Show that the semigroup corresponding to the Yosida approximation is contractive

Question

Let $(\mathcal D(A),A)$ be a closed densely-defined linear operator on a $\mathbb R$-Banach space $E$ such that $(0,\infty)$ is contained in the resolvent set $\rho(A)$ of $(\mathcal D(A),A)$ and $$\left\|\lambda R_\lambda(A)\right\|_{\mathfrak L(E)}\le1\;\;\;\text{for all }\lambda>0\tag1,$$ where $$R_\lambda(A):=B_\lambda^{-1}$$ with $$B_\lambda:=\lambda\operatorname{id}_{\mathcal D(A)}-A$$ for $\lambda\in\rho(A)$. Now, let $$A_n:=nAR_n(A)=n^2R_n(A)-n\operatorname{id}_E$$ and $$T_n(t):=e^{tA_n}\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$.

How can we show that $$\left\|T(t)\right\|_{\mathfrak L(E)}\le 1?\tag2$$

I've seen a proof claiming that $$\left\|T(t)\right\|_{\mathfrak L(E)}\le e^{-nt}e^{\left\|n^2R_n(A)\right\|_{\mathfrak L(E)}t}\tag3\;\;\;\text{for all }n\in\mathbb N,$$ but how do we obtain this? Clearly, $$\left\|e^{tB}\right\|_{\mathfrak L(E)}\le e^{t\left\|B\right\|_{\mathfrak L(B)}}\;\;\;\text{for all }t\ge0\tag4$$ for all $B\in\mathfrak L(E)$, but I don't get why we can pull out the $e^{-nt}$ in $(3)$.

Answer 1

In the Hille-Yosida Theorem, it is assumed that $(\lambda I-A)$ is invertible for all $\lambda > 0$, and that $\|\lambda (\lambda I-A)^{-1}\| \le 1$ for all $\lambda > 0$. Therefore, if $x\in\mathcal{D}(A)$ and $\lambda > 0$, \begin{align} \lambda(\lambda I-A)^{-1}x&=(\lambda I-A)(\lambda I-A)^{-1}x+(\lambda I-A)^{-1}Ax \\ &= x+(\lambda I-A)^{-1}Ax \\ \|\lambda(\lambda I-A)^{-1}x-x\| &\le \frac{1}{\lambda}\|Ax\|. \end{align} Therefore $\lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}x=x$ for all $x\in\mathcal{D}(A)$. Using the density of the domain of $A$ and the resolvent estimate $\|\lambda(\lambda I-A)^{-1}\|$, it follows that $$ \lim_{\lambda\rightarrow\infty}\lambda(\lambda I-A)^{-1}x=x,\;\; \forall x\in X. $$ Therefore, $$ \lim_{\lambda\rightarrow\infty}\lambda (\lambda I-A)^{-1}Ax=Ax,\;\;x\in\mathcal{D}(A) \\ \lim_{\lambda\rightarrow\infty} \lambda(\lambda I-A)^{-1}\{(A-\lambda I)x+\lambda I\}x=Ax \\ \lim_{\lambda\rightarrow\infty}\{-\lambda+\lambda^2(\lambda I-A)^{-1}\}x=Ax. $$ The semigroup for $A$ can be defined as \begin{align} T(t)x & = \lim_{\lambda\rightarrow\infty}\exp\left[t\{-\lambda+\lambda^2(\lambda I-A)^{-1}\}\right]x \end{align} For any $x\in X$ and $\lambda > 0$, $$ \|\exp\left[t\{-\lambda+\lambda^2(\lambda I-A)^{-1}\}\right]x\| \\ \le e^{-t\lambda} \exp\left[t\lambda\|\lambda(\lambda I-A)^{-1}\|\right]\|x\| \\ \le e^{-t\lambda} e^{t\lambda}\|x\|= \|x\|. $$ Therefore $\|T(t)\| \le 1$ for all $t \ge 0$.

Answer 2

How can we show $(2)$?

Following Pazy's argument (which seems to be the standard one, where $T(t)$ is defined by $T(t)x=\lim_{n\to\infty} \mathrm{e}^{t A_n}x$), it is enough to show that $$\|\mathrm{e}^{t A_n}\|_\mathcal{L}\leq 1,\quad\forall\ n\in\mathbb N.$$

For this, note that \begin{equation}\begin{aligned} \|\mathrm{e}^{t A_n}\|_\mathcal{L} &= \|\mathrm{e}^{tn^2(n-A)^{-1}}\mathrm{e}^{-tn I}\|_{\mathcal{L}} = \|\mathrm{e}^{tn^2(n-A)^{-1}}\mathrm{e}^{-tn }I\|_{\mathcal{L}} =\mathrm{e}^{-tn}\|\mathrm{e}^{tn^2(n-A)^{-1}}\|_{\mathcal{L}}\\ &\leq\mathrm{e}^{-tn}\mathrm{e}^{tn^2\|(n-A)^{-1}\|_\mathcal{L}}\leq \mathrm{e}^{-tn}\mathrm{e}^{tn}=1. \end{aligned}\end{equation}