On the adjoint of a densely defined unbounded operator

Question

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.

Answer 1

$D(T^\ast)=K$ 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 $T^{\ast\ast}$.