Lemma 2.1.7.label Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive. If there exists $\lambda > 0$ such that $(T^{*} + \lambda)$ is injective, then $T$ is essentially self-adjoint.
Proof. Let
be the inner product on $D(T^{*})$, then for any $y \in D(T^{*})$ with $y \perp D(T)$,
By density of $D(T)$ and injectivity of $T^{*} + \lambda$, $y = 0$. Hence $D(T)$ is dense in $D(T^{*})$ with respect to the inner product $\dpb{\cdot, \cdot}E$.
Let $S$ be a semipositive self-adjoint extension of $T$, then $D(S) \subset D(T^{*})$, so $D(T)$ is dense in $D(S)$ with respect to the inner product $\dpb{\cdot, \cdot}E$.
Let $E$ be the completion of $D(T^{*})$ with respect to the inner product $\dpb{\cdot, \cdot}E$ and $\iota: E \to H$ be the extension of the inclusion map, then for any $x, y \in D(T) \subset E$,
Hence $S$ and the Friedrichs extension coincide on $D(T)$. By density of $D(T)$ in $D(S)$, they also coincide on $D(S)$.$\square$