Is $x * d/dx$ closed on $L^2(\mathbb{R})$? More formally, sticking to the one-dimensional case for this problem, let $M_x$ be the multiplication-by-$x$ operator,
$$\begin{split} \mathfrak{D}(M_x) &= \lbrace f(x) \in L^2(\mathbb{R}): x f(x) \in L^2(\mathbb{R}) \rbrace \\
M_x [f](x) &= x f(x) \end{split},$$
and let $D$ be the differentiation-by-$x$ operator
$$ \begin{split} \mathfrak{D}(D) & = \lbrace f(x) \in L^2(\mathbb{R}): f^{\prime}(x) \in L^2(\mathbb{R}) \rbrace ( = \mathcal{H}^1(\mathbb{R}))\\
D[f](x) &= f^{\prime}(x) \end{split},$$
where we allow $f^{\prime}(x)$ to be a distributional derivative.  (Of course, thanks to the Fourier Transform $\mathcal{F}$, $$Df(x)  = \mathcal{F}^{-1} \left[i \xi \mathcal{F}[f](\xi) \right](x),$$
perhaps multiplied by a constant depending on the convention on the Fourier Transform.)  Of course, $M_x$ is closed and self-adjoint (e.g., Reed/Simon, Functional Analysis, Chap. VIII, Section 3, Prop. 1), and $D$ is closed by the same reasoning (being a constant multiple of the self-adjoint $i \frac{d}{dx}$).

Question: Is $M_x \circ D$ closed on the natural domain, i.e., $ \mathfrak{D}(M_X \circ D) = \lbrace f(x) \in L^2(\mathbb{R}): f^{\prime}(x) \in L^2(\mathbb{R}), xf^{\prime}(x) \in L^2(\mathbb{R}) \rbrace$ ?


Note: Of course, replacing $\frac{d}{dx}$ by $i \frac{d}{dx}$ may be simpler to allow self-adjointness, but I prefer to state it in the more elementary way.  Of course, if the proof is easier with $i \frac{d}{dx}$, then please just use that.
Work so far: Let us set up a standard closure proof, so suppose that $f_n(x) \in \mathfrak{D}(M_X \circ D)$ and $f(x) \in L^2(\mathbb{R})$ satisfy
\begin{align} f_n(x) &\overset{L^2}{\rightarrow} f(x) \tag{1}\\
x f_n^{\prime}(x) &\overset{L^2}{\rightarrow} g(x);  \tag{2} \end{align}
we wish to show that $f(x) \in \mathfrak{D}(M_X \circ D)$ and $x f_n^{\prime}(x) \overset{L^2}{\rightarrow} x f^{\prime}(x)$.  
Claim: It suffices to show that 
$$f_n^{\prime}(x) \overset{L^2}{\rightarrow} h(x) \tag{3} $$ for some $L^2$-function $h$.  
Proof: Suppose that (1), (2), and (3) hold (for some $L^2$ functions $\left(f_n\right)_{n = 1}^{\infty}$, $f$, $g$, and $h$).  By (1), (3), and the closure of $D$, it follows that $f(x) \in \mathfrak{D}(D)$ and $f_n^{\prime}(x) \overset{L^2}{\rightarrow} f^{\prime}(x)$, i.e., $h(x) = f^{\prime}(x)$.  Letting $h_n(x) = f_n^{\prime}(x)$, (3) now reads
$$h_n(x) \overset{L^2}{\rightarrow} h(x), \tag{3*}$$
and (2) now reads
$$x h_n(x) \overset{L^2}{\rightarrow} g(x), \tag{2*}$$
so by the closure of $M_x$, $h \in \mathfrak{D}(M_x)$ and $x h_n(x) \overset{L^2}{\rightarrow} x h(x)$, i.e., $g(x) = x h(x)$.  Remembering that $h_n(x) = f_n^{\prime}(x)$ and $h(x) = f^{\prime}(x)$, we have that $f^{\prime}(x) \in \mathfrak{D}(M_x)$, i.e., $f(x) \in \mathfrak{D}(M_x \circ D)$, and $x f_n^{\prime}(x) \overset{L^2}{\rightarrow} x f^{\prime}(x)$.  The closure proof is complete. $\square$

