# Ngô Quốc Anh

## April 13, 2011

### How good the Hardy inequality is?

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 3:34

Before going, let us recall the so-called Hardy inequality from this note below

Theorem (Hardy’s inequality). Let $u \in \mathcal D^{1,2}(\mathbb R^n)$ with $n \geqslant 3$. Then

$\displaystyle\frac{{{u^2}}}{{{{\left| x \right|}^2}}} \in {L^1}({\mathbb{R}^n})$

and

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

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

The purpose of this note is to show that the Hardy integral cannot be improved in the usual sense, that is, there are no nontrivial potential $V \geqslant 0$ and no exponent $q>0$ such that, for any function $u$,

$\displaystyle C{\left( {\int_{{\mathbb{R}^n}} {V(x)|u{|^n}dx} } \right)^{\frac{2}{n}}} \leqslant {\left( {\frac{2}{{n - 2}}} \right)^2}\int_{{\mathbb{R}^n}} {{{\left| {\nabla u} \right|}^2}dx} - \int_{{\mathbb{R}^n}} {\frac{{{u^2}}}{{{{\left| x \right|}^2}}}dx}$

for some positive constant $C$.

Indeed, let us construct the following family of test functions $u_\varepsilon$ as follows

$\displaystyle {u_\varepsilon }(x) = \begin{cases}|x{|^{ - \frac{{n - 2}}{2} + \varepsilon }},&|x| \leqslant 1,\\ |x{|^{ - \frac{{n - 2}}{2} - \varepsilon }},&|x| > 1.\end{cases}$

Clearly,

$\displaystyle\int_{{\mathbb{R}^n}} {\frac{{{u_\varepsilon^2}}}{{{{\left| x \right|}^2}}}dx} = \left( {\int_{|x| \leqslant 1} {} + \int_{|x| > 1} {} } \right)\frac{{{u_\varepsilon^2}}}{{{{\left| x \right|}^2}}}dx = \int_{|x| \leqslant 1} {|x{|^{ - n + 2 + 2\varepsilon }}dx} + \int_{|x| > 1} {|x{|^{ - n + 2 - 2\varepsilon }}dx} .$

Now, making use the co-area formula gives us

$\displaystyle\int_{|x| \leqslant 1} {|x{|^{ - n + 2 + 2\varepsilon }}dx} = \int_0^1 {\frac{1}{{{r^{n - 2 - 2\varepsilon }}}}\int_{\partial {B_r}(0)} {d\sigma } dr} = n{\rm vol}({B_1})\int_0^1 {{r^{1 - 2\varepsilon }}dr} = \frac{{n{\rm vol}({B_1})}}{{2 - 2\varepsilon }}.$

Similarly,

$\displaystyle\int_{|x| > 1} {|x{|^{ - n + 2 - 2\varepsilon }}dx} = \int_1^{ + \infty } {\frac{1}{{{r^{n - 2 + 2\varepsilon }}}}\int_{\partial {B_r}(0)} {d\sigma } dr} = \frac{{n{\rm vol}({B_1})}}{{2 + 2\varepsilon }}.$

Therefore,

$\displaystyle\mathop {\lim }\limits_{\varepsilon \to 0 + } \int_{{\mathbb{R}^n}} {\frac{{u_\varepsilon ^2}}{{{{\left| x \right|}^2}}}dx} = n{\rm vol}({B_1}).$

Next, for almost everywhere, it holds

$\displaystyle {\nabla u_\varepsilon }(x) = \begin{cases}\left( { - \frac{{n - 2}}{2} + \varepsilon } \right)|x{|^{ - \frac{{n + 2}}{2} + \varepsilon }}x,&|x| \leqslant 1,\\ \left( { - \frac{{n - 2}}{2} - \varepsilon } \right)|x{|^{ - \frac{{n + 2}}{2} - \varepsilon }}x,&|x| > 1.\end{cases}$

Therefore,

$\displaystyle {|\nabla u_\varepsilon|^2 }(x) = \begin{cases}{\left( { - \frac{{n - 2}}{2} + \varepsilon } \right)^2}|x{|^{ - n + 2\varepsilon }},&|x| \leqslant 1,\\ {\left( { - \frac{{n - 2}}{2} - \varepsilon } \right)^2}|x{|^{ - n - 2\varepsilon }},&|x| > 1.\end{cases}$

