limits_{n \to \infty } \sin \left( { - 3{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \lef...

Older Posts »

Blog at