Problem
Given a Banach space $E$.
Consider a contraction C0-group: $$T:\mathbb{R}\to\mathcal{B}(E):\quad\|T(t)x\|\leq\|x\|$$
Define its generator by: $$Ax:=\lim_{h\to0}\frac{1}{h}(T(h)x-x)\in E$$ (It it is densely defined closed operator.)
Denote the convergence radius by: $$\rho_x:=\left(\limsup_{k\to\infty}\sqrt[k]{\frac{1}{k!}\|A^kx\|}\right)^{-1}$$
Generate a group via Taylor series: $$\rho_x=\infty:\quad e^{tA}x:=\sum_{k=0}^\infty\frac{1}{k!}t^kA^kx$$
Denote for shorthand: $$x(t):=T(t)x$$
Introduce the smoothed elements: $$x_n:=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}ne^{-(ns)^2}x(s)\mathrm{d}s$$
So the closure was all: $$(x_n,e^{tA}x_n)\to(x,e^{tA}x)$$ How to prove this?
Attempt
The intgral exists since: $$x\in\mathcal{C}(E):\quad\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}ne^{-(ns)^2}\|x(s)\|\mathrm{d}s\leq\|x\|<\infty$$
Clearly it is an approximation: $$x\in\mathcal{C}(\mathbb{R}):\quad x_n=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}ne^{-(ns)^2}x(s)\mathrm{d}s\to x(0)=x\quad(\|x\|_\infty=\|x\|<\infty)$$ (See thread on: Mollifiers: Approximation)
Moreover they are smooth since: $$\varphi\in\mathcal{C}^\infty(\mathbb{R})\cap\mathcal{L}(\mathbb{R}):\quad\int_{-\infty}^{\infty}\frac{1}{h}\left(\varphi(\hat{s}-h)-\varphi(\hat{s})\right)x(\tfrac{1}{n}\hat{s})\mathrm{d}\hat{s}\to(-1)\int_{-\infty}^{\infty}\varphi'(\hat{s})x(\tfrac{1}{n}\hat{s})\mathrm{d}\hat{s}$$ (See thread on: Mollifiers: Derivative)
But do the powers grow slow enough: $$\|A^kx_n\|\leq\int_{-\infty}^{\infty}\left|\tfrac{\mathrm{d}^k}{\mathrm{d}\hat{s}^k}e^{-\hat{s}^2}\right|\cdot\|x\|\mathrm{d}\hat{s}\leq\alpha a^n\|x\|$$
Also what about the closure: $$\|e^{tA}x_m-e^{tA}x_n\|\leq\int_{-\infty}^{\infty}e^{e^{-\hat{s}^2}}\|x(\tfrac{1}{m}\hat{s})-x(\tfrac{1}{n}\hat{s})\|\mathrm{d}\hat{s}\to0$$
That proves the assertion!
Reference
This is the close-up to: Semigroups: Entire Vectors (I)
It is taken from: Engel & Nagel, Exercise 3.12, Page 81