For $m$-accretive operators $A$ in a Banach space $(X,\|\cdot\|)$ we define the generalized domain by \begin{align} \hat{D}(A) := \{ x \in X : \exists (x_n,y_n) \in A , n \in \mathbb{N}: x_n \rightarrow x \,\, \text{in} \,\, X, \sup_{n \in \mathbb{N}} \|y_n\| < \infty \}. \end{align} In general we have $D(A) \subseteq \hat{D}(A) \subseteq \overline{D(A)} $. If $X$ is reflexive and $A: D(A) \rightarrow X$ is linear, we have $D(A) = \hat{D}(A)$.
Sketch of proof:
Since $A$ is $m$-accretive, $A$ is closed. Therefore its graph is closed, moreover $A$ is linear, so we get that its graph is convex. Because closed and convex subsets of a Banach space are weakly closed, the graph of $A$ is also weakly closed. So we now have that $A$ is weakly closed. Now let $x \in \hat{D}(A)$ and $(x_n)_{n} \subseteq D(A)$, such that $x_n \rightarrow x$ and $\sup_{n \in \mathbb{N}} \|y_n\| < \infty$. Because $X$ is reflexive we can assume that $Ax_n \rightharpoonup y$ for some $y \in X$ after extraction of a subsequence because of the Eberlein-Shmulyan theorem. Since $A$ is weakly closed we get $x \in D(A)$ and $Ax = y$ and therefore $D(A) = \hat{D}(A)$.
Now assume $A$ is a nonlinear and $m$-accretive operator in the reflexive space $X$. We then do not have the convexity of the graph of $A$ and therefore the proof does not hold.
Is the equation $D(A) = \hat{D}(A)$ still true? If yes, how to proof it?