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);


\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}.


\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

MuPad: Heart in 3D

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<u \in H^1(\mathbb R^n).

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.


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


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


\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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