= Squared Sobolev gradient counterexample
{title2=$w\in H^1\ \not\Longrightarrow\ w\nabla w\in L^2\quad(n\geq2)$}
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 $n\geq3$, 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.
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