= Solution
In dimension one, the <Sobolev embedding theorem> gives $H^1(\Omega)\hookrightarrow L^\infty(\Omega)$. Therefore
$$
|\langle A(w,u),v\rangle|\leq\|w\|_\infty^2\|\nabla u\|_2\|\nabla v\|_2,
$$
and, for fixed $w$,
$$
\langle A(w,u)-A(w,z),u-z\rangle=\int_\Omega w^2|\nabla(u-z)|^2\geq0.
$$
Hence \b[the operator is well-defined and monotone in one dimension].
In every dimension $n\geq2$ it is \b[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 $n\geq3$, set $w=r^{-\alpha}$ with $\alpha=(n-2)/3$. The local <Sobolev norm> is finite because
$$
\int_0^\delta r^{n-1-2\alpha}\,dr<\infty,\qquad\int_0^\delta r^{n-3-2\alpha}\,dr<\infty,
$$
whereas
$$
\int_{B_\delta}w^2|\nabla w|^2\,dx\asymp\int_0^\delta r^{n-3-4\alpha}\,dr=\infty,
$$
since $n-3-4\alpha=(-n-1)/3<-1$.
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.
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