On the existence of the adjoint of a densely defined unbounded operator I am reading proofs on how the adjoint of an unbounded operator $T$ on a Hilbert space $\mathcal{H}$ is well defined provided its domains is a dense linear subspace. The main idea is that if two elements $u,v\in \mathcal{H}$ are such that $$\langle w,u\rangle =\langle w,v\rangle \quad \forall w\in D(T)$$ where $D(T)$ is the (dense) domain of $T$, then $u=v$. This is clear, and clearly is a sufficient argument to show that the adjoint $T^*$ is well-defined, since we define the domain  of $T^*$ as $$D(T^*)=\{ v\in \mathcal{H} \mid \exists \eta \in \mathcal{H}\;  \forall w\in D(T) : \langle Tw,v \rangle=\langle w,\eta \rangle    \} $$ and we define $T^*:D(T^*)\to \mathcal{H}$ as $v\mapsto \eta$, where $\eta$ is as defined above. Our first remark shows that this map is well defined. But my question is: 

Why is $D(T^*)$ non-empty? Under what conditions (other than continuity) can we ensure the existence of such map?

 A: The domain of $\mathcal{D}(T^*)$ always includes $0$, and $T^*0=0$. This is because
$$
        \langle Tx,0\rangle = \langle x,0\rangle,\;\; \forall x\in\mathcal{D}(T).
$$
Another way to write the adjoint relation is as an inner product on the product space $\mathcal{H}\times\mathcal{H}$:
$$
                \langle (x,Tx),(-T^*y,y)\rangle=\langle -x,T^*y\rangle+\langle Tx,y\rangle = 0.   \tag{$\dagger$}
$$
This provides a clue as to how to the define the 
adjoint, and how to prove the existence of it by using orthogonal complements. The graph of the adjoint $T^*$ is the orthogonal complement of the graph of the $T$, but with a negative in one of the coordinates.
This is where it becomes convenient to assume that the graph of $T$, usually denoted by $\mathcal{G}(T)$, is closed in $\mathcal{H}\times\mathcal{H}$, as well as densely-defined. Then an orthogonal complement exists and
$$
       \mathcal{H}\times\mathcal{H}=\mathcal{G}(T)\oplus\mathcal{G}(T)^{\perp}.
$$
That's enough to ensure that $\mathcal{G}(T)^{\perp}$ is the negative transpose of the graph of a closed and densely-defined operator referred to as $T^*$, just as suggested by equation $(\dagger)$. $\mathcal{G}(T)^{\perp}$ is the negative transpose of an operator $T^*$, which is also closed and densely-defined. It is closed because the orthogonal complement is closed.
