Sharp Sobolev threshold for a radial power

For $x\mathrm{\ne }0$, coordinate differentiation gives $v_i(x)=-a{x}_i\mathrm{\mid }x{\mathrm{\mid }}^{-a-2}$ and $\mathrm{\mid }\mathrm{\nabla }u(x)\mathrm{\mid }=a\mathrm{\mid }x{\mathrm{\mid }}^{-a-1}$. On $B\setminus \{0\}$ these functions are continuous; assigning finite values at $0$ and extending by zero off $B$ is finite Borel gluing by [F5]. Hence the representatives, their finite-power integrands, and the radial functions used with [F3] are Borel and Lebesgue measurable under [F1].

Put $I_q(R)={\int }_0^R{r}^{q}dr$. By [F4], $I_q(R)$ is finite exactly for $q>-1$ and then $I_q(R)\le C_q{R}^{q+1}$. Applying the polar formula [F3] to $\mathrm{\mid }u{\mathrm{\mid }}^p$ and to the candidate gradient magnitude gives ${\int }_B\mathrm{\mid }u{\mathrm{\mid }}^p=\sigma (S^{n-1})I_{n-1-ap}(1)$ and ${\int }_B\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^p=a^p\sigma (S^{n-1})I_{n-1-p(a+1)}(1)$. The factor $\sigma (S^{n-1})$ is finite and positive by [F3, F5]; for $0<R\le 1$ the same formulas yield ${\int }_{B(0,R)}\mathrm{\mid }u\mathrm{\mid }\le C{R}^{n-a}$ when $a<n$ and ${\int }_{B(0,R)}\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le C{R}^{n-a-1}$ when $a<n-1$.

The first polar identity in step 2.1 shows that $u\in L^p(B)$ exactly when $ap<n$, since changing $u(0)$ affects only the null singleton [F3]. For every $M>0$, choose $0<R<\min \{1,M^{-1\mathrm{/}a}\}$; then $\mathrm{\mid }u(x)\mathrm{\mid }>M$ on $B(0,R)\setminus \{0\}$, a set of positive measure by [F5] and [F3]. Thus $u$ is essentially unbounded and cannot lie in $L^{\mathrm{\infty }}(B)$. By [F2], it also cannot lie in $W^{1,\mathrm{\infty }}(B)$.

Suppose $1\le p<\mathrm{\infty }$ and $p(a+1)<n$. Then $a<n-1$, so step 2.1 places $u$ and every $v_i$ in $L^p(B)$ and gives the small-ball $L^1$ bounds. Fix one bump $\rho$ from [F6], set ${\eta }_{\epsilon }(x)=\rho (x\mathrm{/}\epsilon )$, and ${\chi }_{\epsilon }=1-{\eta }_{\epsilon }$ for $0<\epsilon <1\mathrm{/}2$. Then ${\chi }_{\epsilon }=0$ on $B(0,\epsilon )$, ${\chi }_{\epsilon }=1$ off $B(0,2\epsilon )$, and $\mathrm{\mid }{\mathrm{\partial }}_i{\chi }_{\epsilon }\mathrm{\mid }\le C\mathrm{/}\epsilon$. The product $w_{\epsilon }={\chi }_{\epsilon }u$, defined as zero near the origin, is $C^1(B)$ and has classical partial ${\mathrm{\partial }}_i{w}_{\epsilon }={\chi }_{\epsilon }{v}_i+u{\mathrm{\partial }}_i{\chi }_{\epsilon }$ away from $0$ by [F6]. For each real test $\phi$, [F7] gives ${\int }_B{w}_{\epsilon }{\mathrm{\partial }}_i\phi =-{\int }_B{\mathrm{\partial }}_i{w}_{\epsilon }\phi$. The left-side error from replacing $w_{\epsilon }$ by $u$ is at most $\mathrm{\parallel }{\mathrm{\partial }}_i\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{B(0,2\epsilon )}\mathrm{\mid }u\mathrm{\mid }=O({\epsilon }^{n-a})$; the missing $v_i$ term is $O({\epsilon }^{n-a-1})$, and the cutoff term is at most $C\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{\epsilon }^{-1}{\int }_{B(0,2\epsilon )}\mathrm{\mid }u\mathrm{\mid }=O({\epsilon }^{n-a-1})$. All tend to zero because $a<n-1$. Thus ${\int }_{B}u{\mathrm{\partial }}_i\phi =-{\int }_B{v}_i\phi$ for real tests. Split a complex test into real and imaginary parts and use [F8] to obtain the same identity for every complex test.

Conversely, suppose $u\in W^{1,p}(B)$ for finite $p$. For each $i$ its weak derivative has an $L^p$ representative. On $B\setminus \{0\}$, locality and classical compatibility in [F7], followed by uniqueness in [F7], identify that representative almost everywhere with $v_i$ from step 1.1. The origin is null by [F3]. Pointwise, $\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^p\le n^{p\mathrm{/}2}{\sum }_{i=1}^n\mathrm{\mid }{D}_{i}u{\mathrm{\mid }}^p$, so the gradient magnitude is in $L^p(B)$. The second polar identity in step 2.1 can be finite only if $p(a+1)<n$; at or above the threshold [F4] gives divergence.

If $p(a+1)<n$, then $ap<p(a+1)<n$, so step 3.1 gives $u\in L^p$; step 3.2 gives all first weak derivatives in $L^p$, and [F2] gives $u\in W^{1,p}$. Step 3.3 proves the converse. Since $a>0$ and $p\ge 1$, the condition is impossible when $n=1$. In the admissible range step 3.2 identifies $D_{i}u=[v_i]$; [F3] makes the chosen value at $0$ irrelevant. Together with the $p=\mathrm{\infty }$ conclusion of step 3.1, this proves every assertion.