 Solution

In dimension one, the Sobolev embedding theorem gives $H^1(\Omega )\hookrightarrow L^{\infty }(\Omega )$. Thereforeand, for fixed $w$,Hence the operator is well-defined and monotone in one dimension.

In every dimension $n\ge 2$ it is not well-defined on all of $H^1\times H^1$. We exhibit $w=u=v\in H^1$ for which the purported pairing $\int w^2|\nabla w{|}^2$ is infinite. Choose a ball compactly contained in $\Omega$, put $r=|x-x_0|$, and multiply each following radial expression by a smooth cutoff function that equals one near $r=0$.

For $n\ge 3$, set $w=r^{-\alpha }$ with $\alpha =(n-2)/3$. The local Sobolev norm is finite becausewhereassince $n-3-4\alpha =(-n-1)/3<-1$.

For $n=2$, take $w=(\log (1/r){)}^{1/3}$ on a sufficiently small ball. With $s=\log (1/r)$, the gradient integral is proportional to ${\int }^{\infty }{s}^{-4/3}ds<\infty$, while the weighted integral is proportional to ${\int }^{\infty }{s}^{-2/3}ds=\infty$. Its $L^2$ integral is finite because the area factor contributes exponential decay in $s$. These singular functions have the indicated weak derivatives; deleting a small central ball and passing to the limit in integration by parts leaves no boundary term.

Thus $w^2\nabla u\cdot \nabla v$ need not be integrable in dimensions at least two. This is the squared Sobolev gradient counterexample. With the additional restriction $w\in L^{\infty }$, the form is well-defined and monotone in any dimension.