"Square root" of multiplication operator on Sobolev space Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a non-negative, smooth, uniformly bounded function with uniformly bounded first derivative. Then $f$ defines a bounded operator on $L^2(\mathbb{R}^n)$ as well as on $H^1(\mathbb{R}^n)$, the first Sobolev space. In $L^2(\mathbb{R}^n)$ the following equality holds: for $e_1,e_2\in C_c^\infty(\mathbb{R}^n)$,
$$\langle fe_1,e_2\rangle_{L^2}=\langle\sqrt{f}e_1,\sqrt{f}e_2\rangle_{L^2}.$$
Question: Is there a way to do this in $H^1$, ie does there exist a function $g_f$ such that
$$\langle fe_1,e_2\rangle_{H^1}=\langle g_f(e_1),g_f(e_2)\rangle_{H^1}?$$
Follow-up question: Is the adjoint $f^*:H^1\rightarrow H^1$ also given by function multiplication?
 A: $\newcommand{\R}{\mathbb R}$
The square root of the multiplication operator does not exist unless the function $f$ is constant (and, obviously, the  square root exists if $f$ is constant). 
Indeed, suppose that there exists a function $g_f$ such that
\begin{equation*}
 \langle fx,y\rangle_{H^1}=\langle g_f(x),g_f(y)\rangle_{H^1} \tag{1}
\end{equation*}
for all $x$ and $y$ in $C_c^\infty(\R^n)$. The simple but crucial observation is that $\langle g_f(x),g_f(y)\rangle_{H^1}=\overline{\langle g_f(y),g_f(x)\rangle_{H^1} }$, so that (1) implies
$$\langle fx,y\rangle_{H^1}=\overline{\langle fy,x\rangle_{H^1}},
$$
which can be rewritten as
\begin{equation*}
 \int_{\R^n}\sum_{j=1}^n (D_jf)\,(x\,D_jz-z\,D_jx)=0, \tag{2} 
\end{equation*}
where $z:=\bar y$ and $D_j$ denotes the partial derivative with respect to the $j$th argument. 
Take any $x\in C_c^\infty(\R^n)$ and any $a=(a_1,\dots,a_n)\in\R^n$, and then let $z:=xe_a$, where the function $e_a$ is defined by the formula 
$e_a(t_1,\dots,t_n):=e^{i(a_1t_1+\dots+a_nt_n)}$ for $t=(t_1,\dots,t_n)\in\R^n$. Then $x\,D_jz-z\,D_jx=ia_jx^2e_a$ for all $j=1,\dots,n$. So, (2) yields $\sum_{j=1}^n a_jD_jf=0$ for all $(a_1,\dots,a_n)\in\R^n$, which means that $f$ is constant. 
