TITLE: On the adjoint of a densely defined unbounded operator
QUESTION [0 upvotes]: Let $H,K$ be Hilbert spaces, $D(T)\subset H$ be a dense vector subspace.
Suppose $T: D(T) \rightarrow K$ and $D(T^*)=K$. What can you say about $T$?
By definition, we have $D(T^*)= \{k \in K: \exists x\in H , \langle k, Th \rangle= \langle x, h \rangle, \forall h \in D(T) \}$.
I know that since $D(T)$ is dense and such $x$ exists, we can define the adjoint of $T$ by $T^*: D(T^*) \rightarrow H$ by $T^*(h)=x$.
But not sure what I can say about $T$.
Thank you.

REPLY [2 votes]: $D(T^\ast)=K$ is equivalent to the boundedness of $T$. I assume you already know that bounded operators have an everywhere defined adjoint, so I will only prove the converse implication.
First recall that the adjoint of a densely defined operator is always closed. Since $D(T^\ast)=K$ is complete, this implies that $T^\ast$ is bounded by the closed graph theorem. Thus also its adjoint $T^{\ast\ast}$ is bounded. Moreover, if $\xi\in D(T)$ and $\eta\in D(T^\ast)$, then
$$
\langle \xi,T^\ast \eta\rangle=\langle T\xi,\eta\rangle.
$$
Thus $\xi \in D(T^{\ast\ast})$ and $T^{\ast\ast}\xi=T\xi$. Therefore $T$ is bounded as the restriction of the bounded operator $T^{\ast\ast}$.
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