Assume we have a positive (so that we can take the square-root by functional calculus) self-adjoint operator $H: D(H) \subset \mathcal{H} \rightarrow \mathcal{H},$ then we can define $V:=D(H^{\frac{1}{2}})$ and denote by $V^*$ the dual to $V$ w.r.t. to the $\mathcal{H}-$ inner-product, i.e. $V \subset \mathcal{H} \subset V^*.$ Now, I read that one can show that $H$ extends to a continuous operator $L(V,V^*).$
But I don't fully see how this construction works, can anybody give me a view details on this?
$V^{*}=$ completion of $D\left(H\right)$ w.r.t. \begin{eqnarray*} \left\Vert x\right\Vert _{V^{*}} & := & \left\Vert H^{-1/2}x\right\Vert ,\;x\in D\left(H\right). \end{eqnarray*} For $x\in D\left(H\right)$, \begin{eqnarray*} \left\Vert x\right\Vert _{V} & = & \left\Vert H^{1/2}x\right\Vert \\ \left\Vert Hx\right\Vert _{V^{*}} & = & \left\Vert H^{-1/2}Hx\right\Vert =\left\Vert H^{1/2}x\right\Vert , \end{eqnarray*} i.e., isometric; and $D\left(H\right)$ is dense in $V$ and $V^{*}$. So $H$ extends to $L\left(V,V^{*}\right)$.