# Show that the semigroup corresponding to the Yosida approximation is contractive

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 $$(3)$$.

• $e^{A+B}=e^{A}e^{B}$ if $A$ and $B$ commute. In your case, $-nI$ and $n^2 R_n(A)$ commute. Nov 9, 2018 at 0:43
• @DisintegratingByParts Do you have a reference for that? Nov 9, 2018 at 9:23
• What is $A_1$? $I-A$ or $A(I-A)^{-1}$? Nov 9, 2018 at 15:36
• A power series argument works the same for real or complex numbers $A,B$ as it does for bounded commuting operators. In either case, $e^{A+B}=e^{A}e^{B}$. Nov 9, 2018 at 18:14
• @Pedro $A_1=\operatorname{id}_{\mathcal D(A)}-A$. Why do you think it could be something else? Nov 10, 2018 at 11:27

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 $$t \ge 0$$.
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{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}