Is $(A v_\lambda)_{\lambda>0}$ a Cauchy sequence in $H$?

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 an unbounded linear operator that is densely defined. 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 $A^*=A$.

I'm trying to prove Proposition 7.6 in Brezis' Functional Analysis, i.e.,

If $A$ is maximal monotone and symmetric, then $A$ is self-adjoint.

In below failed attempt, I'm unable to prove that $(A v_\lambda)_{\lambda>0}$ is a Cauchy sequence in $H$. Could you please elaborate on how to fix it?

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• 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 $A$.

• We define a new inner product on $D(A)$ by $$ \langle u, v \rangle_{D(A)} := \langle u, v \rangle + \langle Au, Av \rangle. $$

I already proved that

• Lemma 1 $A$ is symmetric IFF $A \subset A^*$, i.e., $D(A) \subset D(A^*)$ and $A^*=A$ on $D(A)$.
• Lemma 2 $D(A)$ together with $\langle \cdot, \cdot \rangle_{D(A)}$ is a Hilbert space
• Lemma 3 $A_\lambda$ is symmetric on $H$ for all $\lambda>0$.

By Lemma 1, it remains to prove that $D(A^*) \subset D(A)$. Let $v \in D(A^*)$. For $\lambda >0$, let $v_\lambda := J_\lambda v \in D(A)$. Then $\lim_{\lambda \to 0^+} v_\lambda =v$ by Proposition 7.2(c) (in the same book). By Lemma 2, it remains to prove that $(A v_\lambda)_{\lambda>0}$ is a Cauchy sequence in $H$. Let $B_H$ be the closed unit ball of $H$. Notice that $D(A)$ is dense in $H$. For $\lambda, \mu>0$, we have $$ \begin{align*} |A v_\lambda - A v_\mu|^2 &= |(A_\lambda - A_\mu)v|^2 \\ &= \sup_{u \in B_H \cap D(A)} \langle (A_\lambda - A_\mu) v, u \rangle \\ &= \sup_{u \in B_H \cap D(A)} \langle v, (A_\lambda - A_\mu) u \rangle \quad \text{by Lemma 3} \\ &= \sup_{u \in B_H \cap D(A)} \langle v, A(J_\lambda - J_\mu) u \rangle. \end{align*} $$

Because $v\in D(A^*)$, there is $c>0$ such that $$ |\langle v, Au \rangle| \le c |u| \quad \forall u \in D(A). $$

Then $$ \begin{align*} |A v_\lambda - A v_\mu|^2 \le c \sup_{u \in B_H \cap D(A)} |(J_\lambda - J_\mu) u|. \end{align*} $$

Answer 1

I am not sure if $Av_\lambda$ is Cauchy. I tried to come up with a counterexample but I haven't succeded yet. However if you want a proof of $D(A^*) \subset D(A)$, let $v\in D(A^*)$, and $v_\lambda = v + \lambda A^* v$ then

$$\left\langle v_\lambda, u\right\rangle = \left\langle v, \left(I + \lambda A\right)u\right\rangle,\; \forall u \in D(A)$$

then, $$\left\langle v_\lambda, J_{\lambda}u\right\rangle = \left\langle v, u\right\rangle,\; \forall u \in H$$

Since $J_\lambda$ is symmetric, then $v = J_\lambda v_\lambda\in D(A)$