Let $H$ together with an inner product $\langle \cdot, \cdot \rangle$ be a real Hilbert space. Let $|\cdot|$ be the induced norm. I'm reading Theorem 7.4 (Hille-Yosida) in Brezis' Functional Analysis, i.e.,
Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. Then, given any $u_0 \in D(A)$ there exists a unique function $$ u \in C^1([0,+\infty) ; H) \cap C([0,+\infty) ; D(A)) $$ satisfying $$ (6) \quad \begin{cases} \frac{d u}{d t}+A u&=0 \quad \text{on} \quad [0,+\infty), \\ u(0)&=u_0. \end{cases} $$ Moreover, $$ |u(t)| \leq\left|u_0\right| \quad \text { and } \quad\left|\frac{d u}{d t}(t)\right|=|A u(t)| \leq\left|A u_0\right| \quad \forall t \geq 0. $$
My question is about the last step of the proof, i.e.,
We now turn to the proof of Theorem 7.4. Given $u_0 \in D(A)$ we construct (using Lemma 7.2) a sequence $\left(u_{0 n}\right)$ in $D\left(A^2\right)$ such that $u_{0 n} \rightarrow u_0$ and $A u_{0 n} \rightarrow A u_0$. By Step 5 we know that there is a solution $u_n$ of the problem $$ (19) \quad \begin{cases} \frac{d u_n}{d t}+A u_n &= 0 \quad \text { on }[0,+\infty), \\ u_n(0) &= u_{0 n} . \end{cases} $$ We have, for all $t \geq 0$, $$ (20) \quad\begin{cases} \left|u_n(t)-u_m(t)\right| & \leq\left|u_{0 n}-u_{0 m}\right| &\xrightarrow{m, n \rightarrow \infty} 0, \\ \left|\frac{d u_n}{d t}(t)-\frac{d u_m}{d t}(t)\right| & \leq\left|A u_{0 n}-A u_{0 m}\right| &\xrightarrow{m, n \rightarrow \infty} 0 . \end{cases} $$ Therefore $$ \begin{align*} u_n(t) & \rightarrow u(t) & \text { uniformly on }[0,+\infty), \\ \frac{d u_n}{d t}(t) & \rightarrow \frac{d u}{d t}(t) & \text { uniformly on }[0,+\infty). \end{align*} $$ with $u \in C^1([0,+\infty) ; H)$. Passing to the limit in (19) - using the fact that $A$ is a closed operator-we see that $u(t) \in D(A)$ and $u$ satisfies (6). From (6) we deduce that $u \in C([0,+\infty) ; D(A))$.
Could you please explain how to obtain the inequalities in (20)?
Update I have figured how to prove the first inequality, but I'm stuck at the second one.
We have $$ \left ( \frac{d u_n}{d t}+A u_n \right ) - \left ( \frac{d u_m}{d t}+A u_m \right ) = 0, $$ which implies $$ \begin{align*} \frac{1}{2} \frac{d}{dt} |u_n -u_m|^2 &= - \langle A (u_n - u_m), u_n-u_m \rangle \\ & \le 0 \quad \text{because } A \text{ is monotone}. \end{align*} $$ Then the map $t \mapsto |u_n (t) -u_m (t)|^2$ is non-increasing. Then $$ \left|u_n(t)-u_m(t)\right| \leq\left|u_{n} (0) - u_{m} (0)\right| = \left|u_{0 n}-u_{0 m}\right|. $$
I was silly. Those inequalities are actually trivial.
Notice that the solution of problem (19) will satisfy the upper bounds $$ |u_n(t)| \leq\left|u_{0n}\right| \quad \text {and} \quad\left|\frac{d u_n}{d t}(t)\right|=|A u_n(t)| \leq\left|A u_{0n}\right| \quad \forall t \geq 0. $$
The two desired inequalities then follow by noticing that if $u_n, u_m$ are solutions of (19), then so are $u_n \pm u_m$ with the corresponding initial data $u_{0n} \pm u_{0m}$.