# Ngô Quốc Anh

## 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.

Proof. For all $a,b\in \mathbb R$ and $u \in C^\infty_0(\mathbb R^N\backslash\{0\})$ one has, for all $t$, the following $\displaystyle\int_{{\mathbb{R}^N}} {{{\left| {\frac{{\nabla u}}{{|x{|^a}}} + t\frac{x}{{|x{|^{b + 1}}}}\Delta u} \right|}^2}dx} \geqslant 0.$

Expanding the above yields $\displaystyle\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} + {t^2}\int_{{\mathbb{R}^N}} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} + 2t\int_{{\mathbb{R}^N}} {\frac{{\Delta u}}{{|x{|^{a + b + 1}}}}(x \cdot \nabla u)dx} \geqslant 0.$

If we denote the last integral by $B$ and write it as $\displaystyle B = \int_{{\mathbb{R}^N}} {\big({\rm div}(\nabla u)\big)\left( {\frac{x}{{|x{|^\gamma}}} \cdot \nabla u} \right)dx}$

an integration by parts gives $\displaystyle B = - \frac{1}{2}\int_{{\mathbb{R}^N}} {\frac{x}{{|x{|^\gamma}}} \cdot \nabla (|\nabla u{|^2})dx} - \int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx + \gamma } \int_{{\mathbb{R}^N}} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} .$

Keep in mind that $\displaystyle\frac{\partial }{{\partial {x_i}}}\left( {\frac{{{x_i}}}{{|x{|^\gamma }}}} \right) = \frac{1}{{|x{|^\gamma }}} - \gamma \frac{{x_i^2|x{|^{\gamma - 2}}}}{{|x{|^{2\gamma }}}}.$

A second integration by parts on the first integral above yields $\displaystyle - \frac{1}{2}\int_{{\mathbb{R}^N}} {\frac{x}{{|x{|^\gamma }}} \cdot \nabla (|\nabla u{|^2})dx} = \frac{{N - \gamma }}{2}\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx}$

so that $B$ becomes $\displaystyle B = \left( {\frac{{N - 2 - \gamma }}{2}} \right)\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{{\mathbb{R}^N}} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} .$

Therefore, we have $At^2+2Bt+C \geqslant 0$

for any $t \geqslant 0$ where $B$ is given above and $\displaystyle A = \int_{{\mathbb{R}^N}} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} , \quad C = \int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} .$

This is equivalent to $B^2 - AC \leqslant 0$, i.e. $\displaystyle {\left( {\frac{{N - 2 - \gamma }}{2}\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{{\mathbb{R}^N}} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} } \right)^2} \leqslant \left( {\int_{{\mathbb{R}^N}} {\frac{{|\Delta u{|^2}}}{{|x{|^{2b}}}}dx} } \right)\left( {\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^{2a}}}}dx} } \right).$

This completes the first inequality. On the other hand, since $\displaystyle 0 \leqslant \frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^2}}} \leqslant |\nabla u{|^2}$

we know that $\displaystyle\frac{{N + \gamma - 2}}{2}\int_{{\mathbb{R}^N}} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx} \leqslant \frac{{N - 2 - \gamma }}{2}\int_{{\mathbb{R}^N}} {\frac{{|\nabla u{|^2}}}{{|x{|^\gamma }}}dx} + \gamma \int_{{\mathbb{R}^N}} {\frac{{{{(x \cdot \nabla u)}^2}}}{{|x{|^{\gamma + 2}}}}dx}$

provided $\gamma \leqslant N-2$. This proves the second inequality.

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