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 $k=2$ 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. Now I want to tackle the case $k=2$. There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
It remains to prove $u \in C^2 ((0, +\infty); H)$. Consider the Hilbert space $H_1 = D(A)$ equipped with the inner product $\langle \cdot, \cdot \rangle_{D(A)}$. Then the operator $A_1 :D(A_1) \subset H_1 \to H_1$ defined by $$ \begin{cases} D(A_1) &= D(A^2), \\ A_1 u &= Au, \end{cases} $$ is maximal monotone and self-adjoint 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 $$ u \in C ([0, +\infty); H_1) \cap C^1 ((0, +\infty); H_1) \cap C((0, +\infty); D(A^2)) $$ that satisfies $(*)$. Since $A \in \mathcal L(H_1, H)$ and $u \in C^1 ((0, +\infty); H_1)$, we get $Au \in C^1 ((0, +\infty); H)$ and $$ \frac{d}{dt} (Au) = A \left (\frac{du}{dt} \right ) \quad \text{on} \quad (0, +\infty). $$
Then $\frac{d u}{d t} \in C^1 ((0, +\infty); H)$ and thus $u \in C^2 ((0, +\infty); H)$. Moreover, $$ \frac{d}{dt} \left (\frac{d u}{d t} \right )+A \left (\frac{du}{dt} \right )=0 \quad \text{on} \quad (0,+\infty). $$
Update @Kroki has found a critical mistake in my above attempt. Below is my fix.
Let $t_0 >0$. From the case $k=1$, we get $u(t_0) \in H_1$. We apply the case $k=1$ to the operator $A_1$ in the space $H_1$, we see that there is a unique function $$ v \in C ([t_0, +\infty); H_1) \cap C^1 ((t_0, +\infty); H_1) \cap C((t_0, +\infty); D(A^2)) $$ that satisfies $$ \begin{cases} \frac{d v}{d t}+A v&=0 \quad \text{on} \quad (t_0,+\infty), \\ v(t_0)&=u(t_0). \end{cases} $$
Since $A \in \mathcal L(H_1, H)$ and $v \in C^1 ((t_0, +\infty); H_1)$, we get $Av \in C^1 ((t_0, +\infty); H)$ and $$ \frac{d}{dt} (Av) = A \left (\frac{dv}{dt} \right ) \quad \text{on} \quad (t_0, +\infty). $$
Then $\frac{d v}{d t} \in C^1 ((t_0, +\infty); H)$ and thus $v \in C^2 ((t_0, +\infty); H)$. Moreover, $$ \frac{d}{dt} \left (\frac{d v}{d t} \right )+A \left (\frac{dv}{dt} \right )=0 \quad \text{on} \quad (t_0,+\infty). $$
By uniqueness, $u$ and $v$ coincide on $(t_0, +\infty)$. Because $t_0>0$ is arbitrary, we get $u \in C^2 ((0, +\infty); H)$ and $$ \frac{d}{dt} \left (\frac{d u}{d t} \right )+A \left (\frac{du}{dt} \right )=0 \quad \text{on} \quad (0,+\infty). $$
The only small mistake that you have is when you construct your solution using $A_1$, you are assuming that $u_0\in H_1$ which is not correct. Can you fix it by taking another initial value for example $u(t_0)$ for $t_0>0$?