Reed and Simon, Fourier Analysis and Self-Adjointness, second corollary to Theorem X.$25$: how to show that $D(A^2)$ is dense in $D(A)$ for its norm?

Question

This question arose while trying to figure out the proof of the second corollary to Theorem X.$25$ in Reed and Simon's Fourier Analysis, Self-Adjointness, stated as follows:

Theorem X.$25$: Let $A : D(A) \subset H \to H$, $H$ Hilbert space, be a closed densely defined operator. Then $A^* \circ A$ is self-adjoint.

Corollary: Let $A$ be a closed operator. Then any core for $A$ is a form core for $A^* \circ A$.

Corollary: If $A$ is symmetric and $A^2$ is densely defined, then $A^* \circ A$ is the Friedrichs extension of $A^2$.

I had no major problem with the proof of the Theorem itself nor with the first corollary.

By the proof of Theorem X.$25$, I did recognise that $D(A)$ is the form domain of $A^* \circ A$ with the complete norm below: $$\| \cdot \|_A : \varphi \mapsto \sqrt{\|A(\varphi)\|^2 + \|\varphi\|^2}$$ (which coincides with the norm $\|\cdot\|_{+1}$, used for a form domain of a positive form, in this particular case) hence if $B$ denotes the Friedrichs extension of $A^2$, then if we do have $B = A^* \circ A$, the form domain $Q(B)$ of $B$ needs to be $D(A)$ too.
On the other hand, the form domain of $B$ is the completion of $D(A^2)$ for $\|\cdot\|_{+1} = \|\cdot\|_A$, hence it is equal to the $\|\cdot\|_A$-closure of $D(A^2)$ in $D(A)$.
Lastly, it is easy to see that $A^* \circ A$ does extend $A^2$ as $A^*$ extends $A$, and the domain of $A^* \circ A$ is contained in $D(A)$, yet the Friedrichs extension $B$ of $A^2$ is the only self-adjoint extension whose domain is contained in the form domain of $B$, therefore if we can show that $Q(B) = D(A)$ then we'll have $B = A^* \circ A$ with no issue.

My question today is thus:

How does one show that $D(A^2)$ is $\|\cdot\|_A$-dense in $D(A)$? Alternatively, if that claim is not actually true, what did I misunderstand?

What I've tried: Since $D(A)$ is a Hilbert space for $\|\cdot\|_A$, I've tried looking at $Q(B)^{\perp_A} = D(A^2)^{\perp_A}$, where $\perp_A$ means the orthogonal for $\|\cdot\|_A$ in contrast with $\perp$ which I'll use for the usual orthogonal in $H$. Assume the same about other notations involving $A$.
We can see that $D(A^2)^{\perp_A} = \operatorname{Ran}\left(A^2 + \operatorname{Id}\right)^\perp \cap D(A)$ thanks to the following relation for all $\varphi \in D(A^2)$ and $\psi \in D(A)$: $$\langle \varphi, \psi\rangle_A = \langle A(\varphi), A(\psi)\rangle + \langle \varphi, \psi\rangle = \left\langle A^2(\varphi) + \varphi, \psi\right\rangle$$ Now, we have $\overline{\operatorname{Ran}\left(A^2 + \operatorname{Id}\right)} = \operatorname{Ker}\left((A^2)^* + \operatorname{Id}\right)^\perp$ due to usual properties of adjoints, hence: $$D(A^2)^{\perp_A} = \operatorname{Ker}\left((A^2)^* + \operatorname{Id}\right) \cap D(A)$$ Take $\psi$ in the above space. What I would like to do is the following manipulation, with an exclamation point where the outlier equality lies: $$0 = \langle (A^2)^*(\psi) + \psi, \psi\rangle \overset{!}{=} \langle A(\psi), A(\psi) \rangle + \langle \psi, \psi\rangle = \|\psi\|^2_A$$ which would tell me that $\psi = 0$, and thus would let me conclude that, $D(A)$ being complete, $Q(B)$ is indeed the whole of $D(A)$.
However I don't really know how to proceed directly since it's not clear to me that we should have $\psi \in D\left((A^*)^2\right)$ in order to be able to write $(A^2)^*(\psi) = (A^*)^2(\psi)$ which would justify that equality, so I tried taking out a sequence in $D(A^2)$ converging to $\psi$ in $H$.

