Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. We define by induction the subspaces $$ D(A^{k+1}) := \{ v \in D(A^k) :Av \in D(A^k)\} \quad \forall k \in \mathbb N^*. $$ Then $D(A^k)$ is a Hilbert space for the inner product $$ \langle u, v \rangle_{D(A^k)} := \sum_{j=0}^k \langle A^j u, A^j v \rangle. $$
Let $|\cdot|_{D(A^k)}$ be the induced norm of $\langle \cdot, \cdot \rangle_{D(A^k)}$. I'm trying to prove the case $k\ge 3$ in Theorem 7.7 in Brezis' Functional Analysis, i.e.,
Let $A$ be a self-adjoint maximal monotone (unbounded linear) operator. Then, given any $u_0 \in H$ there exists a unique function $$ u \in C([0,+\infty) ; H) \cap C^1((0,+\infty) ; H) \cap C((0, +\infty); D(A)) $$ such that $$ (*) \quad\begin{cases} \frac{d u}{d t}+A u&=0 \quad \text{on} \quad (0,+\infty), \\ u(0)&=u_0. \end{cases} $$ Moreover, we have
- $|u(t)| \leq\left|u_0\right|$ and $\left|\frac{d u}{d t}(t)\right|=|Au(t)| \leq \frac{1}{t} \left|u_0\right|$ for all $t > 0$.
- $u \in C^{k-\ell} ((0,+\infty) ; D(A^\ell))$ for all $k \in \mathbb N^*$ and $\ell=0, \ldots, k$.
I have proved for the case $k=1$ here and for the case $k=2$ here. There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
We argue by induction on $k \ge 2$: assume that the result holds up to order $(k-1)$. We want to prove $$ u \in C^{k-\ell} ((0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k. $$
From the case $k=2$, we know that $$ u \in C^2 ((0, +\infty); H) \cap C^1((0, +\infty); D(A)), $$ and that $u$ satisfies $$ \frac{d}{dt} \left (\frac{d u}{d t} \right )+A \left (\frac{du}{dt} \right )=0 \quad \text{on} \quad (0,+\infty). $$
Let $$ v :=\frac{d u}{d t}. $$
Then $v \in C^1 ((0, +\infty); H) \cap C((0, +\infty); D(A))$. Let $t_0>0$. Then $$ \begin{cases} \frac{d v}{d t}+A v &= 0 \quad \text{on} \quad (t_0,+\infty), \\ v(t_0) &= -A u(t_0). \end{cases} $$
Then the restriction of $v$ to $[t_0, +\infty)$ is the solution of $(*)$ with the initial data $v(t_0) = -A u(t_0)$. By inductive hypothesis, $$ v \in C^{k-1-\ell} ((t_0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k-1. \tag{1} $$
This implies $$ u \in C^{k-\ell} ((t_0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k-1. $$
It remains to check that $$ u \in C ((t_0,+\infty) ; D(A^k)). \tag{2} $$
Applying (1) with $\ell=k-1$, we get $$ \frac{du}{dt} \in C ((t_0,+\infty) ; D(A^{k-1})). $$
Then by $(*)$ $$ Au \in C ((t_0,+\infty) ; D(A^{k-1})). $$
Notice that $$ |x|_{D(A^k)}^2 = |x|^2 + |Ax|_{D(A^{k-1})}^2 \quad \forall x \in D(A^k). $$
Then (2) follows. Because $t_0>0$ is arbitrary, the proof is completed.
As suggested by @Kroki in a comment, we can treat the case $k=2$ and $k\ge 3$ simultaneously.
We argue by induction on $k \ge 2$: assume that the result holds up to order $(k-1)$. In particular, $$ u \in C ((0,+\infty) ; D(A^{k-1})). \tag{1} $$
We want to prove $$ u \in C^{k-\ell} ((0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k. $$
Consider the Hilbert space $H_1 = D(A^{k-1})$ equipped with the inner product $\langle \cdot, \cdot \rangle_{D(A^{k-1})}$. Then the operator $A_1 :D(A_1) \subset H_1 \to H_1$ defined by $$ \begin{cases} D(A_1) &= D(A^k), \\ A_1 u &= Au, \end{cases} $$ is maximal monotone and self-adjoint.
Fix $t_0 >0$. By (1), $u(t_0) \in H_1$. We apply the case $k=1$ to the operator $A_1$ in the space $H_1$, we see that there is a unique function $$ \overline u \in C ([t_0, +\infty); H_1) \cap C^1 ((t_0, +\infty); H_1) \cap C((t_0, +\infty); D(A_1)) $$ that satisfies $$ \begin{cases} \frac{d \overline u}{d t}+A \overline u &= 0 \quad \text{on} \quad (t_0,+\infty), \\ \overline u(t_0) &= u(t_0). \end{cases} \tag{2} $$
Since $A \in \mathcal L(H_1, H)$ and $\overline u \in C^1 ((t_0, +\infty); H_1)$, we get $A\overline u \in C^1 ((t_0, +\infty); H)$ and $$ \frac{d}{dt} (A\overline u) = A \left (\frac{d\overline u}{dt} \right ) \quad \text{on} \quad (t_0, +\infty). $$
By (2), $\frac{d \overline u}{d t} \in C^1 ((t_0, +\infty); H)$ and thus $\overline u \in C^2 ((t_0, +\infty); H)$. Also, $$ \frac{d}{dt} \left (\frac{d \overline u}{d t} \right )+A \left (\frac{d\overline u}{dt} \right )=0 \quad \text{on} \quad (t_0,+\infty). $$
By uniqueness, $u$ and $\overline u$ coincide on $[t_0, +\infty)$. Let $$ v :=\frac{d u}{d t}. $$
We have just shown above that $$ v \in C^1 ((t_0, +\infty); H) \cap C((t_0, +\infty); H_1), $$ and $$ \begin{cases} \frac{d v}{d t}+A v &= 0 \quad \text{on} \quad (t_0,+\infty), \\ v(t_0) &= -A u(t_0). \end{cases} $$
If $t_1 > t_0$, then $v \in C ([t_1, +\infty); H_1)$. So $v \in C ([t_0, +\infty); H_1)$. Notice that $H_1 \subset D(A) \subset H$. Thus $$ v \in C ([t_0, +\infty); H) \cap C^1 ((t_0, +\infty); H) \cap C((t_0, +\infty); D(A)). $$
Then the restriction of $v$ to $[t_0, +\infty)$ is the solution of $(*)$ with the initial data $v(t_0) = -A u(t_0)$. By inductive hypothesis, $$ v \in C^{k-1-\ell} ((t_0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k-1. \tag{3} $$
This implies $$ u \in C^{k-\ell} ((t_0,+\infty) ; D(A^\ell)) \quad \forall \ell = 0, \ldots,k-1. $$
It remains to check that $$ u \in C ((t_0,+\infty) ; D(A^k)). \tag{4} $$
Applying (3) with $\ell=k-1$, we get $$ \frac{du}{dt} \in C ((t_0,+\infty) ; D(A^{k-1})). $$
Then by $(*)$ $$ Au \in C ((t_0,+\infty) ; D(A^{k-1})). $$
Notice that $$ |x|_{D(A^k)}^2 = |x|^2 + |Ax|_{D(A^{k-1})}^2 \quad \forall x \in D(A^k). $$
Then (4) follows. Because $t_0>0$ is arbitrary, the proof is completed.