Adjoint of multiplication operator on Sobolev space This is sort of an idle question, and I'll admit I didn't think very hard about it.
Let $H^1 = H^1(\mathbb{R}^n)$ be the Sobolev space with norm $||f||_{H^1}^2 = ||f||_{L^2(\mathbb{R}^n)}^2 + ||\nabla f||_{L^2(\mathbb{R}^n)}^2$.  Suppose $\psi \in C^1(\mathbb{R}^n)$ is a bounded function with bounded gradient.  Then the multiplication operator $Af = \psi f$ is a bounded operator on $H^1$ (with norm at most $\sqrt{||\psi||_\infty^2 + ||\nabla \psi ||_\infty^2}$ ).

What is the adjoint of $A$?

That is, given $g \in H^1$ we wish to find $h$ such that
$$(\psi f, g) + (\nabla(\psi f),\nabla  g) = (f,h) + (\nabla f,\nabla h)$$
for all $f \in H^1$.  I tried some simple manipulations on this expression but didn't get anywhere.
 A: First assume that $g\in H^2(\mathbb{R}^n)$ and $f\in C_c^\infty(\mathbb{R}^n)$. The equation for the adjoint implies
$$
\int f\psi (g-\Delta g)\,dx=\int(f-\Delta f)h\,dx.
$$
Thus $\Delta h\in L^2$ and $(1-\Delta)h=\psi(1-\Delta)g$, i.e., $h=(1-\Delta)^{-1}(\psi(g-\Delta g))$.
To see that this extends continuously to $H^1$, first note that $(1-\Delta)^{-1}$ is continuous from $H^{-1}$ to $H^1$. Moreover,
\begin{align*}
\left|\int\psi(g-\Delta g)\phi\,dx\right|&\leq\left|\int g\psi \phi\,dx\right|+\left|\int\nabla g\cdot\nabla (\phi\psi)\right|\\
&\leq \|\psi\|_\infty\|g\|_2\|\phi\|_2+\|\nabla g\|_2(\|\psi\|_\infty\|\nabla \phi\|_2+\|\nabla \psi\|_\infty\|\phi\|_2)\\
&\leq\max\{\|\psi\|_\infty,\|\nabla \psi\|_\infty\}\|g\|_{H^1}\|\phi\|_{H^1}
\end{align*}
for $\phi\in H^1$, hence $\|\psi(g-\Delta g)\|_{H^{-1}}\leq \max\{\|\psi\|_\infty,\|\nabla \psi\|_\infty\}\|g\|_{H^1}$.
Therefore the map $g\mapsto (1-\Delta)^{-1}(\psi(g-\Delta g))$ can be extended to a bounded linear operator on $H^1$.
