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?



"Square root" of multiplication operator on Sobolev space
$\newcommand{\R}{\mathbb R}$
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. 


