Adjoint of Symmetric Densely-Defined Linear Operator Extends Operator

Theorem
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.

Let $\struct {\map D T, T}$ be a symmetric densely-defined linear operator.

Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $T$.

Then $\map D T \subseteq \map D {T^\ast}$ and:


 * $T x = T^\ast x$ for each $x \in \map D T$.

Proof
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 $x \in \map D T$.

We show that for $y \in \map D T$, $f_y$ is bounded.

Let $y \in \map D T$, then:


 * $\innerprod {T x} y = \innerprod x {T y}$ for each $x \in \map D T$.

Then we have:

So $f_y$ is bounded and $\map D T \subseteq \map D {T^\ast}$.

Now, for $x, y \in \map D T$, we have:


 * $\innerprod {T x} y = \innerprod x {T y}$

since $\struct {\map D T, T}$ is symmetric and:


 * $\innerprod {T x} y = \innerprod x {T^\ast y}$

from the definition of the adjoint.

So we have:


 * $\innerprod x {T y - T^\ast y} = 0$

for each $x \in \map D T$.

Taking a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $x_n \to T y - T^\ast y$.

Since:


 * $\innerprod {x_n} {T y - T^\ast y} = 0$

for each $n \in \N$, we have:


 * $\innerprod {T y - T^\ast y} {T y - T^\ast y} = \norm {T y - T^\ast y}^2 = 0$

from Inner Product is Continuous.

So $T y = T^\ast y$ for each $y \in \map D T$.