Consider $(\psi_n)_n \subset D(A^2)$ converging to $\psi$ for $\|\cdot\|_H$. This sequence exists by denseness of $D(A^2)$ in $H$.
We obtain: $$0 = \langle (A^2)^*(\psi) + \psi, \psi_n\rangle = \left\langle \psi, A^2(\psi_n) \right\rangle + \langle \psi, \psi_n\rangle = \langle A(\psi), A(\psi_n) \rangle + \langle \psi, \psi_n\rangle = \langle \psi, \psi_n\rangle_A$$ It's almost what I need yet I don't see how to go forward from there as the function $\langle \psi, \cdot\rangle_A$ might not be $\|\cdot\|_H$-continuous.

Feel free to edit and/or re-tag if necessary or appropriate.

Answer 1

$$ \DeclareMathOperator{\im}{im} $$

It is a second attempt.

Friedrichs extension

Let $S$ is symmetric operator with positive lower bound. Consider $$ H_s = \overline{\left\{ (Sf, g): f, g\in H, f,g\in D(S) \right\}}, $$ and $\psi: x \mapsto x: H_s\to H$ is continuous one-to-one (see citation below) with dense image (trivial), $\psi^t$ (continuous adjoint) is also one-to-one with dense image. $$ (f, x)_H = (\psi f, x)_H = (f, \psi^tx)_s, f\in H_s, x\in H $$ In particular, for $f\in D(S),$ $(f, x)_H = (Sf, \psi^tx)_H.$

Let $g\in \im \psi^t = D((\psi^t)^{-1}) \subset H_s,$ then $$ (f, (\psi^t)^{-1}g)_H = (Sf, g)_H, f\in D(S) $$

It shows that $g\in D(S^*)$ and $(\psi^t)^{-1}g = S^*g,$ i.e. $(\psi^t)^{-1} = D(S^*)|H_s$ is Friedrichs extension of $S.$

One-to-one (Weidmann "Linear Operators in Hilbert Spaces") If $S$ is semi-bounded symmetric operator with lower bound $\gamma > 0$ then $s(f, g) = (Sf, g)_H, f,g\in D(S)$ defines a semi-bounded sesquilinear form $s$ on $D(S).$ Then $H_s\to H$ is one-to-one. Take $f_n$ $\left\| \cdot \right\|_s$-Cauchy sequence such that $f_n\to 0$ in $H:$ $$ \begin{split} \left\| f_n \right\|^2_s \le \left| (f_n, f_n-f_m)_s + (f_n, f_m)_s\right| \le\\ \left\| f_n \right\|_s \left\| (f_n-f_m) \right\|_s + \left\| S f_n \right\| \left\| f_m \right\| \end{split} $$ Make first summand small by $\left\| \cdot \right\|_s$-Cauchy and second by $f_n\to0.$

Conclusion $$ H(A^2)\xrightarrow{i} H(A^*\circ A)\xrightarrow{\Psi:x \mapsto x} H. $$ are continuous for norms $(A^2\varphi, \varphi) + (\varphi, \varphi), \varphi\in D(A^2)$ and $(A^*A\varphi', \varphi') + (\varphi', \varphi'), \varphi'\in D(A^*\circ A),$ where $i$ is an isometry and $i^t$ is orthogonal projection on $H(A^2)$ in $H(A^* \circ A).$

Friedrichs extensions are: \begin{align} F(I+A^2) &= ((\Psi \circ i)^t)^{-1} &\subseteq& 1+(A^2)^* \\ F(I+A^* \circ A) &= (\Psi^t)^{-1} &\subseteq& 1+(A^* \circ A)^* \end{align} $D(F(I+A^2)) \subseteq H(A^2)$ and $D(F(I+A^* \circ A)) \subseteq H(A^* \circ A).$

$$ (\Psi \circ i)^t = i^t \circ \Psi^t, $$ If $\im \Psi^t \cap \ker i^t \neq 0$ then there would be no one-to-one $F(I + A^2),$ so $\im \Psi^t \subseteq \im i$ and $i^t \circ \Psi^t = \Psi^t.$ That is $F(I + A^2) = F(I + A^* \circ A).$