Subcritical $W^{1,p}$ is not closed under multiplication

The choice $\alpha <n\mathrm{/}p-1$ gives $p(\alpha +1)<n$, so by [F2] the power function $g$ with $g(0)=0$ and $g(x)=\mathrm{\mid }x{\mathrm{\mid }}^{-\alpha }$ for $x\mathrm{\ne }0$ lies in $W^{1,p}(B;\mathbb{R})$. The other inequality $\alpha \ge (n\mathrm{/}p-1)\mathrm{/}2$ gives $p(2\alpha +1)\ge n$, equivalently the radial exponent $q=n-1-p(2\alpha +1)$ satisfies $q\le -1$. Since $p<n$ and $p\ge 1$ we have $n\mathrm{/}p-1>0$, so the displayed interval for $\alpha$ is nonempty and contained in $(0,\mathrm{\infty })$.

Fix $\chi$ as in [F6] and $u=\chi g$ on $B$; then $u$ agrees with the definition of the Statement off the origin and $u(0)=0$. Since $\chi \in C_c^{\mathrm{\infty }}(B)$ and $g\in W^{1,p}(B;\mathbb{R})$, [F7] gives $u\in W^{1,p}(B;\mathbb{R})$, with weak gradient represented by the Leibniz formula ${\mathrm{\partial }}_i(\chi g)=({\mathrm{\partial }}_i\chi )g+\chi {\mathrm{\partial }}_{i}g$.

Suppose for contradiction that $u^2\in W^{1,p}(B;\mathbb{R})$, and let $w\in L^p(B)$ be a representative of its weak gradient. On the punctured ball ${\mathrm{\Omega }}^{\circ }=B\setminus \{0\}$ the function $u^2={\chi }^2\mathrm{\mid }x{\mathrm{\mid }}^{-2\alpha }$ is $C^1$, with classical gradient $z_i={\mathrm{\partial }}_i({\chi }^2\mathrm{\mid }x{\mathrm{\mid }}^{-2\alpha })=2\chi ({\mathrm{\partial }}_i\chi )\mathrm{\mid }x{\mathrm{\mid }}^{-2\alpha }-2\alpha {\chi }^2{x}_i\mathrm{\mid }x{\mathrm{\mid }}^{-2\alpha -2}\mathrm{.}$ By [F8] the classical gradient $z$ is the weak gradient of $u^2{\mathrm{\mid }}_{{\mathrm{\Omega }}^{\circ }}$, while restricting the global weak gradient $w$ to the open subset ${\mathrm{\Omega }}^{\circ }$ gives another weak gradient of the same restriction; uniqueness almost everywhere on ${\mathrm{\Omega }}^{\circ }$ therefore gives $w=z$ almost everywhere on ${\mathrm{\Omega }}^{\circ }$.

On $B(0,1\mathrm{/}2)$ the cutoff satisfies $\chi =1$ and ${\mathrm{\partial }}_i\chi =0$, so step 3.1 gives $\mathrm{\mid }z(x)\mathrm{\mid }=2\alpha \mathrm{\mid }x{\mathrm{\mid }}^{-2\alpha -1}$ there. Since $\{0\}$ is null, [F5] and [F4] yield ${\int }_B\mathrm{\mid }w{\mathrm{\mid }}^p={\int }_{{\mathrm{\Omega }}^{\circ }}\mathrm{\mid }z{\mathrm{\mid }}^p\ge (2\alpha {)}^p\sigma (S^{n-1}){\int }_0^{1\mathrm{/}2}{r}^{n-1-p(2\alpha +1)}dr=+\mathrm{\infty },$ because the exponent $q=n-1-p(2\alpha +1)$ satisfies $q\le -1$ by step 1.1 and the surface measure $\sigma (S^{n-1})$ is finite and positive. This contradicts $w\in L^p(B)$, so $u^2\mathrm{\notin }{W}^{1,p}(B;\mathbb{R})$.

The counterexample is therefore complete: $u$ is a $W^{1,p}$ function with $u\cdot u\mathrm{\notin }{W}^{1,p}$, in the range $n\ge 2$, $1\le p<n$. The endpoint $p=n$ is excluded by the hypothesis, and the construction degenerates at $\alpha =0$ in dimension $n=1$, where Sobolev functions are continuous and multiplication is well behaved; neither case is claimed here. The only choice principle used is Countable Choice [F1], inherited from the sharp radial example and spent through the polar-coordinate interface [F5]; no full Axiom of Choice is used. $\mathrm{\square }$