Ngô Quốc Anh

December 31, 2011

A Hardy-Moser-Trudinger inequality: A conjecture by Wang and Ye

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 21:48

Let B denote the standard unit disk in \mathbb R^2. The famous Moser–Trudinger inequality says that

\displaystyle\int_B {\exp \left( {\frac{{4\pi {u^2}}}{{\left\| {\nabla u} \right\|_2^2}}} \right)dx} \leqslant C < \infty ,\quad\forall u \in H_0^1(B)\backslash \{ 0\}

holds. There is another important inequality in analysis, the Hardy inequality which claims that

\displaystyle H(u) = \int_B {|\nabla u{|^2}dx} - \int_B {\frac{{{u^2}}}{{{{(1 - |x{|^2})}^2}}}dx} \geqslant 0,\quad\forall u \in H_0^1(B)

holds. The one H is usuall called the Hardy functional. One can immediately see that

\displaystyle\frac{{4\pi {u^2}}}{{\left\| {\nabla u} \right\|_2^2}} \leqslant \dfrac{{4\pi {u^2}}}{{\displaystyle\int_B {|\nabla u{|^2}dx} - \int_B {\frac{{{u^2}}}{{{{(1 - |x{|^2})}^2}}}dx} }}

for any u \in H_0^1(B)\backslash \{ 0\}. Recently, in a paper accepted in Advances in Mathematics journal, Wang and Ye proved that there exists a constant C_0 >0 such that the following

\displaystyle\int_B {\frac{{4\pi {u^2}}}{{H(u)}}dx} \leqslant C_0 < \infty ,\quad\forall u \in \mathcal H(B^n)\backslash \{ 0\}

where B^n is the unit ball in \mathbb R^n, n \geqslant 2 and \mathcal H=\mathcal H(B^n) is the complement of C_0^\infty(B^n) with respect to the following norm \|u\|_{\mathcal H}=\sqrt{H(u)}.

Let us go back to the case n=2. They then defined

\displaystyle {H_d}(u) = \int_\Omega {|\nabla u{|^2}dx} - \frac{1}{4}\int_\Omega {\frac{{{u^2}}}{{d{{(x,\partial \Omega )}^2}}}dx} > 0,\quad \forall u \in H_0^1(\Omega )\backslash \{ 0\}

where \Omega is a regular, bounded and convex domain sitting in \mathbb R^2. They then conjectured that the following

\displaystyle\int_\Omega {\frac{{4\pi {u^2}}}{{{H_d}(u)}}dx} \leqslant C(\Omega ) < \infty ,\quad\forall u \in {\mathcal H_d}(\Omega )\backslash \{ 0\}

still holds for some constant C(\Omega)>0 where {\mathcal H_d}(\Omega ) denotes the completion of C_0^\infty (\Omega) with the corresponding norm associated with H_d. Apparently, the conjecture holds true for \Omega = B.

April 22, 2011

On Costa-Hardy-Rellich inequalities

Filed under: Giải Tích 2, Giải Tích 6 (MA5205) — Tags: , , , — Ngô Quốc Anh @ 15:44

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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April 13, 2011

How good the Hardy inequality is?

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 3:34

Before going, let us recall the so-called Hardy inequality from this note below

Theorem (Hardy’s inequality). Let u \in \mathcal D^{1,2}(\mathbb R^n) with n \geqslant 3. Then

\displaystyle\frac{{{u^2}}}{{{{\left| x \right|}^2}}} \in {L^1}({\mathbb{R}^n})

and

\displaystyle {\left( {\frac{{n - 2}}{2}} \right)^2}\int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^2}}}dx} \leqslant \int_{{\mathbb{R}^n}} {{{\left| {\nabla u} \right|}^2}dx}.

The constant {\left( {\frac{{n - 2}}{2}} \right)^2} is the best possible constant.

The purpose of this note is to show that the Hardy integral cannot be improved in the usual sense, that is, there are no nontrivial potential V \geqslant 0 and no exponent q>0 such that, for any function u,

\displaystyle C{\left( {\int_{{\mathbb{R}^n}} {V(x)|u{|^n}dx} } \right)^{\frac{2}{n}}} \leqslant {\left( {\frac{2}{{n - 2}}} \right)^2}\int_{{\mathbb{R}^n}} {{{\left| {\nabla u} \right|}^2}dx} - \int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^2}}}dx}

for some positive constant C.

