# Domain of $T^\star T$ in Von Neumman's Theorem for closed operators

If $$H$$ is a separable Hilbert complex or real space and $$T: D(T) \subseteq H \to H$$ is a closed densely defined operator then $$T^\star T$$ is densely defined and self adjoint.

Does $$D(T^\star T) = D(T)$$?

For this to hold then $$D(T^\star) \subseteq Ran(T)$$ which is not clear to me.

By definition $$D(T^*T) = \{x \in D(T): Tx \in D(T^*)\}$$. In general you don't have an equality $$D(T^*T) = D(T)$$ (I'd say that equality never happens, except some trivial cases). You are only guaranteed the inclusion $$D(T^*T) \subset D(T)$$.
Also, as you mentioned, it's possible to prove that $$D(T^*T)$$ is densely-defined and self-adjoint for arbitrary closed densely-defined operator $$T$$. See for example Rudin's FA theorem 13.13.
For example consider operator $$T = M_f:L_2(X, \mu) \rightarrow L_2(X, \mu)$$ where $$X$$ is a set, $$\mu$$ is a positive measure, $$f$$ - $$\mu$$-measurable function. $$M_f$$ is defined as follows: $$D(M_f) = \{u \in L_2: fu \in L_2\}$$, $$M_f(u) = fu$$. It is easy to see that $$M_f^* = M_{\overline{f}}$$ and $$M_f^*M_f = M_{|f|^2}$$ and, obviously, $$D(M_f) \ne D(M_f^* M_f)$$, except some trivial cases such as bounded $$f$$. I think if $$f$$ is essentially unbounded ($$f \notin L_\infty$$) then equality never holds.