Ladyzhenskaya type inequality when d=2

Problem. If $u\in C_{0}^{1}\left(\mathbb{R}^{2}\right)$ , then we have

$\norm u{\Leb 4\left(\mathbb{R}^{2}\right)}^{2}\le\sqrt{2}\norm u{\Leb 2\left(\mathbb{R}^{2}\right)}\norm{\nabla u}{\Leb 2\left(\mathbb{R}^{2}\right)}.$

This is the Ladyzhenskaya’s inequality.

To prove this, we need the following lemma.

Lemma. For $f\in C_{0}^{\infty}\left(\mathbb{R}^{d}\right)$ , we have

$\norm f{\Leb{p^{*}}\left(\mathbb{R}^{d}\right)}\le\left(\prod_{j=1}^{d}\norm{\frac{\partial f}{\partial x_{j}}}{\Leb 1\left(\mathbb{R}^{d}\right)}\right)^{\frac{1}{d}}.$

Here $\frac{1}{p^{*}}=1-\frac{1}{d}$ .

Proof. e proceed by induction on $d$ . The case $d=1$ is trivial since $f\left(x\right)=\int_{-\infty}^{x}f^{\prime}\left(t\right)dt$ and this leads to

$\norm f{\Leb{\infty}}\le\norm{f^{\prime}}{\Leb 1}.$

We assume that the inequality holds for $d-1$ . Write $x=\left(x_{1},x^{\prime}\right)$ with $x_{1}\in\mathbb{R}$ , $x^{\prime}\in\mathbb{R}^{d-1}$ . Set

$I_{j}\left(x_{1}\right)=\int_{\mathbb{R}^{d-1}}\left|\frac{\partial f}{\partial x_{j}}\left(x_{1},x^{\prime}\right)\right|dx^{\prime},\quad j=2,\dots d$

and

$I_{1}\left(x^{\prime}\right)=\int_{\mathbb{R}^{1}}\left|\frac{\partial f}{\partial x_{1}}\left(x_{1},x^{\prime}\right)\right|dx_{1}.$

For temporary set $q=\frac{d}{d-1}$ , $q^{\prime}=\frac{d-1}{d-2}$ . Then

$\left(\int_{\mathbb{R}^{d-1}}\left|f\left(x_{1},x^{\prime}\right)\right|^{q^{\prime}}dx^{\prime}\right)^{\frac{1}{q^{\prime}}}\le\left(\prod_{j=2}^{d}I_{j}\left(x_{1}\right)\right)^{\frac{1}{d-1}}.$

Since $\left|f\left(x\right)\right|\le I_{1}\left(x^{\prime}\right)$ , we have

$\left|f\right|^{q}\le I_{1}\left(x^{\prime}\right)^{q-1}\left|f\right|$

because $q=\frac{1}{d-1}+1$ .

Thus,

$\begin{align*} \left(\int_{\mathbb{R}^{d-1}}\left|f\right|^{q}dx^{\prime}\right) & \le\int_{\mathbb{R}^{d-1}}I_{1}\left(x^{\prime}\right)^{q-1}\left|f\right|dx^{\prime}\\ & \le\left(\int_{\mathbb{R}^{d-1}}I_{1}\left(x^{\prime}\right)dx^{\prime}\right)^{\frac{1}{d-1}}\left(\int_{\mathbb{R}^{d}}\left|f\right|^{q^{\prime}}dx^{\prime}\right)^{\frac{1}{q^{\prime}}} \end{align*}$

by Hölder inequality. So

$\left(\int_{\mathbb{R}^{d-1}}\left|f\right|^{q}dx^{\prime}\right)\le\left(\int_{\mathbb{R}^{d-1}}I_{1}\left(x^{\prime}\right)dx^{\prime}\right)^{\frac{1}{d-1}}\left(\prod_{j=2}^{d}I_{j}\left(x_{1}\right)\right)^{\frac{1}{d-1}}.$

By integrating this with respect to $x_{1}$ and multiple Hölder inequality, we get

