A boundary term in a Hardy inequality

Question

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?

Answer 1

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.