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.

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February 23, 2017

In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose (X, \|\cdot\|) is a normed space and \{x_1,...,x_n\} is a set of linearly independent elements in X. Then \{x_1,...,x_n\} is stable under a small perturbation in the sense that there exists some small number \varepsilon>0 such that for any \|y_i\| < \varepsilon with 1 \leqslant i \leqslant n, the all elements of \{x_1+y_1,...,x_n+y_n\} are also linearly independent.

We prove this result by way of contradiction. Indeed, for any \varepsilon>0, there exist n elements y_i \in X with \|y_i\| < \varepsilon such that all elements of \{x_1+y_1,...,x_n+y_n\} are linearly dependent, that is, there exist real numbers \alpha_i with 1 \leqslant i \leqslant n such that

\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0

with

\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.

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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.

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January 6, 2011

The Alexandrov-Bol inequality

Filed under: Giải Tích 6 (MA5205) — Tags: , — Ngô Quốc Anh @ 19:55

In the literature, there is an inequality called the Alexandrov-Bol inequality which is frequently used in partial differential equations. Here we just recall its statement without any proof.

Theorem. Let \Omega be a good domain in \mathbb R^2. Assume p \in C^2(\Omega)\cap C^0(\overline \Omega) be a positive function satisfying the elliptic inequality

\displaystyle -\Delta \log p \leqslant p

in \Omega. Then it holds

\displaystyle l^2(\partial\Omega) \geqslant \frac{1}{2} \big(8\pi-m(\Omega)\big)m(\Omega)

where

\displaystyle l(\partial\Omega)=\int_{\partial\Omega}\sqrt{p}ds

and

\displaystyle m(\Omega)=\int_\Omega pdx.

An analytic proof was given by C. Bandle aroud 1975 when she assumed p to be real analytic. The above version was due to Suzuki in an elegant paper published in the Ann. Inst. H. Poincare in 1992 [here]. The proof is mainly depended on the isoperimetric inequality for the flat Riemannian surfaces. We refer the reader to the paper by Suzuki for the proof.

September 26, 2010

Subharmonic functions

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 1:53

In this entry, we shall discuss a geometric meaning of subharmonic functions. This will help us to easily remember the definition of subharmonic functions.

In mathematics, a harmonic function is a twice continuously differentiable function f : U\to \mathbb R (where U is an open subset of \mathbb R^n) which satisfies Laplace’s equation, i.e.

\displaystyle\frac{\partial^2f}{\partial x_1^2} + \frac{\partial^2f}{\partial x_2^2} + \cdots + \frac{\partial^2f}{\partial x_n^2} = 0

everywhere on U. This is usually written as

\textstyle \Delta f = 0.

In 1D, this condition is about to say that f is harmonic if and only if f is linear. Concerning to the case of functions with one-variable, we have the s0-called convexity saying that function f is convex if and only if the function lies below or on the straight line segment connecting two points, for any two points in the interval. Mathematically, a function f is said to be convex if

\textstyle \Delta f \geqslant 0.

In higher dimension, the notion of linearity and convexity become harmonicity and subharmonicity. Precisely, two points mentioned above become a hyper-surface, for e.g. like a curve in 2D and a straight line becomes a graph of harmonic function. In practice, the closed interval connecting those two points will be replaced by a closed ball. Therefore, we have

Definition. A C^2 function that satisfies \Delta f \ge 0 is called subharmonic. More generally, a function is subharmonic if and only if, in the interior of any ball in its domain, its graph lies below that of the harmonic function interpolating its boundary values on the ball.

Let us consider several examples in 2D.

  • \log functions.

It is well-known that in 2D function \log|z|, where z=(x,y), is harmonic. Therefore, every functions lying below the graph of \log|z| turns out to be subharmonic.

  • \sin functions.

Again, one can easily show that e^x \sin y is harmonic.

September 14, 2010

Asympotic behavior of integrals, 2

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 15:07

We now prove the following result

Theorem. Let u and f be two smooth functions on \mathbb R^2 satisfying

\Delta u(x)=f(x), \quad x \in \mathbb R^2.

Suppose that f is bounded and also f \in L^1(\mathbb R^2) and

|u(x)| \leqslant o(|x|), \quad |x| \to \infty.

Then

\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = \frac{1}{{2\pi }}\int_{{\mathbb{R}^2}} {f(y)dy} .

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September 7, 2010

Asympotic behavior of integrals

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 10:52

Long time ago, we studied [here] the following fact

Suppose f \in L^1(\mathbb R^n) \cap L_{loc}^\infty (\mathbb R^n) with f \geq 0. Define

\displaystyle Sf\left( x \right) = \int_{\mathbb{R}^n } {\log \frac{{\left| y \right|}}{{\left| {x - y} \right|}}f\left( y \right)dy}.

