Assume that T be a self-adjoint operator on the Hilbert space $L^2(0,1)=\{f:\int_0^1 |f(x)|^2 dx<\infty\}$ with the domain $D(T)$ satisfying $C^\infty_c(0,1)\subset D(T)$ (here $C^\infty_c(0,1)$ denotes the space of smooth functions with a compact support on $(0,1)$). Let $M$ denote the multiplication operator
$$
Mf(x)=\frac{f(x)}{x^2}, \qquad x\in(0,1),
$$
defined on its natural domain $D(M)=\{f\in L^2(0,1): \int_0^1\Big|\frac{f(x)}{x^2}\Big|^2 dx\}$.
Assume that there exists $C>0$ such that \begin{equation}\label{a1} (*)\qquad \int_0^1 |Mu(x)|^2 dx \le C \int_0^1 |Tu(x)|^2 dx, \qquad u\in C_c^\infty(0,1). \end{equation} Question: Does it follow from $(*)$ that $D(T)\subset D(M)$?
My attempt: Take $f\in D(T)$. Let $(\varphi_n)_{n=1}^\infty$ be a family of smooth functions satisfying: \begin{align*} &(1)\quad 0\le \varphi_1\le\varphi_2\le\dots\le 1, \\ &(2)\quad supp\, \varphi_n\subset(1/n,1-1/n), \\ &(2)\quad \lim_{n\rightarrow \infty}\varphi_n(x)=1, \qquad x\in(0,1), \end{align*} and consider $f_n:=\varphi_n f$. Then $(f_n)\subset C_c^\infty(0,1)$ and it follows from the Lebesgue Monotone Convergence Theorem and our assumption $(*)$ that \begin{align*} \int_0^1 |Mf(x)|^2 dx&=\int_0^1 |f(x)/x^2|^2 dx=\lim_{n\infty} \int_0^1 |f_n(x)/x^2|^2 dx= \lim_{n\rightarrow\infty} \int_0^1 |Mf_n(x)|^2 dx \\ &\le C \limsup_{n\rightarrow\infty} \int_0^1 |Tf_n(x)|^2 dx. \end{align*} I can't see how to use the assumption $f\in D(T)$ to argue that the expression on the right is finite. Do unbounded operators have some kind of continuity on their domains? That would allow us to say that the last expression equals $ C\int_0^1 |Tf(x)|^2 dx$.
It's at least true if we assume that $\mathcal{C}^\infty_c(0,1)$ is a core for $T$, i.e. if it is dense in $D(T)$ equipped with the graph norm defined by $\|f\|_T^2 := \|f\|^2 + \|T(f)\|^2$, where I denote the $L^2$ norm by $\|\cdot\|$.
As a reminder, a self-adjoint operator is in particular closed (because an adjoint is always closed), which is equivalent to saying that its domain is complete for the aforementioned graph norm.
Indeed, consider $f \in D(T)$, and pick $(f_n)_n \subset \mathcal{C}^\infty_c(0,1)$ converging towards $f$ for $\|\cdot\|_T$. This comes down to saying that $f_n \to f$ and $T(f_n) \to T(f)$ in $L^2$.
Up to the choice of a subsequence, we can assume that $(f_n)_n$ converges pointwise to $f$. Thus, we get, using Fatou's lemma: $$\|M(f)\|^2 \leq \liminf_{n \to \infty} \|M(f_n)\|^2 \leq C \liminf_{n \to \infty} \|T(f_n)\|^2 = C \|T(f)\|^2 < \infty$$ Hence: $D(T) \subset D(M)$.