Calculating the form domain of an operator

Question

I am reading the book "Mathematical Methods in Quantum Mechanics" by Gerald Teschl and just came across the concept of a form domain.
It is defined for non-negative operators i.e $<\phi,A \phi> \geq 0$ by taking the completion of $D(A)$ w.r.t the scalar product $<\phi,\psi>_A := <\phi,(A+1)\psi>$ and then by identifying the limit of a Cauchy sequence w.r.t that scalar product with the limit of the same sequence w.r.t the ordinary one. (I hope I have made myself clear...)
Now it is possible to extend the quadratic form of $A$ to this new domain.
Next up comes an example with a multiplication operator on $L^2$ with a non-negative function $A(x)$ and it is stated, that the form domain of $A$ is $D(A^{\frac{1}{2}})$.
This however with a "show this" note I'm being unable to follow.
Somehow it makes sense to me, when looking at the quadratic form, but i fail to prove it.

Answer 1

The form is $<\phi,\phi>_{A}=\|A^{1/2}\phi\|^{2}_{L^{2}}+\|\phi\|^{2}_{L^{2}}$ is defined on $\mathcal{D}(A)$, a subspace consisting of all $\phi \in L^{2}$ such that $A\phi \in L^{2}$. Notice that the form norm is the same as the graph norm of $A^{1/2}$. So you want to show that the restriction $A^{1/2}|_{\mathcal{D}(A)}$ of $A^{1/2}$ has closure equal to $A^{1/2}$. If this isn't true, then there is a continuous linear functional $\Phi$ on the graph of $A^{1/2}$ which vanishes on the domain of the restriction. That is, there exists $g \in \mathcal{D}(A^{1/2})$ such that $$ \Phi(f) = (f,g)_{L^{2}}+(A^{1/2}f,A^{1/2}g)_{L^{2}}=0,\;\;\; f\in\mathcal{D}(A). $$ This may be rewritten as $$ (Af,g)_{L^{2}}=-(f,g)_{L^{2}},\;\;\; f \in \mathcal{D}(A). $$ Hence, $g \in \mathcal{D}(A)$ and $Ag=-g$ because $\mathcal{D}(A)$ is dense. But that means $(A+1)g=0$ and, hence, $g=0$ because $A+1 \ge 1$. Thus, the closure of $A^{1/2}|_{\mathcal{D}(A)}$ is $A^{1/2}$. Hence, the form domain is $\mathcal{D}(A^{1/2})$.