$\begin{align*} \int_{\mathbb{R}^{d}}\left|f\right|^{q}dx & \le\left(\int_{\mathbb{R}^{d-1}}I_{1}\left(x^{\prime}\right)dx^{\prime}\right)^{\frac{1}{d-1}}\prod_{j=2}^{d}\left(\int_{\mathbb{R}^{1}}I_{j}\left(x_{1}\right)dx_{1}\right)^{\frac{1}{d-1}}\\ & =\left(\prod_{j=1}^{d}\norm{\frac{\partial f}{\partial x_{j}}}{\Leb 1}\right)^{\frac{1}{d-1}}. \end{align*}$

Since $q=p^{*}$ , we get

$\norm f{\Leb{p^{*}}}\le\left(\prod_{j=1}^{d}\norm{\frac{\partial f}{\partial x_{j}}}{\Leb 1}\right)^{\frac{1}{d}}.\qedhere$

Proof of the Ladyzhenskaya’s inequality. For $d=2$ , the result of Lemma is

$\left(\int_{\mathbb{R}^{2}}\left|u\right|^{2}dx\right)^{\frac{1}{2}}\le\left(\int_{\mathbb{R}^{2}}\left|D_{1}u\right|dx\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{2}}\left|D_{2}u\right|dx\right)^{\frac{1}{2}}.$

Put $u^{2}$ instead of $u$ . Then we have

$\left(\int_{\mathbb{R}^{2}}\left|u\right|^{4}dx\right)^{\frac{1}{2}}\le\left(\int_{\mathbb{R}^{2}}\left|D_{1}u^{2}\right|dx\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{2}}\left|D_{2}u^{2}\right|dx\right)^{\frac{1}{2}}.$

Note that for $j=1,2$ we have

$\int_{\mathbb{R}^{2}}\left|D_{j}u^{2}\right|dx=2\int_{\mathbb{R}^{2}}\left|uD_{j}u\right|dx.$

By Cauchy-Schwarz inequality, we have

$\int_{\mathbb{R}^{2}}\left|D_{j}u^{2}\right|dx\le2\left(\int_{\mathbb{R}^{2}}\left|u\right|^{2}dx\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{2}}\left|D_{j}u\right|^{2}dx\right)^{\frac{1}{2}}$

for $j=1,2$ .

Hence,

$\begin{align*} & \relphantom{=}\int_{\mathbb{R}^{2}}\left|u\right|^{4}dx\\ & \le4\left(\int_{\mathbb{R}^{2}}\left|u\right|^{2}dx\right)\left(\int_{\mathbb{R}^{2}}\left|D_{1}u\right|^{2}dx\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^{2}}\left|D_{2}u\right|^{2}dx\right)^{\frac{1}{2}}\\ & =4\norm u{\Leb 2}^{2}\norm{D_{1}u}{\Leb 2}\norm{D_{2}u}{\Leb 2}\\ & \le2\norm u{\Leb 2}^{2}\left(\norm{D_{1}u}{\Leb 2}^{2}+\norm{D_{2}u}{\Leb 2}^{2}\right). \end{align*}$

$\norm u{\Leb 4\left(\mathbb{R}^{2}\right)}^{2}\le\sqrt{2}\norm u{\Leb 2\left(\mathbb{R}^{2}\right)}\norm{\nabla u}{\Leb 2\left(\mathbb{R}^{2}\right)}.$

Reference
1. Elias M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton University Press, 1971.

2 thoughts on “Ladyzhenskaya type inequality when d=2”

Leun Kim February 11, 2016

이 부등식이 이름이 있었네요 ㅋㅋ 어디서 많이 봤나 했더니 Burgers Equation 레포트 쓸 때 3차원 버전이 자주 나와서 냈던 기억이…

Reply ↓
1. Will Kwon Post authorFebruary 11, 2016
  
  아하 그렇군요ㅋㅋ Ladyzhenskaya inequality에 대해서 교수님께 잠시 여쭤보니 저걸로 2차원 Navier-Stokes의 중요한 결과를 얻어냈다고 하네요.
  
  일반화에 대해서 여쭤보니 조금 더 큰 이론이 필요한거 같고..
  
  저 레포트 잘 읽어보겠습니다 ㅎㅎ
  
  Reply ↓

2 thoughts on “Ladyzhenskaya type inequality when d=2”

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