For $T$ a densely defined linear, not necessarily bounded operator on a Hilbert space $\mathscr{H}$, and $T^{**}$ the adjoint of $T$'s adjoint, I read somewhere that $\text{ran}(T)=\text{ran}(T^{**})$. Is that true? I know that $\text{ran}(T)\subset \text{ran}(T^{**})$ because of $\text{dom}(T)\subset \text{dom}(T^{**})$, which also gives $\ker(T)\subset \ker(T^{**})$, but I wouldn't know how to prove $\text{ran}(T^{**})\subset \text{ran}(T)$.
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3$\begingroup$ Simple example, $T = \operatorname{id}\lvert_D$ for a dense subspace $D\subset \mathscr{H}$. If $D$ isn't the whole space, we have $\operatorname{Ran} T \subsetneqq \operatorname{Ran}(T^{\ast\ast})$. $\endgroup$– Daniel FischerCommented Sep 24, 2015 at 12:04
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$\begingroup$ @DanielFischer As $\text{ran}(T) = D$ and $T^* = \text{id}_{\mathcal{H}}$ and thus $T^{**} = \text{id}_{\mathcal{H}}$, too, yielding $\text{ran}(T^{**}) = \mathcal{H}$, right? $\endgroup$– ViktorSteinCommented Dec 23, 2019 at 21:04
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$\begingroup$ @ViktorGlombik Yup. Generally, taking a closable but not closed operator has a very good chance of producing a $T$ with $\operatorname{ran} T \subsetneqq \operatorname{ran}(T^{\ast\ast})$. $\endgroup$– Daniel FischerCommented Dec 25, 2019 at 17:11
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Your result is true if $T$ is a closed operator (i.e. the graph of $T$ is a closed subspace of $H \times H$) because we can prove that the closure of $T$ (ie: the smallest closed extension of $T$), $\overline T = T^{**}$ ($\overline T = T$ if $T$ is a closed operator). For a proof (easy only) you may see any standard book which deals with unbounded operators for eg: Reed and Simon, Functional Analysis Vol 1, Theorem VIII.1 (revised and enlarged version).
