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. Let $I:H \to H$ be the identity map.
For every $\lambda>0$, we define the resolvent $J_\lambda$ and the Yosida approximation $A_\lambda$ of $A$ by $$ J_\lambda=(I+\lambda A)^{-1} \quad \text { and } \quad A_\lambda=\frac{1}{\lambda}\left(I-J_\lambda\right). $$
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 $$ \langle u, v \rangle_{D(A^k)} := \sum_{j=0}^k \langle A^j u, A^j v \rangle. $$
I'm reading Theorem 7.7 at page 194 of 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 $$ \begin{cases} \frac{d u}{d t}+A u&=0 \quad \text{on} \quad (0,+\infty), \\ u(0)&=u_0. \end{cases} $$ Moreover, we have $$ |u(t)| \leq\left|u_0\right| \quad \text { and } \quad\left|\frac{d u}{d t}(t)\right|=|Au(t)| \leq \frac{1}{t} \left|u_0\right| \quad \forall t > 0, $$ and $$ u \in u \in C^k ((0,+\infty) ; D(A^\ell)) \quad \forall k, \ell \in \mathbb N. $$
Assume that $u_0 \in D(A^2)$. For $\lambda>0$, let $u_\lambda$ be the solution of the problem $$ \begin{cases} \frac{d u_\lambda}{d t}+A_\lambda u_\lambda&=0 \quad \text{on} \quad [0,+\infty), \\ u_\lambda(0)&=u_0. \end{cases} $$
The author then goes on to prove that $$ \left |\frac{d u_\lambda}{dt} (T) \right| \le\frac{|u_0|}{T} \quad \forall T>0, \quad (*) $$ by using the following properties
- $A_\lambda$ is self-adjoint.
- $u_\lambda \in C^2([0, +\infty) ; H) \cap C^1 ([0, +\infty) ; D(A)) \cap C([0, +\infty); D(A^2))$.
- $\frac{d}{dt} (A_\lambda u_\lambda) = A_\lambda (\frac{d u_\lambda}{dt})$.
- the map $t \mapsto |\frac{d u_\lambda}{dt} (t)|$ is non-increasing.
His proof of $(*)$ is non-trivial by integrating and then combining inequalities. It's like magic coming out of nowhere. I could not get a feeling of how such ideas arise.
Could you shed some light on how you approach to prove $(*)$?
We start by rewriting the inequality as:
$$\left\|u_0\right\|^2 - T^2 \left\|\frac{\mathrm d u_\lambda}{\mathrm dt}(T)\right\|^2 \ge 0$$
This transformation allows us to use the scalar product and it properties. Now this inequality you can transform it to one of the following inequlities:
$$\left\|u_0\right\|^2 - T\int_{0}^{T} \left\|\frac{\mathrm d u_\lambda}{\mathrm dt}(t)\right\|^2 \mathrm dt \ge 0$$ or
$$\left\|u_0\right\|^2 - \int_{0}^{T} 2t\left\|\frac{\mathrm d u_\lambda}{\mathrm dt}(t)\right\|^2 \mathrm dt \ge 0$$
The first one is problematic since you have the $T$ outside the integral and if you continue in that road you will probably get stuck. For the second one everything is inside the integral, so continue transforming the inequality by replacing $\frac{\mathrm d u_\lambda}{\mathrm dt} = -A_\lambda u_\lambda$,
\begin{align} \left\|u_0\right\|^2 - \int_0^T 2t \left\|\frac{\mathrm d u_\lambda}{\mathrm dt}(t)\right\|^2 \mathrm dt &= \left\|u_0\right\|^2 + \int_0^T 2t \left\langle\frac{\mathrm d u_\lambda}{\mathrm dt}(t), A_\lambda u_\lambda(t)\right\rangle \mathrm dt \end{align}
The $t$ inside the integral is a little bit not comfortable to have so let's remove it. How? You can use a integration by parts using the fact that, $\frac{\mathrm d}{\mathrm d t}\left\langle u_\lambda(t), A_\lambda u_\lambda(t)\right\rangle = 2 \left\langle\frac{\mathrm d u_\lambda}{\mathrm dt}(t), A_\lambda u_\lambda(t)\right\rangle$
\begin{align} \left\|u_0\right\|^2 - 2\int_0^Tt \left\|\frac{\mathrm d u_\lambda}{\mathrm dt}(t)\right\|^2 \mathrm dt &= \left\|u_0\right\|^2 + \left[t\left\langle u_\lambda(t), A_\lambda u_\lambda(t)\right\rangle\right]_0^T - \int_0^T \left\langle u_\lambda(t), A_\lambda u_\lambda(t)\right\rangle \mathrm dt\\ &= \left\|u_0\right\|^2 + T\left\langle u_\lambda(T), A_\lambda u_\lambda(T)\right\rangle + \int_0^T \left\langle u_\lambda(t), \frac{\mathrm d u_\lambda}{\mathrm dt}(t)\right\rangle \mathrm dt\\ &= \left\|u_0\right\|^2 + T\left\langle u_\lambda(T), A_\lambda u_\lambda(T)\right\rangle + \left[\frac12 \left\|u_\lambda(t)\right\|^2\right]_0^T\\ &= \left\|u_0\right\|^2 + T\left\langle u_\lambda(T), A_\lambda u_\lambda(T)\right\rangle + \frac12 \left\|u_\lambda(T)\right\|^2 - \frac12 \left\|u_\lambda(0)\right\|^2 \end{align}