Positive self-adjoint operators and norm resolvent convergence I recently came across a reference to the following Theorem (Simon/Reed, Methods of Modern Mathematical Physics, viii.25) and am now trying to figure out a proof for it:
If $A_n$ and $A$ are positive self-adjoint Operators with a common form domain and 
$$ \sup_{0 \neq \psi \in D} \frac{|\langle \psi, (A-A_n)\psi\rangle|}{\langle \psi, (A+1)\psi \rangle} \to 0, \quad n \to \infty.$$
Then $A_n \to A$ in the norm resolvent sense.
Unfortunately the book doesn't give a complete proof of this statement. As a rough guide it states that one should proof
$$ (A+I)^{-1/2} (A_n - A) (A+I)^{-1/2} \to 0 \quad \text{ in the norm sense} $$
and
$$ (A_n + I)^{-1}= (A+I)^{-1/2} (I + (A+I)^{-1/2} (A_n - A) (A+I)^{-1/2})^{-1}(A+I)^{-1/2} $$
and continue from this. I tried to mess around with the first and second resolvent identity, but since I'm not particularly well trained with these things in general, I didn't really get anywhere.
If would appreciate if anyone could add some more hints on how a proof for this may look like.
 A: The form domain of $A$ is $(A+I)^{-1/2}$ because $A$ is positive. This is because
$$
    (Ax,x)+(x,x) = ((A+I)x,x) =  ((A+I)^{1/2}x,(A+I)^{1/2}x)=\|(A+I)^{1/2}x\|^2.
$$
Because all of the form domains are the same, then $(A_n+I)^{1/2}(A+I)^{-1/2}$ must be a bounded operator for all $n$. Therefore, on the domain of $A$, the following is bounded: 
$$
       (A+I)^{-1/2}(A_n+I)(A+I)^{-1/2}.
$$
Therefore it has a bounded extension $B_n$ to all of $X$ from its dense domain $\mathcal{D}((A+I)^{1/2})$.
You were given that the following is finite and tends to $0$ as $n\rightarrow\infty$:
$$
             \sup_{0\ne\psi\in\mathcal{D}}\frac{|((A-A_n)\psi,\psi)|}{\|(A+I)^{1/2}\psi\|^2} \\
   = \sup_{0\ne\varphi\in X}\frac{|((A-A_n)(A+I)^{-1/2}\varphi,(A+I)^{-1/2}\varphi)|}{\|\varphi\|^2} \\
   = \sup_{0\ne\varphi\in X}\frac{|((A+I)^{-1/2}(A-A_n)(A+I)^{-1/2}\varphi,\varphi)|}{\|\varphi\|^2} \\
   = \|(A+I)^{-1/2}(A-A_n)(A+I)^{-1/2}\| = \|B_n\|.
$$
For large enough $n$, $I+(A+I)^{-1/2}(A_n-A)(A+I)^{-1/2}$ is invertible because the operator norm of the second term being subtracted from $I$ tends to $0$. Therefore, the following converges in operator norm:
$$
      (\;\;I+(A+I)^{-1/2}(A_n-A)(A+I)^{-1/2}\;\;)^{-1}\rightarrow I,\;\;\; n\rightarrow\infty.
$$
Therefore, with a few details,
$$
   (A+I)^{-1/2}\{\;(I+(A+I)^{-1/2}(A_n-A)(A+I)^{-1/2})^{-1}\}(A+I)^{-1/2}\rightarrow (A+I)^{-1} \\
   (\;(A+I)^{1/2}(I+(A+I)^{-1/2}(A_n-A)(A+I)^{-1/2})(A+I)^{1/2}\;)^{-1}\rightarrow (A+I)^{-1} \\
   ( (A+I)+(A_n-A) )^{-1} \rightarrow (A+I)^{-1} \\
   ( I+A_n )^{-1} \rightarrow (A+I)^{-1}.
$$
