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.


1 Answer 1


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.


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