Ngô Quốc Anh

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})


\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].

Proof. Assuming without loss of generality that u \in C_0^\infty(\mathbb R^n), we define a vector field by setting

\displaystyle {h_\varepsilon }(x) = \left( {\frac{{{x_1}}}{{\varepsilon + {{\left| x \right|}^2}}}{u^2},...,\frac{{{x_n}}}{{\varepsilon + {{\left| x \right|}^2}}}{u^2}} \right), \quad \varepsilon > 0,

and use the divergence theorem. Then

\displaystyle\begin{gathered} \int_{{\mathbb{R}^n}} {\frac{{n\varepsilon + (n - 2){{\left| x \right|}^2}}}{{{{\left( {\varepsilon + {{\left| x \right|}^2}} \right)}^2}}}u{{(x)}^2}dx} = - 2\int_{{\mathbb{R}^n}} {\frac{{x \cdot \nabla u(x)}}{{\varepsilon + {{\left| x \right|}^2}}}u(x)dx} \hfill \\ \qquad\qquad\leqslant 2\int_{{\mathbb{R}^n}} {\frac{{\left| x \right|\left| {u(x)} \right|}}{{\varepsilon + {{\left| x \right|}^2}}}\left| {\nabla u(x)} \right|dx} \hfill \\ \qquad\qquad\leqslant 2\sqrt {\int_{{\mathbb{R}^n}} {\frac{{{{\left| x \right|}^2}}}{{{{\left( {\varepsilon + {{\left| x \right|}^2}} \right)}^2}}}u{{(x)}^2}dx} } \sqrt {\int_{{\mathbb{R}^n}} {{{\left| {\nabla u(x)} \right|}^2}dx} } \hfill \\ \end{gathered}


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

Letting \varepsilon \to 0, we obtain the required result.

It follows from this proof that if, instead of h_\varepsilon vector field he defined above, we consider the vector field

\displaystyle {k_\varepsilon }(x) = \left( {\frac{{{x_1}}}{{\varepsilon + {{\left| x \right|}^2}}}{{\left| u \right|}^{p - 1}}u,...,\frac{{{x_n}}}{{\varepsilon + {{\left| x \right|}^2}}}{{\left| u \right|}^{p - 1}}u} \right), \quad \varepsilon > 0,

where p > 1, then we can establish tile following well-known inequality.

Theorem. Let u \in  \mathcal D^{1,p}(\mathbb R^n) with n>p. Then

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


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

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

An example of an inequality that can be proved by using this simple method is the following version of Hardy’s inequality obtained by F. Rellich in 1953.

Theorem (Rellich’s inequality). Let n > 4. Then for any u \in W^{2,2}(\mathbb R^n) the following inequality holds

\displaystyle {\left( {\frac{{n(n - 2)}}{4}} \right)^2}\int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^4}}}dx} \leqslant \int_{{\mathbb{R}^n}} {{{\left| {\Delta u} \right|}^2}dx}.

The constant {\left({\frac{{n(n - 2)}}{4}}\right)^2} is the best possible constant.

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