Is it true that $\|J_{\lambda_n} - J_\lambda\|_{\mathcal L(H)} \to 0$ as $\lambda_n \to \lambda$?

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. 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 $\lambda>0$.

Let $\lambda_n, \lambda >0$ such that $\lambda_n \to \lambda$. I would like to ask if the stronger conclusion that $\|J_{\lambda_n} - J_\lambda\|_{\mathcal L(H)} \to 0$ is true. Thank you so much for your elaboration!

Answer 1

\begin{align} \left\|J_{\lambda_n} - J_{\lambda}\right\|_{\mathcal L(H)} &= \left\|\left(I + \lambda_n A\right)^{-1} - J_{\lambda}\right\|_{\mathcal L(H)}\\ &= \left\|\left(I + \lambda A + \left(\lambda_n - \lambda\right)A\right)^{-1} - J_{\lambda}\right\|_{\mathcal L(H)}\\ &= \left\|J_{\lambda} \left(I + \left(\lambda_n - \lambda\right)AJ_{\lambda}\right)^{-1} - J_{\lambda}\right\|_{\mathcal L(H)}\\ &= \left\|J_{\lambda} \left(I + \left(\lambda_n - \lambda\right)A_{\lambda}\right)^{-1} - J_{\lambda}\right\|_{\mathcal L(H)}\\ &= \left\|J_{\lambda}\sum_{k=1}^{\infty} (-1)^k\left(\lambda_n - \lambda\right)^k A_{\lambda}^{k}\right\|_{\mathcal L(H)}\\ &\le \sum_{k=1}^{\infty} \left|\lambda_n - \lambda\right|^k \frac1{\lambda^k}= \frac{\left|\lambda_n - \lambda\right|}{\lambda - \left|\lambda_n - \lambda\right|} \to 0 \end{align}