Let $H$ together with an inner product $\langle \cdot, \cdot \rangle$ be a real Hilbert space. Let $|\cdot|$ be the induced norm of $\langle \cdot, \cdot \rangle$. Let $A: D(A) \subset H \to H$ be an unbounded linear operator with closed graph. We define by induction the set $$ D(A^{k+1}) := \{ v \in D(A^k) :Av \in D(A^k)\} \quad \forall k \in \mathbb N^*. $$
I would like to verify a claim at page 191 in Brezis' Functional Analysis, i.e.,
For $k \ge 1$, the set $D(A^k)$ is a Hilbert space for the inner product $$ \langle u, v \rangle_{D(A^k)} := \sum_{j=0}^k \langle A^j u, A^j v \rangle; $$ the corresponding norm is $$ |u|_{D(A^k)} := \left ( \sum_{j=0}^k |A^ju|^2 \right )^{1/2}. $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Fix $k \in \mathbb N^*$. It is clear that $D(A^{k})$ is the maximal vector subspace of $H$ on which $A^k$ is well-defined. Let $(u_n)$ be a Cauchy sequence in $D(A^{k})$ for the norm $|\cdot|_{D(A^k)}$. For each $j \in \{0, \ldots,k\}$, we have $(A^j u_n)_n$ is a Cauchy sequence in $H$ for the norm $|\cdot|$.
For each $j \in \{0, \ldots,k\}$, there is $v_j \in H$ such that $|A^j u_n - v_j| \xrightarrow{n \to \infty} 0$. For each $j \in \{0, \ldots,k-1\}$, $|A^ju_n - v_j| \xrightarrow{n \to \infty} 0$ and $|A(A^j u_n) - v_{j+1}| \xrightarrow{n \to \infty} 0$. Because $A$ has a closed graph, $v_j \in D(A)$ and $Av_j = v_{j+1}$ for all $j \in \{0, \ldots,k-1\}$.
Then for $j \in \{0, \ldots,k\}$, we have $v_0 \in D(A^j)$ and $|A^j u_n - A^jv_0| \xrightarrow{n \to \infty} 0$. Then $v_0 \in D(A^k)$ and $|u_n-v_0|_{D(A^k)} \xrightarrow{n \to \infty} 0$.