Squared Sobolev gradient counterexample ($w\in H^1/⟹w\nabla w\in L^2(n\ge 2)$)

In dimensions at least two a function in the Sobolev space $H^1$ can have finite norm while $\int w^2|\nabla w{|}^2=\infty$. Localize the following radial examples with a smooth cutoff function in an interior ball. For $n\ge 3$, set $w(r)=r^{-\alpha }$, $\alpha =(n-2)/3$. Its gradient energy has radial exponent $n-3-2\alpha >-1$, but its weighted gradient energy has exponent $n-3-4\alpha <-1$. For $n=2$, take $w(r)=(\log (1/r){)}^{1/3}$ near zero. Substituting $s=\log (1/r)$ makes the two gradient integrals proportional to ${\int }^{\infty }{s}^{-4/3}ds$ and ${\int }^{\infty }{s}^{-2/3}ds$, respectively. The first converges and the second diverges. In dimension one the Sobolev embedding theorem gives $H^1\subset L^{\infty }$, preventing this failure. These examples show why a nonnegative coefficient $w^2$ need not even define a bounded gradient form when only $w\in H^1$ is assumed.