Ngô Quốc Anh

April 14, 2019

Extending functions between metric spaces: Continuity, uniform continuity, and uniform equicontinuity

Filed under: Giải Tích 3, Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:02

This topic concerns a very classical question: extend of a function f : X \to Y between two metric spaces to obtain a new function \widetilde f : \overline X \to Y enjoying certain properties. I am interested in the following three properties:

  • Continuity,
  • Uniformly continuity,
  • Pointwise equi-continuity, and
  • Uniformly equi-continuity.

Throughout this topic, by X and Y we mean metric spaces with metrics d_X and d_Y respectively.

CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function f : X \to Y is only assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function f to obtain a new continuous function \widetilde f : \overline X \to Y. The following counter-example demonstrates this:

Let X = [0,\frac 12 ) \cup (\frac 12, 1] and let f be any continuous function on X such that there is a positive gap between f(\frac 12+) and f(\frac12-). For example, we can choose

\displaystyle f(x)=\begin{cases}x^2&\text{ if } x<\frac 12,\\x^3 & \text{ if } x>\frac 12.\end{cases}

Since f is monotone increasing, we clearly have

\displaystyle f(\frac12-)-f(\frac 12+)=\frac18.

Hence any extension \widetilde f of f cannot be continuous because \widetilde f will be discontinuous at x =\frac 12. Thus, we have just shown that continuity is not enough. For this reason, we require f to be uniformly continuous.

SIMPLE OBSERVATIONS. We start with the following basic results.


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 )}}


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.

September 4, 2010

CE: Convergence of improper integrals does not imply the integrands go to zero

Filed under: Counter-examples, Giải Tích 3 — Ngô Quốc Anh @ 12:31

This entry devotes a similar question that raises during a course of series. We all know that for a convergent series of (positive) real number

\displaystyle \sum_{n=1}^\infty a_n

it is necessary to have

\displaystyle\mathop {\lim }\limits_{n \to \infty } {a_n} = 0.

This is the so-called n-th term test. A natural extension is the following question

Question. Suppose f(x) is positive on [0,\infty) and

\displaystyle\int_0^{ + \infty } {f(x)dx}

exists. Must f(x) tend to zero as x \to +\infty?


July 27, 2010

The implicit function theorem: How to prove a continuously dependence on parameters for solutions of ODEs

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 0:25

It is clear that the implicit function theorem plays an important role in analysis. From now on, I am going to demonstrate this significant matter from the theory of differential equations, both ODE and PDE, point of view.

Let us start with the following ODE

-u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}=0

on some domain \Omega \subset \mathbb R^n with \alpha \not\equiv 0 and \beta \not\equiv 0. We assume the existence result on W_+^{2,p} is proved for some p>1. We prove the following

Theorem. The solution u \in W_+^{2,p} depends continuously on (\alpha, \beta) \in L^\infty \times L^\infty.

Proof. Consider the map

\mathcal N : W_+^{2,p} \times (L^\infty \times L^\infty) \to L^p


(u,\alpha,\beta) \mapsto -u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}.

This map is evidently continuous (since W_+^{2,p} is an algebra). One readily shows that its Fréchet derivative at (u, \alpha, \beta) with respect to u in the direction h is

\mathcal N'[u,\alpha ,\beta ]h = - h'' + \left[ {(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}} \right]h.

The continuity of the map

(u,\alpha,\beta) \mapsto \mathcal N'[u,\alpha ,\beta ]

follows from the fact that W_+^{2,p} is an algebra continuously embedded in C^0(\Omega).

Since \alpha \not\equiv 0 and \beta \not\equiv 0, the potential

V={(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}}

is not identically zero. Thus it is well-known that the map

-\Delta +V : W^{2,p} \to L^p

is an isomorphism.

The implicit function theorem then implies that if u_0 is a solution for data (\alpha_0, \beta_0), there is a continuous map defined near (\alpha_0, \beta_0) taking (\alpha, \beta) to the corresponding solution of the ODE. This establishes the conclusion.

For the more details, we prefer the reader to this preprint.

September 17, 2008

L^2 differentiable?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2, Giải Tích 3 — Ngô Quốc Anh @ 13:04

Let and define .

a) Must be differentiable at 0?
b) Must have any differentiable points?
c) Let , show that exists and determine what it is.


a) No. For example, let for so that near zero. This but does not exist.

b) Yes. In fact, must be differentiable almost everywhere, by the Lebesgue theorem on the differentiation of the integral. This theorem requires only that which is true.

c) By the Schwarz inequality,

f^2(x) = \left(\int_0^x}g(t)\,dt\right)^2\le \left(\int_0^x 1\,dt\right) \left(\int_0^xg^2(t)\,dt\right).

At least that’s it for Being careful about the other side, we determine that

0\le\phi(x) = f^2(x)\le|x|\left|\int_0^xg^2(t)\,dt\right|.

But since is an integrable function we have (by an argument that uses the Dominated Convergence Theorem) that

\lim_{x\to 0}\int_0^xg^2(t)\,dt = 0.

Hence \lim_{x\to0}\frac {\phi(x)}{x} = 0, so

April 19, 2008

Bài tập hay về metric

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 0:00

Let (X,d) be a metric space and f: \mathbb R \to \mathbb R a function with f(0)=0, f'(t) >0, f''(t) \leq 0 for t>0. Prove that f \circ d is a metric on X.

Solution. The function f is increasing and concave down. Let d'=fd

  1. Since d'(x,y)=0 then d(x,y)=0 (because f is increasing), so x=y. Of course, d'(x,y)>0 for x\ne y (because f is increasing) and d'(x,x)=0 because f(0)=0.
  2. Symmetry is easy because d is so.
  3. d'(x,z)=f(d(x,z)) \leq f(d(x,y)+d(y,z)) (because f is increasing) \leq f(d(x,y))+f(d(y,z)) (because f(0)=0 and f is concave down), so triangle inequality is verified.

Remark. You want to prove f(a+b)\leq f(a)+f(b) (then you can substitute a = d(x,y), b = d(y,z)).

By concavity,

\displaystyle f(a) \ge \frac{a}{a+b} f(a+b) + \frac{b}{a+b} f(0)


\displaystyle f(b) \ge \frac{b}{a+b} f(a+b) + \frac{a}{a+b}f(0).

Sum them up, you are done.

March 30, 2008

Một số bài tập Giải tích của Khoa Toán đại học Florida

Filed under: Giải Tích 2, Giải Tích 3, Giải Tích 4, Giải Tích 5 — Ngô Quốc Anh @ 12:04


Click to access fall2007.pdf

Click to access fall2007.pdf

January 5, 2008

GT3 – Bài Tập 08

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 20:57

Lấy bài tập ở đây: gt3-08.pdf

GT3 – Bài Tập 07

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 2:22

Lấy bài tập ở đây: gt3-07.pdf

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