Adjoint of the multiplication operator on a Sobolev space

Question

Let $f\colon\mathbb{R}^n\rightarrow\mathbb{C}$ be a bounded function with a bounded first derivative. Then the multiplication operator $H^1(\mathbb{R}^n)\ni x\mapsto A_f x:=fx\in H^1(\mathbb{R}^n)$ is bounded, where $H^1(\mathbb{R}^n)=W^{1,2}(\mathbb{R}^n)$, a Sobolev space. In a comment, user ougoah asked whether the adjoint operator $A_f^*$ is a multiplication operator, too. Here this question will be answered.

Answer 1

$\newcommand{\R}{\mathbb R}$ The answer is: $A_f^*$ is a multiplication operator iff the function $f$ is constant.

Indeed, recall that for all $x$ and $y$ in $H^1:=H^1(\mathbb R^n)$ \begin{equation*} \langle x,y\rangle:=\langle x,y\rangle_{H^1} =\int\Big(x\bar y+\sum_{j=1}^n(D_jx)\,(D_j\bar y)\Big), \end{equation*} where $\int:=\int_{\mathbb R^n}$ and $D_j$ denotes the partial derivative with respect to the $j$th argument.

The condition that $A_f^*$ is a multiplication operator $A_{\bar g}$ for some bounded function $g\colon\mathbb{R}^n\rightarrow\mathbb{C}$ with a bounded first derivative means that \begin{equation*} \langle fx,\bar z\rangle=:L(x,z)=R(x,z):=\langle x,\bar g\bar z\rangle \tag{1} \end{equation*} for all $x$ and $z$ in $H^1$.

So, the "if" part of our iff claim is obvious: if $f$ is constant, then $A_f^*=A_{\bar f}$.

To prove the "only if" part, take any $c=(c_1,\dots,c_n)$ and $a=(a_1,\dots,a_n)$ in $\R^n$ and let \begin{equation*} x(t):=e^{ia\cdot t-|t|^2/2}\quad\text{and}\quad z(t):=e^{i(c-a)\cdot t-|t|^2/2} \end{equation*} for $t=(t_1,\dots,t_n)\in\R^n$, where $a\cdot t:=\sum_1^n a_j t_j$ and $|t|:=\sqrt{t\cdot t}$. Then \begin{align*} R(x,z)&=\int\Big(gzx+\sum_{j=1}^n D_j(gz)\,D_jx\Big) \\ &=\int gz\Big(x-\sum_{j=1}^n D_j^2x\Big) \\ & =\int dt\,g(t)e^{ic\cdot t-|t|^2/2}\sum_{j=1}^n(a_j^2+2ia_jt_j+\tfrac{n+1}n-t_j^2) \end{align*} and, similarly, \begin{equation*} L(x,z)=\int dt\,f(t)e^{ic\cdot t-|t|^2/2}\sum_{j=1}^n((c_j-a_j)^2+2i(c_j-a_j)t_j+\tfrac{n+1}n-t_j^2). \end{equation*} So, in view of (1), $L(x,z)$ and $R(x,z)$ are equal quadratic polynomials in $a_1$, for each $c\in\R^n$. Equating the coefficients of $a_1^2$ in these two polynomials, we see that $$\int dt\,f(t)e^{ic\cdot t-|t|^2/2}=\int dt\,g(t)e^{ic\cdot t-|t|^2/2} $$ for all $c\in\R^n$, which implies $f=g$, which in turn means that (1) can be rewritten as (2) in the previous answer, whence, according to what follows formula (2) in that answer, $f$ is indeed constant.