Ladyzhenskaya type inequality when d=2

By | February 11, 2016

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*}

So

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

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

    1. Will Kwon Post author

      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

Leave a Reply

Your email address will not be published. Required fields are marked *