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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

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C-star-W-star
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  • You need to explicit that $t\rightarrow T(t)x$ is continuous in $t$ as a map from $\mathbb{R}$ to $X$ for each fixed $t$. That's the $C^{0}$ property. Otherwise what you wrote is not true. – Disintegrating By Parts Jan 09 '15 at 13:14
  • Yes I implicitely assumed so. Btw I was also gonna ask you: What does the zero stand for in $\mathcal{C}_0$? – C-star-W-star Jan 09 '15 at 13:27
  • $C^{1}$ is continuously differentiable, $C^{2}$ is twice continuously differentiable. $C^{0}$ is continuous. You cannot implicitly assume $C^{0}$. It is an independent assumption, and critical for what you're doing. – Disintegrating By Parts Jan 09 '15 at 13:28
  • So $\mathcal{C}_0$ means the same as $\mathcal{C}^0$? (It's just so weird to write the regularity as subscript.) – C-star-W-star Jan 09 '15 at 13:31
  • Yes, $C_{0}$ means that $t\mapsto T(t)x$ is $C^{0}$ for all $x$. The proper assumption is that it is continuous from the right at $t=0$, but it ends up implying continuity everywhere for a semigroup of bounded linear operators. – Disintegrating By Parts Jan 09 '15 at 13:33
  • Ok, nice. :) Do you think it's worth it to open a new thread specially on that smoothness property claimed above in general setting with lebesgue-integral tag? – C-star-W-star Jan 09 '15 at 13:36
  • Ah so the zero actually stands for continuous in zero (from the right)? That makes sense as minimal requirement for the generator to exist. – C-star-W-star Jan 09 '15 at 13:38
  • $C_{0}$ semigroup specifically means $T(t)$ is a bounded linear operator for $t \ge 0$, $T(0)=I$, $T(s+t)=T(s)T(t)$ for $t, s \ge 0$ and $\lim_{h\downarrow 0}T(h)x=x$ for all $x \in X$. But $C_{0}$ ends up meaning $t\mapsto T(t)x$ is $C^{0}$ for all $x$. – Disintegrating By Parts Jan 09 '15 at 13:43
  • Your definition of $e^{tA}$ is not an appropriate one. – Disintegrating By Parts Jan 10 '15 at 00:54
  • @T.A.E.: What do you mean? – C-star-W-star Jan 10 '15 at 00:56
  • Actually, what do you mean? The point of semigroup theory is to define operators with an exponential property; one deliberately avoids an exponential series for an unbounded operator $A$ because it may not defined for anything except $0$. – Disintegrating By Parts Jan 10 '15 at 00:59
  • Yes, for the general study of semigroup theory this here might be even useless. So one is even forced to avoid the approach by series. But for the study of thermodynamical states it is extremely practical to know that the entire elements establish a core for $\mathcal{C}_0$-groups. So that's why I'm doing this here. – C-star-W-star Jan 10 '15 at 01:22
  • @T.A.E.: Besides, do you have an example at hand where this happens for $\mathcal{C}_0$-semigroups? :) – C-star-W-star Jan 10 '15 at 01:23

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