# A boundary term in a Hardy inequality

If we look at the Hardy inequality in A technical step in proving Hardy's inequality, the answer gets to a point where we almost have the inequality except for a boundary term $$\frac{C}{r}\int_{\partial B(0, r)} u^2dS$$ and we want to obtain $$\frac{C}{r^2}\int_{B(0, r)} u^2dx$$.

But I can't see how the two are comparable: if $$r = 1$$, so that we are comparing $$\int_{\partial B} u^2dS$$ and $$\int_B u^2dx$$ where $$B$$ is the unit ball, then take $$u^2$$ radial and equal to $$u^2(r) = r^\alpha$$. Then the first integral is $$\omega_{n-1}$$, the volume of the $$(n-1)$$-sphere, since $$u^2\rvert_{\partial B} \equiv u^2(1) = 1$$, but the second integral is $$\omega_{n-1}\int_0^1 r^{\alpha}r^{n-1}dr = \frac{1}{\alpha+n}\omega_{n-1}$$ using spherical coordinates/the coarea formula. If $$\alpha$$ is large enough then there is no constant for which we find $$\int_{\partial B} u^2dS \leq C\int_B u^2dx$$. What am I missing?

Your example is correct, but the issue is with what we want to show. The version of the inequality with $$r$$ that is claimed is $$r\int_{\partial B(0,r)} u^2\,dS ≤ C\int_{B(0,r)} u^2 + r^2|Du|^2\,dx,\tag{1}\label{one}$$ instead of $$r\int_{\partial B(0,r)}u^2\,dS ≤ C\int_{B(0,r)}u^2\,dx,$$ which is false, as you showed. The inequality \eqref{one} is obtained by applying the divergence theorem and noticing that $$r\int_{\partial B(0,r)}u^2\,dS = \int_{B(0,r)}\mathrm{div}(\vec x u^2)\,dx = \int_{B(0,r)} nu^2 + 2uDu\cdot \vec x\,dx.$$ If we add the term $$\int_B|Du|^2\,dx$$ for the example you gave, $$|Du(r)|^2\sim (\alpha r^{\frac\alpha2-1})^2 = \alpha^2r^{\alpha-2}$$. For large $$\alpha$$, the right-hand side of \eqref{one} is bounded by the second term, which is approximately $$\int_0^1 \alpha^2r^{\alpha-2}r^{n-1}\,dr \sim \alpha,$$ while the left-hand side is $$\sim 1$$, as you showed, and these are consistent.
• Thanks for the answer - I didn't have access to the book so it just seemed like the boundary term was being compared in that way. Just to confirm, is the inequality (1) obtained by the fundamental theorem of calculus? It looks like $u^2 + \lvert{Du\rvert}^2 \geq 2\lvert{u\rvert}\lvert{Du\rvert} = \lvert{\nabla u^2\rvert}$. May 15 '21 at 6:07
• @meaninglesspen: That's correct—the left-hand side of (1) is equal to $∫_{B(0,r)}\mathrm{div}(\vec{x}u^2)\,dx = ∫_{B(0,r)}nu^2 + 2uDu\cdot \vec{x}\,dx$. May 15 '21 at 19:18