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 first part of 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 $\ell=0, \ldots, k$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Uniqueness
Let $u, v \in C([0,+\infty) ; H)$ be two functions that are differentiable on $(0, +\infty)$ and that are solutions of $(*)$. By monotonicity of $A$, we see that the map $\varphi (t) = |u(t) - v(t)|^2$ is non-increasing on $(0, +\infty)$. On the other hand, $\varphi (t)$ is continuous on $[0, +\infty)$ with $\varphi (0)=0$. By mean value theorem, $\varphi \equiv 0$.
Existence
Let $u_0 \in H$. By Proposition 7.1 (in the same book), $D(A)$ is dense in $(H, |\cdot|)$. By Lemma 7.2, $D(A^2)$ is dense in $(D(A), |\cdot|_{D(A)})$. So there is a sequence $(u_{0n})$ in $D(A^2)$ such that $|u_{0n} - u_0| \xrightarrow{n \rightarrow \infty} 0$. By Theorem 7.4, there is a unique function $$ u_n \in C^1 ([0, +\infty); H) \cap C([0, +\infty); D(A)) $$ such that $$ \begin{cases} \frac{d u_n}{d t}+A u_n&=0 \quad \text{on} \quad [0,+\infty), \\ u_n(0)&=u_{0n}. \end{cases} $$
Moreover, $|u_n (t)| \le |u_{0n}|$ and $\left|\frac{d u_n}{d t}(t)\right| = |A u_n(t)| \le |A u_{0n}|$ for all $t \ge 0$. This implies $$ |u_n(t)-u_m(t)| \le |u_{0n} - u_{0m}| \quad \forall m,n \in \mathbb N^*, \forall t \ge 0. $$
Because $A$ is self-adjoint and $u_{0n} \in D(A^2)$, we can show that $\left |\frac{d u_n}{dt} (t) \right| \le\frac{|u_{0n}|}{t}$ for all $t>0$. This implies $$ \left | \frac{d u_n}{dt} (t) - \frac{d u_m}{dt} (t) \right| \le\frac{|u_{0n} - u_{0m}|}{t} \quad \forall m,n \in \mathbb N^*, \forall t > 0. $$
Then
- $(u_n)$ converges uniformly on $[0, +\infty)$ to some limit $u \in C ([0, +\infty); H) \cap C^1 ((0, +\infty); H)$.
- $(\frac{d u_n}{dt})$ converges locally uniformly on $(0, +\infty)$ to $\frac{du}{dt}$.
In particular, $Au$ is continuous on $(0, +\infty)$. Because $A$ has a closed graph, we get
- $u(t) \in D(A)$ for all $t>0$
- $u \in C ((0, +\infty); D(A))$.
- $\frac{d u}{d t}+A u=0$ on $(0, +\infty)$.