# Domain of double adjoint

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 $$\text{ran}(T^{**})\subset \text{ran}(T)$$.

• Simple example, $T = \operatorname{id}\lvert_D$ for a dense subspace $D\subset \mathscr{H}$. If $D$ isn't the whole space, we have $\operatorname{Ran} T \subsetneqq \operatorname{Ran}(T^{\ast\ast})$. Commented Sep 24, 2015 at 12:04
• @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? Commented Dec 23, 2019 at 21:04
• @ViktorGlombik Yup. Generally, taking a closable but not closed operator has a very good chance of producing a $T$ with $\operatorname{ran} T \subsetneqq \operatorname{ran}(T^{\ast\ast})$. Commented Dec 25, 2019 at 17:11

## 1 Answer

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).