# Hardy-type inequality for point boundary

Let $f$ be in $W^{2,p}(\mathbb{R}^n)$ for $n\geq 3$ and $p>n/2$, with $f=0$ at the origin. I want to show that the integral $$\int_{B(0,r)} (f |x|^{-2})^p dV <\infty$$ for some small $r>0$. A quick application of Sobolev embedding gives $f\leq C |x|^{2-n/p}$, which then immediately shows $$\int_{B(0,r)} (f|x|^{-2+\epsilon})^p dV<\infty$$ for any $\epsilon>0$.

A Hardy-type inequality seems like it should work in this situation. For instance, from this (though this is hardly the original reference), if $p>n$, Theorem 2 holds for this instance. Let $\Omega$ be an open domain, $d(x)$ be the distance to $\partial \Omega$, and $\Omega_r = \{x\in \Omega: d(x)<r\}$. Then we get $$\int_{\Omega_r} (f d(x)^{-2})^p dV \leq c \int_{\Omega_r} |\nabla^2 f|^p dx$$ for any open domain and $r>0$ small enough. Since all we care about is near the origin, taking $\Omega = B(0,r_0)\setminus\{0\}$ gives the inequality I want.

My question is essentially if that last inequality holds more generally for $p\in (n/2,n]$. Obviously it doesn't hold for some domains, but I only care about my specific circumstance. And really, I don't care if it is exactly of that form. For instance, in Evans' PDE book, 5.8.4 Theorem 7, he proves for $n\geq 3$ and $f\in H^1(B(0,r))$ that $$\int_{B(0,r)} f^2 |x|^{-2} dV \leq C \int_{B(0,r)} |\nabla f|^2 + f^2r^{-2} dV.$$ Something like that would be more than sufficient.

If $p > n/2$ and if $f \in C^2_c (\mathbb{R}^n \setminus \{0\})$ (twice continuously differentiable functions whose support is compact in $\mathbb{R}^n \setminus \{0\}$), then the weighted Hardy inequality to $f$ says that $$\int_{\mathbb{R}^n} \frac{| f (x) |^p}{| x |^{2 p}} \,dx \le \Bigl(\frac{p}{2 p - n}\Bigr)^p \int_{\mathbb{R}^n} \frac{| \nabla f (x) |^p}{| x |^{p}} \,dx$$ Next an application of the classical Hardy inequality gives you, if $p \ne n$, $$\int_{\mathbb{R}^n} \frac{| \nabla f (x) |^p}{| x |^{p}} \,dx \le \Bigl(\frac{p}{p - n}\Bigr)^p \int_{\mathbb{R}^n} |D^2 f (x) |^p \,dx.$$
If $p \in (n/2, n)$, then such functions are dense in the subspace of $W^{2, p} (\mathbb{R}^n)$ of functions that vanish at $0$, and this proves the desired inequality.
(Edit) For a reference for the Hardy inequality, see for example M. Willem, Functional analysis: Fundamentals and applications, 2013, Theorem 6.4.10; the proof adapts immediately to the weighted case and $p > N$.
If $p > n$, then the inequality $W^{2, p} (\mathbb{R}^n)$ of functions that vanish at $0$ together with their derivative. It is important that the derivative vanishes as it can be checked that if $f \in C^2 (\mathbb{R}^n)$ and $f (0) = 0$, then $$\frac{| f (x) |^p}{| x |^{2 p}} \simeq \frac{\rvert \nabla f (0)\lvert}{|x|^p},$$ which is not integrable near the origin $0$ when $p > n$.
(Edit) For $p = n$, the inequality fails. In that space the condition that the derivative vanishes at the origin does not make any sense. A counterexample that vanishes at the origin is obtained by considering the family of functions $f_\alpha$ define by $$f_\alpha (x) = |x| \Bigl(\log \frac{1}{|x|}\Bigr)^\alpha \eta (x),$$ where $\eta$ is a cutoff function that is a nonzero constant in a neighbourhood of the origin and $-\frac{1}{p} \le \alpha < 1 - \frac{1}{p}$.
• Thanks! For $p>n$, in the situation where I want to apply this, I think I could probably get the first derivatives to be zero, since they exist and are continuous. However, what about $p=n$? What happens there? – James Dilts Sep 19 '14 at 16:43