Proof. For any $f, g \in D(T)$, let

then $\dpb{\cdot, \cdot}{E}$ is an inner product on $D(T)$. Let $E$ be the completion of $D(T)$ with respect to the norm induced by $\dpb{\cdot, \cdot}{E}$. Since $\norm{\cdot}_{E} \ge \norm{\cdot}_{H}$, the inclusion $\iota: D(T) \to H$ admits a continuous extension to $E$, which maps it into $H$ as a dense subspace.

To this end, let $\seq{x_n}\subset D(T)$ be a Cauchy sequence in $E$ such that $\norm{x_n}_{H} \to 0$ as $n \to \infty$, then

Hence $x_{n} \to 0$ in $E$, and the extension $\iota: E \to H$ is injective.

Since $\iota: E \to H$ is bounded and injective, it admits a bounded adjoint $\iota^{*}: H \to E$. Let $B = \iota \circ \iota^{*}: H \to H$, then $B$ is a bounded, injective, and self-adjoint.

Now, let $A = B^{-1}: B(H) \to H$, then $A$ is self-adjoint by Lemma 2.1.4. Moreover, for any $x, y \in B(H)$,

So $(A - 1): B(H) \to H$ is the self-adjoint extension of $T$.$\square$