If $A$ is maximal monotone and symmetric, then its resolvent and Yosida approximation are symmetric

Question

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. One says that

• $A$ is symmetric IFF $\langle Au, v \rangle = \langle u, A v \rangle$ for all $u,v\in D(A)$.
• $A$ is self-adjoint IFF $D(A) =D(A^*)$ and $A^*=A$.

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 $\left\|J_\lambda\right\|_{\mathcal{L}(H)} \leq 1$. We have from Brezis' Functional Analysis, i.e.,

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 $\lambda>0$,
• (a2) $A_\lambda v = J_\lambda A v$ for all $v \in D(A)$ and $\lambda>0$,
• (b) $|A_\lambda v| \le |Av|$ for all $v \in D(A)$ and $\lambda>0$,
• (c) $\lim_{\lambda \to 0^+} J_\lambda v = v$ for all $v \in H$,
• (c) $\lim_{\lambda \to 0^+} A_\lambda v = Av$ for all $v \in D(A)$,
• (e) $\langle A_\lambda v, v \rangle \ge 0$ for all $v \in H$ and $\lambda>0$,
• (f) $|A_\lambda v| \le \frac{1}{\lambda} |v|$ for all $v \in H$ and $\lambda>0$.

In order to prove Proposition 7.6 (in the same book), I would like to verify that

If $A$ is maximal monotone and symmetric, then

1. $I + \lambda A$ is symmetric on $D(A)$ for all $\lambda>0$,
2. $J_\lambda$ is symmetric on $H$ for all $\lambda>0$,
3. $A_\lambda$ is symmetric on $H$ for all $\lambda>0$,
4. $J_\lambda A$ is symmetric on $D(A)$ for all $\lambda>0$.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?

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Notice that $(I+\lambda A)$ is bijective from $D(A)$ onto $H$, so $J_\lambda$ is bijective from $H$ onto $D(A)$. It suffices to consider $\lambda=1$.

1. Let $u,v \in D(A)$. Then $$ \begin{align*} \langle (I + A)u, v \rangle &= \langle u, v \rangle + \langle Au, v \rangle \\ &= \langle u, v \rangle + \langle u, Av \rangle \quad \text{because} \quad A \text{ is symmetric}\\ &= \langle u, (I + A) v \rangle. \end{align*} $$

2. Let $u,v \in H$. Then $$ \begin{align*} \langle J_1u, v \rangle &= \langle (I+A)^{-1} u, (I+A)(I+A)^{-1}v \rangle \\ &= \langle (I+A) (I+A)^{-1} u, (I+A)^{-1}v \rangle \quad \text{by} \quad (1.) \\ &= \langle u, (I+A)^{-1}v \rangle \\ &= \langle u, J_1 v \rangle. \end{align*} $$

3. Let $u,v \in H$. Because $D(A)$ is dense in $H$ and $A_1$ is continuous, it suffices to restrict ourself to $u,v \in D(A)$. We have $$ \begin{align*} \langle A_1u, v \rangle &= \langle AJ_1u, v \rangle\\ &= \langle J_1u, Av \rangle \quad \text{because} \quad A \text{ is symmetric}\\ &= \langle u, J_1 Av \rangle \quad \text{by} \quad (2.) \\ &= \langle u, A_1 v \rangle. \end{align*} $$

4. Let $u,v \in D(A)$. Then $$ \begin{align*} \langle J_1 Au, v \rangle &= \langle Au, J_1 v \rangle \quad \text{by} \quad (2.) \\ &= \langle u, A J_1 v \rangle \quad \text{because} \quad A \text{ is symmetric}\\ &= \langle u, J_1 A v \rangle. \end{align*} $$

This completes the proof.