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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