Hardy inequality in Sobolev spaces

Remarks

For $p = 2$ and $n \ge 3$ there is a short proof. Let $a = (n - 2)/2$ and expand $0 \le \int \abs{\nabla u + a u \, x / \abs{x}^2}^2 \, dx$. The cross term is $a \int \nabla (u^2) \cdot x / \abs{x}^2 \, dx = -a (n - 2) \int u^2 / \abs{x}^2 \, dx$ by integration by parts, since $\operatorname{div}(x / \abs{x}^2) = (n - 2) / \abs{x}^2$. Collecting terms gives $\frac{(n-2)^2}{4} \int u^2 / \abs{x}^2 \, dx \le \int \abs{\nabla u}^2 \, dx$, which is the statement.

The constant $\big( \frac{p}{n - p} \big)^p$ is the best possible, but there is no nonzero function for which equality holds. By density, the inequality extends to every $u$ in the Sobolev space $W^{1,p}(\R^n)$. It is the $n$-dimensional analogue of the one-dimensional Hardy's inequality.

The hypothesis $p < n$ is needed. For $p = n$, take $u$ smooth with compact support and $u(0) \ne 0$: then $\abs{u}^n / \abs{x}^n$ is not integrable near $0$, while the right side is finite.

Going deeper

For a student

Hardy's inequality controls singular potentials by kinetic energy. In quantum mechanics it shows that the operator $-\Delta - c / \abs{x}^2$ on $\R^n$ is bounded below when $c \le (n-2)^2 / 4$. In partial differential equations it is used to treat lower-order terms with singular coefficients and to study equations with critical singular weights.