Theorem: Existence and Uniqueness of Adjoint of Densely-Defined Linear Operator

Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\Bbb F$.
Let $\struct {\map D T, T}$ be a densely defined linear operator on $\HH$.
For each $y \in \HH$, define the linear functional $f_x : \map D T \to \Bbb F$ by:
:$\map {f_y} x = \innerprod {T x} y$ for each $x \in \map D T$.
Define:
:$\map D {T^\ast} = \set {y \in H : f_y \text { is a bounded linear functional}}$
Then there exists a unique linear transformation $T^\ast : \map D {T^\ast} \to \HH$ with:
:$\innerprod {T x} y = \innerprod x {T^\ast y}$ for all $x \in \map D T$ and $y \in \map D {T^\ast}$.

Proof:
We first check that $\map D {T^\ast}$ is a linear subspace of $\HH$, so that the question whether $T^\ast$ be a linear transformation is well-posed.
Let $u, v \in \map D {T^\ast}$ and $\alpha \in \Bbb F$. 
Then there exists real numbers $M_1, M_2 > 0$ such that: 
:$\cmod {\map {f_u} x} \le M_1 \norm x$
and:
:$\cmod {\map {f_v} x} \le M_2 \norm x$
for each $x \in \map D T$.
Then, we have:
{{begin-eqn}}
{{eqn	| l = \cmod {\map {f_{\alpha u + v} } x}
		| r = \cmod {\innerprod {T x} {\alpha u + v} }
}}
{{eqn	| r = \cmod {\overline \alpha \innerprod {T x} u + \innerprod {T x} v}
}}
{{eqn	| o = \le
		| r = \cmod {\overline \alpha} \cmod {\innerprod {T x} u} + \cmod {\innerprod {T x} v}
		| c = Triangle Inequality
}}
{{eqn	| r = \cmod \alpha \cmod {\innerprod {T x} u} + \cmod {\innerprod {T x} v}
		| c = Complex Modulus equals Complex Modulus of Conjugate
}}
{{eqn	| o = \le
		| r = \paren {M_1 \cmod \alpha + M_2} \norm x
}}
{{end-eqn}}
for each $x \in \map D T$.
So, $f_{\alpha u + v}$ is a bounded linear functional.
So $\alpha u + v \in \map D {T^\ast}$.
So from One-Step Vector Subspace Test, we have that $\map D {T^\ast}$ is a linear subspace of $\HH$.
Now let $y \in \map D {T^\ast}$.
Then:
:$\map {f_y} x = \innerprod {T x} y$ defines a bounded linear functional $\map D T \to \Bbb F$
So, from Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain, there exists a bounded linear functional $F_y : \HH \to \Bbb F$ such that: 
:$\map {F_y} x = \innerprod {T x} y$ for all $x \in \map D T$.
Then from the Riesz representation theorem there exists a unique $z_y \in \HH$ such that: 
:$\map {F_y} x = \innerprod x {z_y}$ for all $x \in \map D T$.
For each $y \in \map D {T^\ast}$, let: 
:$T^\ast y = z_y$
Then: 
:$\innerprod {T x} y = \innerprod x {T^\ast y}$ for all $x \in \map D T$ and $y \in \map D {T^\ast}$.
It remains to verify that $T^\ast$ is linear and that it is unique. 
Note that uniqueness is not immediate, since $z_y$ is the unique element of $\HH$ such that $\innerprod {T x} y = \innerprod x z$ for ''all'' $x \in \HH$, there may exist other $z$ for which this only holds for $\map D T$, and we need to verify that this is not the case.
Let $u, v \in \map D {T^\ast}$ and $\alpha \in \Bbb F$. 
Then for all $x \in \map D T$ we have: 
{{begin-eqn}}
{{eqn	| l = \innerprod x {\map {T^\ast} {\alpha u + v} }
		| r = \innerprod {T x} {\alpha u + v}
}}
{{eqn	| r = \overline \alpha \innerprod {T x} u + \innerprod {T x} v
		| c = Inner Product is Sesquilinear
}}
{{eqn	| r = \overline \alpha \innerprod x {T^\ast u} + \innerprod x {T^\ast v}
}}
{{eqn	| r = \innerprod x {\alpha T^\ast u + T^\ast v}
		| c = Inner Product is Sesquilinear
}}
{{end-eqn}}
Since $\map D T$ is dense in $\HH$, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ converging to $\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v}$.
We have: 
:$\innerprod {x_n} {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } = 0$ for each $n \in \N$.
So, from Inner Product is Continuous, we have: 
:$\innerprod {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } {\alpha T^\ast u + T^\ast v - \map {T^\ast} {\alpha u + v} } = 0$
So, from the definition of the inner product, we have: 
:$\map {T^\ast} {\alpha u + v} = \alpha T^\ast u + T^\ast v$
Now, we look to uniqueness.
Suppose for some $y \in \map D {T^\ast}$ that $z_y^{(1)}, z_y^{(2)}$ are such that: 
:$\innerprod {T x} y = \innerprod x {z_y^{(1)} } = \innerprod x {z_y^{(2)} }$ for $y \in \map D {T^\ast}$ and all $x \in \map D T$.
Then, we would have: 
:$\innerprod x {z_y^{(1)} - z_y^{(2)} } = 0$ for each $x \in \map D T$.
Since $\map D T$ is dense, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ converging to $z_y^{(1)} - z_y^{(2)}$. 
Then: 
:$\innerprod {x_n} {z_y^{(1)} - z_y^{(2)} } = 0$ for each $n \in \N$.
Then, taking $n \to \infty$, we have: 
:$\innerprod {z_y^{(1)} - z_y^{(2)} } {z_y^{(1)} - z_y^{(2)} } = 0$ 
from Inner Product is Continuous.
So, from the definition of the inner product, we have: 
:$z_y^{(1)} = z_y^{(2)}$
justifying that $T^\ast$ is unique.
{{qed}}