Show that Sf(x) is finite for all x \in \mathbb R^n and Sf \in L_{loc}^1(\mathbb R^n).

In this entry, from now on we continue to prove several useful results appearing in PDE. We shall prove the following

Theorem. Assume u is a solution to

\displaystyle (-\Delta)^\frac{3}{2} u(x)=-2e^{3u(x)}, \quad x \in \mathbb R^3

with finite energy

\displaystyle \int_{{\mathbb{R}^3}} {{e^{3u(x)}}dx} < \infty.

Then

\displaystyle\mathop {\lim }\limits_{|x| \to \infty } \frac{{u(x)}}{{\log |x|}} = - \frac{1}{{{\pi ^2}}}\int_{{\mathbb{R}^3}} {{e^{3u(y)}}dy} .

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September 1, 2010

The inverse of the Laplace transform by contour integration

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 17:23

Usually, we can find the inverse of the Laplace transform \mathcal L[\cdot](s) by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.

Consider the piece-wise differentiable function f(x) that vanishes for x < 0. We can express the function e^{-cx}f(x) by the complex Fourier representation of

\displaystyle f(x){e^{ - cx}} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{i\omega x}}\left[ {\int_0^\infty {{e^{ - ct}}f(t){e^{ - i\omega t}}dt} } \right]d\omega }

for any value of the real constant c, where the integral

\displaystyle I = \int_0^\infty {{e^{ - ct}}|f(t)|dt}

exists. By multiplying both sides of first equation by e^{cx} and bringing it inside the first integral

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{(c + i\omega )x}}\left[ {\int_0^\infty {f(t){e^{ - (c + i\omega )t}}dt} } \right]d\omega } .

With the substitution z = c+\omega i, where z is a new, complex variable of integration,

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{c - \infty i}^{c + \infty i} {{e^{zx}}\left[ {\int_0^\infty {f(t){e^{ - zt}}dt} } \right]d\omega }.

The quantity inside the square brackets is the Laplace transform \mathcal L[f](z). Therefore, we can express f(t) in terms of its transform by the complex contour integral

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August 29, 2010

Achieving regularity results via bootstrap argument, 4

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 2:36

Let us consider the following equation

\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} , \quad x \in {\mathbb{R}^n}

for n\geqslant 1 and 0<\alpha<n. In this entry, by using boothstrap argument, we show that

Theorem. If positive function u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n) solves the equation, then u \in C^\infty(\mathbb R^n).

In the process of proving the result, we need the following result

Proposition. Let V \in L^\frac{n}{\alpha}(B_3) be a non-negative function and set

\displaystyle \delta(V)=\|V\|_{L^\frac{n}{\alpha}(B_3)}.

For \nu >r>\frac{n}{n-\alpha}, there exist positive constants \overline \delta<1 and C \geqslant 1 depending only on n, \alpha, r and \nu such that for any 0 \leqslant V \in L^\frac{n}{\alpha}(B_3) with \delta(V) \leqslant \overline \delta, h \in L^\nu(B_2) and 0 \leqslant u \in L^r(B_3) satisfying

\displaystyle u(x) \leqslant \int_{{B_3}} {\frac{{V(y)u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} + h(x), \quad x \in {B_2}

we have

\displaystyle {\left\| u \right\|_{{L^\nu }({B_{1/2}})}} \leqslant C\left( {{{\left\| u \right\|}_{{L^r}({B_3})}} + {{\left\| h \right\|}_{{L^\nu }({B_2})}}} \right).

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August 22, 2010

Liouville’s theorem and related problems

Filed under: Giải tích 7 (MA4247), PDEs — Tags: — Ngô Quốc Anh @ 6:35

The following theorem is well-known

Theorem (Liouville). Let \Omega be a simply connected domain in \mathbb R^2. Then all real solutions of

\displaystyle \Delta u +2Ke^u=0

in \Omega where K a constant, are of the form

\displaystyle u=\log\frac{|f'|^2}{\left(1+\frac{K}{4}|f|^2\right)^2}

where f is a locally univalent meromorphic function in \Omega.

In geometry, our PDE

\displaystyle \Delta u +2Ke^u=0

says that under the case \Omega=\mathbb R^2, it holds

e^u|dz|^2=f^*g_K

where g_K denotes the standard metric on \mathbb S^2 with constant curvature K. Thus we have

Corollary. All solutions of the PDE in \mathbb R^2 with K>0 and

\displaystyle \int_{\mathbb R^2} e^u<\infty

are of the form

\displaystyle u(x)=\log\frac{16\lambda^2}{\left(4+\lambda^2K|x-x_0|^2\right)^2},\quad \lambda>0, \quad x_0 \in \mathbb R^2.

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