Thus, the problem reduces to proving the claim.  Certainly, for any $\delta > 0$, we can define the bounded measurable function 
$$\varphi_{\delta}(x) = \begin{cases} \frac{1}{x}, & |x| \geq \delta \\ 0, & |x| < \delta \end{cases},$$
which as a multiplier of course defines a bounded (hence continuous) linear operator on $L^2(\mathbb{R})$, so applying to (2),
$$\varphi_{\delta}(x) \cdot x f_n^{\prime}(x) \overset{L^2}{\rightarrow} \varphi_{\delta}(x) g(x). $$
Of course, $\varphi_{\delta}(x) \cdot x $ is the characteristic function $\mathbb{1}_{|x| \geq \delta}(x)$, so
$$ \mathbb{1}_{|x| \geq \delta}(x) \cdot f_n^{\prime}(x) \overset{L^2}{\rightarrow} \varphi_{\delta}(x) g(x). \tag{4} $$
Thus, we seem to have two options: (A) show for some small $\delta$ that $\mathbb{1}_{|x| < \delta}(x) f_n^{\prime}(x)$ also converges to some $L^2$ function, (B) carefully take the limit in (4) as $\delta \to 0$. [I regard (B) as nonobvious: clearly as $\delta \to 0$, for each $n$ separately, the left-hand side tends to $f_n^{\prime}(x)$ in $L^2$-norm, but without any uniformity in $n$, it would be difficult.  Certainly, the right-hand-side will get worse as $\delta \to 0$, without an additional observation.]  Here is where I am stuck.
 A: If the domain of $M_x\circ D$ should contain functions with weak derivative in $L^2$, the claim is not true:
On $\Omega =(-1,1)$ take 
$$
f_n(x) = \begin{cases} 0 & x<0 \\ \min(nx,1) & x\ge 0
\end{cases}
$$
Then $f_n$ converges in $L^2$ to the step function, $xf_n'$ converges in $L^2$ to zero, but the limit does not have a locally integrable weak derivative. To extend this to $\Omega=\mathbb R$, multiply with a smooth cut-off function.
If one understands the expression $g=xf'$ in a distributional sense, then the claim is true: Let $f,g\in L^2(\mathbb R)$ be given. Then we say $g=xf'$ iff it holds
$$
\int_{\mathbb R} f(x) \cdot (x \psi(x))'dx =
\int_{\mathbb R} f(x) \cdot (x \psi'(x) + \psi(x) )dx = -\int_{\mathbb R} g(x)\psi(x)dx \quad \forall \psi\in C_c^\infty(\mathbb R).
$$
Now if $f_n\to f$ and $xf_n'\to g$ in $L^2(\mathbb R)$ then we can pass to the limit in the integrals to obtain $g=xf'$.
A: The operator on your domain is not closed. Consider $f_{\alpha}(x) = e^{-x^2}(x^{2})^{\alpha}$ for $0 < \alpha < 1$. The function $f_{1/4}$ is not in the domain of the operator as you have described it, because
$$
         f_{1/4}'(x)=-2xe^{-x^2}|x|^{1/2}+\frac{1}{2}e^{-x^2}\frac{|x|^{1/2}}{x}\notin L^2(\mathbb{R}), \\
       x f_{1/4}'(x)=-2x^2e^{-x^2}|x|^{1/2}+\frac{1}{2}e^{-x^2}|x|^{1/2}\in L^2(\mathbb{R}).
$$
However, $f_{\alpha}(x)$ is in the domain of your operator for any $0 < \alpha < 1/4$. And,
$$
        \lim_{\alpha\uparrow 1/4}\|f_{\alpha}-f_{1/4}\|=0\\
        \lim_{\alpha\uparrow 1/4}\|xf_{\alpha}'(x)-xf_{1/4}'(x)\|_{L^2}=0.
$$
So your operator $M_x\circ D$ is not closed.
