Brezis' theorem 7.5: how is $A_1$ maximal monotone in $H_1$?

Question

Let $H$ together with an inner product $\langle \cdot, \cdot \rangle$ be a real Hilbert space. Let $|\cdot|$ be the induced norm of $\langle \cdot, \cdot \rangle$. Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. Let $I:H \to H$ be the identity map.

We define by induction the set $$ D(A^{k+1}) := \{ v \in D(A^k) :Av \in D(A^k)\} \quad \forall k \in \mathbb N^*. $$

For $k \ge 1$, the set $D(A^k)$ is a Hilbert space for the inner product $$ \langle u, v \rangle_{D(A^k)} := \sum_{j=0}^k \langle A^j u, A^j v \rangle. $$

It is mentioned in the proof of Theorem 7.5 in Brezis' Functional Analysis that

Consider the Hilbert space $H_1=D(A)$ equipped with the scalar product $\langle \cdot, \cdot \rangle_{D(A)}$. It is easy to check that the operator $A_1: D\left(A_1\right) \subset$ $H_1 \rightarrow H_1$ defined by $$ \begin{cases} D(A_1) &= D(A^2) \\ A_1 u &= A u \quad \text {for} \quad u \in D(A_1), \end{cases} $$ is maximal monotone in $H_1$.

Could you please explain how $A_1$ is maximal monotone in $H_1$, i.e., the image of $D(A^2)$ through $I+A_1$ is $D(A)$?

Answer 1

We will prove that $(I+A) (D(A^2)) = D(A)$. We need the following result, i.e.,

Definition. Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator. For every $\lambda>0$, set $$ J_\lambda=(I+\lambda A)^{-1} \quad \text { and } \quad A_\lambda=\frac{1}{\lambda}\left(I-J_\lambda\right) ; $$ $J_\lambda$ is called the resolvent of $A$, and $A_\lambda$ is the Yosida approximation (or regularization) of $A$. Keep in mind that $\left\|J_\lambda\right\|_{\mathcal{L}(H)} \leq 1$.

Proposition 7.2. Let A be a maximal monotone operator. Then

• (a1) $A_\lambda v = A (J_\lambda v)$ for all $v \in H$ and for all $\lambda>0$,
• (a2) $A_\lambda v = J_\lambda (A v)$ for all $v \in D(A)$ and for all $\lambda>0$.

1. Let $u \in D(A^2)$. Then $u \in D(A)$ and $Au \in D(A)$. Then $(I+A)u =u+Au\in D(A)$. Then $(I+A) (D(A^2)) \subset D(A)$.

2. Let $u \in D(A)$. Let $v := (I+A)^{-1} u =J_1 u$. Because $A$ is maximal monotone in $H$, we have $J_1$ is bijection from $H$ onto $D(A)$. Then $v\in D(A)$. We have $(I+A)v=u$. It remains to prove $Av \in D(A)$. We have $Av=AJ_1u$. By Proposition 7.2 (a1, a2), $AJ_1u = J_1 Au$. Clearly, $J_1Au \in D(A)$ and thus $Av\in D(A)$. Hence $D(A) \subset (I+A) (D(A^2))$.