Stone's Theorem Integral: Bad Example!

Question

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Problem

Given a finite Borel measure $\mu(\mathbb{R})<\infty$ and a Banach space $E$.

Construct a strongly continuous group $T(t)$.

Find a trajectory that is not uniform limit: $$x(t):=T(t)x_0:\quad s_n\nrightarrow x$$ so that integrals must be not taken as uniform but as Bochner or Riemann.

freishahiri · Accepted Answer · 2014-12-12T10:09:34.167

1

Consider the Banach space $\ell^\infty(\mathbb{N})$.

Define the group on the basis: $$T(t)\chi_{\{n\}}:=e^{it/n!}\chi_{\{n\}}$$ and extended by linearity and continuity.

Then it is a group: $$T(s+t)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)=\sum_{k\in\mathbb{N}}e^{i(s+t)/k!}a_k\chi_{\{k\}}=T(t)T(s)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)$$ $$T(0)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)=\sum_{k\in\mathbb{N}}e^{i0/k!}a_k\chi_{\{k\}}=1\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)$$ of bounded even isometric operators: $$\left\|T(t)\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)\right\|=\sup_{k\in\mathbb{N}}\left|e^{it/k!}\right|\cdot|a_k|=\sup_{k\in\mathbb{N}}1\cdot|a_k|=\left\|\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right\|$$ that is even uniformly continuous: $$\left\|(T(h)-T(0))\left(\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right)\right\|=\sup_{k\in\mathbb{N}}\left|e^{ih/k!}-1\right|\cdot|a_k|\leq\left|e^{ih/1!}-1\right|\cdot\left\|\sum_{k\in\mathbb{N}}a_k\chi_{\{k\}}\right\|$$ So it satisfies all requirements.

Consider a finite Borel measure $\mu(\mathbb{R})<\infty$.

Regard the trajectory: $$x_0:=\sum_{k\in\mathbb{N}}1\chi_{\{k\}}:\quad x(t):=T(t)x_0$$

Then its range is bounded but not precompact as: $$\|x((2M)!\pi)-x((2N)!\pi)\|=\sup_{k\in\mathbb{N}}\left|e^{i(2M)!\pi/k!}-e^{i(2N)!\pi/k!}\right|\geq\left|e^{i(2M)!\pi/(2N)!}-e^{i(2N)!\pi/(2N)!}\right|\stackrel{M<N}{=}2$$ so that it can't be uniform limit of simple functions $s_n\nrightarrow x$.

edited Dec 12 '14 at 10:09

answered Nov 30 '14 at 23:03

freishahiri

17,045

What does your $T(t)$ do to basis elements? I don't seen an actual definition, and what I'm thinking does not have the property $T(t)T(t')=T(t+t')$. – Disintegrating By Parts Dec 01 '14 at 10:44
@T.A.E.: Oh there was missing an $e_n$. Does it make more sense now? – freishahiri Dec 01 '14 at 10:47
That's what I was thinking. For $0 \le t+t' \le 1$, you see that $T(t+t')e_{n}$ involves only $e_{n}$, $e_{n+1}$, but that's not the case for $T(t)T(t')e_{n}$. – Disintegrating By Parts Dec 01 '14 at 10:53
Aaah, right. Good that you double checked it! Do you have an idea how to adjust it appropriately? – freishahiri Dec 01 '14 at 10:55
I don't any idea how to adjust it. – Disintegrating By Parts Dec 01 '14 at 10:58
But do you guess there is one? – freishahiri Dec 01 '14 at 11:19
You forgot to ping me with @ . Remember, this is your answer and I don't get the comments. No, I don't think you can salvage this definition. – Disintegrating By Parts Dec 02 '14 at 00:47
@T.A.E.: My fault. But do you guess there's any such example at all or are all orbits of elements under strongly continuous semigroups necessarily uniform limits of Banach-space-valued simple functions and thus uniformly integrable? Besides, may you do a quick double check on the other main ones? It would be a mess if there's flaw at crucial points, too. – freishahiri Dec 02 '14 at 00:56
@T.A.E.: Hehe I think I got it. Anyway can you have a quick check on it, please? – freishahiri Dec 12 '14 at 02:22

Stone's Theorem Integral: Bad Example!

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