Indeed, let us construct the following family of test functions u_\varepsilon as follows

\displaystyle {u_\varepsilon }(x) = \begin{cases}|x{|^{ - \frac{{n - 2}}{2} + \varepsilon }},&|x| \leqslant 1,\\ |x{|^{ - \frac{{n - 2}}{2} - \varepsilon }},&|x| > 1.\end{cases}

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May 27, 2010

A Simple Approach to the Hardy and Rellich inequalities

Filed under: Giải Tích 6 (MA5205) — Tags: , — Ngô Quốc Anh @ 16:07

The classical Hardy inequality in \mathbb R^n, n \geqslant 3, is stated as follows

Theorem (Hardy’s inequality). Let u \in \mathcal D^{1,2}(\mathbb R^n) with n \geqslant 3. Then

\displaystyle\frac{{{u^2}}}{{{{\left| x \right|}^2}}} \in {L^1}({\mathbb{R}^n})

and

\displaystyle {\left( {\frac{{n - 2}}{2}} \right)^2}\int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^2}}}dx} \leqslant \int_{{\mathbb{R}^n}} {{{\left| {\nabla u} \right|}^2}dx}.

The constant {\left( {\frac{{n - 2}}{2}} \right)^2} is the best possible constant.

I suddenly found a very simple proof due to E. Mitidieri [here].

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June 7, 2008

Some important functional inequalities

Filed under: Nghiên Cứu Khoa Học — Tags: , , , , , , , — Ngô Quốc Anh @ 13:55

Hardy’s inequality: Nếu p>1, f(x) \geq 0 and F(x) = \int_0^x f(t) dt, thì

\displaystyle\int_0^{+\infty} \left( \frac{F(x)}{x} \right)^p dx < \left( \frac{p}{p - 1} \right)^p \int_0^{+\infty} f^p( t )dt

trừ trường hợp hàm f(x) \equiv 0. Hằng số ở vế phải là tốt nhất.

Opial‘s inequality: Giả sử y(x) thuộc lớp C^1 trên đoạn [0, h] với y(0)=y(h)=0 and y(x) >0 với mọi 0<x<h. Khi đó ta có

\displaystyle\int_0^h {|y(x)y'(x)|dx} \leqq \frac{h}{4}\int_0^h {|y'(x){|^2}dx}.

Hằng số \frac{h}{4} ở đây là tốt nhất.

Rellich‘s inequality: Giả sử hàm u khả vi vô hạn với giá compắc trong \mathbb R^N trừ điểm gốc. Khi đó ta có bất đẳng thức

\displaystyle\int_{\mathbb{R}^N } {\left| {\Delta u} \right|^2 dx} \geqq \frac{{n^2 \left( {n - 4} \right)^2 }} {{16}}\int_{\mathbb{R}^N } {\left| x \right|^{ - 4} \left| u \right|^2 dx} , \quad n \ne 2.

Serrin‘s inequality: Giả sử hàm u khả vi vô hạn với giá compắc triệt tiêu trên biên \Omega, khi đó

\displaystyle\left( {\int_\Omega {u^{\frac{n} {{n - 1}}} dx} } \right)^{\frac{{n - 1}} {n}} \leqq \frac{1} {{\sqrt {4n} }}\int_\Omega {\left| {\nabla u} \right|dx} .

Caffarelli–Kohn–Nirenberg‘s inequality: Giả sử hàm u khả vi vô hạn với giá compắc trong \mathbb R^N trừ điểm gốc. Khi đó ta có bất đẳng thức

\displaystyle\frac{{\left| {N - \left( {a + b + 1} \right)} \right|}} {2}\int_{\mathbb{R}^N } {\frac{{\left| u \right|^2 }} {{\left| x \right|^{a + b + 1} }}dx} \leqq \sqrt {\int_{\mathbb{R}^N } {\frac{{\left| u \right|^2 }} {{\left| x \right|^{2a} }}dx} } \sqrt {\int_{\mathbb{R}^N } {\frac{{\left| u \right|^2 }} {{\left| x \right|^{2b} }}dx} } .

Gagliardo-Nirenberg-Sobolev‘s inequality: Giả sử hàm u khả vi liên tục với giá compắc trong \mathbb R^N và 1 \leq p < N. Khi đó ta có bất đẳng thức

\displaystyle\left( {\int_{\mathbb{R}^N } {\left| u \right|^{\frac{{Np}} {{N - p}}} dx} } \right)^{\frac{{N - p}} {{Np}}} \leqq C\left( {p,N} \right)\left( {\int_{\mathbb{R}^N } {\left| {\nabla u} \right|^p dx} } \right)^{\frac{1} {p}} .

Horgan‘s inequality: Giả sử hàm u trơn, khi đó với miền đang xét là bị chặn với biên đủ trơn thì

\displaystyle\int_\Omega {\left| u \right|^3 dx} \leqq \frac{1} {{\sqrt {4\pi } }}\left( {\int_\Omega {\left| u \right|^2 dx} } \right)^{\frac{3} {4}} \left( {\int_\Omega {\left| {\nabla u} \right|^2 dx} } \right)^{\frac{3} {4}} .

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