Let $H$ together with an inner product $\langle \cdot, \cdot \rangle$ be a real Hilbert space. Let $A: D(A) \subset H \to H$ be a maximal monotone (unbounded linear) operator.
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 $\left\|J_\lambda\right\|_{\mathcal{L}(H)} \leq 1$.
I'm reading a proposition in Brezis' Funtional Analysis
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 $\lambda>0$.
The proof is given as follows.
(a1) can be written as $v=\left(J_\lambda v\right)+\lambda A\left(J_\lambda v\right)$, which is just the definition of $J_\lambda v$.
(a2) By (a1), we have $$ A_\lambda v+A\left(v-J_\lambda v\right)=A v, $$ i.e., $$ A_\lambda v+\lambda A\left(A_\lambda v\right)=A v, $$ which means that $A_\lambda v=(I+\lambda A)^{-1} A v$.
Let $R(A_\lambda)$ be the range of $A_\lambda$. Could you please explain how $R(A_\lambda) \subset D(A)$ so that the writing $A\left(A_\lambda v\right)$ makes sense?
As @JohnDoe mentioned in a comment, I interpreted the text incorrectly. It's not necessarily true that $R(A_\lambda) \subset D(A)$. However, it's true that the image of $D(A)$ through $A_\lambda$ is a subspace of $D(A)$, i.e., $$ A_\lambda v \in D(A) \quad \forall v \in D(A). $$