So

$\displaystyle\begin{gathered} \int_{{\mathbb{R}^n}} {{{\left| {\nabla {u_\varepsilon }} \right|}^2}dx} = {\left( { - \frac{{n - 2}}{2} + \varepsilon } \right)^2}\left( {\int_0^1 {\frac{1}{{{r^{n - 2\varepsilon }}}}\int_{\partial {B_r}(0)} {d\sigma } dr} + \int_1^{ + \infty } {\frac{1}{{{r^{n + 2\varepsilon }}}}\int_{\partial {B_r}(0)} {d\sigma } dr} } \right) \hfill \\ \qquad\qquad\qquad= {\left( { - \frac{{n - 2}}{2} + \varepsilon } \right)^2}\left( {\frac{{n{\rm vol}({B_1})}}{{2 - 2\varepsilon }} + \frac{{n{\rm vol}({B_1})}}{{2 + 2\varepsilon }}} \right). \hfill \\ \end{gathered}$

Thus

$\displaystyle\mathop {\lim }\limits_{\varepsilon \to 0 + } \int_{{\mathbb{R}^n}} {{{\left| {\nabla {u_\varepsilon }} \right|}^2}dx} = {\left( {\frac{{n - 2}}{2}} \right)^2}n{\rm vol}({B_1}).$

Hence, in order to complete the proof, we need to show that

$\displaystyle\mathop {\lim }\limits_{\varepsilon \to 0 + } {\left( {\int_{{\mathbb{R}^n}} {V(x)|{u_\varepsilon }{|^n}dx} } \right)^{\frac{2}{n}}} > 0.$

Fortunately, this is trivial. We fix some $R>0$ large enough such that

$\displaystyle\int_{|x| \leqslant R} {V(x)dx} > 0.$

Then within the ball $B_R(0)$, it is easy to see that

$\displaystyle |{u_\varepsilon }|(x) \geqslant \frac{1}{{{R^{\frac{{n - 2}}{2} + \varepsilon }}}}.$

Therefore,

$\displaystyle\begin{gathered} \mathop {\lim }\limits_{\varepsilon \to 0 + } {\left( {\int_{{\mathbb{R}^n}} {V(x)|{u_\varepsilon }{|^n}dx} } \right)^{\frac{2}{n}}} > \mathop {\lim }\limits_{\varepsilon \to 0 + } {\left( {\int_{|x| \leqslant R} {V(x)|{u_\varepsilon }{|^n}dx} } \right)^{\frac{2}{n}}} \hfill \\ \qquad\qquad> \mathop {\lim }\limits_{\varepsilon \to 0 + } {\left( {\frac{1}{{{R^{\frac{{n - 2}}{2} + \varepsilon }}}}\int_{|x| \leqslant R} {V(x)dx} } \right)^{\frac{2}{n}}} \hfill \\ \qquad\qquad= {R^{ - \frac{{n - 2}}{n}}}{\left( {\int_{|x| \leqslant R} {V(x)dx} } \right)^{\frac{2}{n}}} > 0. \hfill \\ \end{gathered}$

The proof is now complete.

For more details, we refer the reader to a paper due to de Pino et al. published in J. Funct. Anal. in 2010 [here].

1. Another improved version: the Hardy inequality

$\displaystyle-\Delta - \frac{(n-2)^2}{4}|x|^{-2} \ge 0$

(as operators in $L^2$) can be improved to

$\displaystyle -\Delta - \frac{(n-2)^2}{4}|x|^{-2} \ge -\Delta^{s}-{\rm const}$

for any $0, see http://arxiv.org/abs/0809.3797. Thus the new "kinetic operator"

$\displaystyle -\Delta - \frac{(n-2)^2}{4}|x|^{-2}$

still has the compact embedding properties as in $-\Delta^{s}$ (which is "almost" as good as $-\Delta$).

Comment by hoadai — April 16, 2011 @ 16:16

• Thanks Nam, the preprint you mention has already published at CMP [here].

Comment by Ngô Quốc Anh — April 16, 2011 @ 20:49