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Disclaimer

This thread has been renewed: Stone's Theorem Integral: Advanced Integral

Problem

Given a finite Borel measure $\mu$ and a Hilbert space $\mathcal{H}$.

Consider a strongly continuous unitary group $U:\mathbb{R}\to\mathcal{B}(\mathcal{H})$.

Take the time evolution $\varphi(t):=U(t)\varphi$.

In worst case it is locally lipschitz: $$\|\varphi(t)-\varphi(s)\|\leq L_\varphi|t-s|$$ But the real line is not compact.

So it may not be approximated uniformly by simple functions!

Thus the uniform integral makes no sense in this case: $$\varphi=\lim_n\sigma_n:\quad\int_\mathbb{R}\varphi(s)\mathrm{d}s:=\lim_n\int_\mathbb{R}\sigma_n(s)\mathrm{d}s$$

How to solve this debacle?

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freishahiri
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2 Answers2

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Hilbert space integrals are easiest to define using weak arguments, $$ B_{U}(x,y) = \int (U(t)x,y)\,d\mu(t),\\ |B_{U}(x,y)| \le \mu(\mathbb{R})\|x\|\|\|y\|, \\x,y\in\mathcal{H}. $$ Therefore, there exists a unique bounded linear operator, that we'll write as $\int U(t)\,d\mu(t)$, such that $$ \int (U(t)x,y)\,d\mu(t) = \left(\int U(t)\,d\mu(t)x,y\right), \\x,y\in\mathcal{H}. $$ Automatically, $$ \left\|\int U(t)\,d\mu(t)\right\| \le \mu(\mathbb{R}). $$ What's nice is that everything reduces to integration of scalar functions $(U(t)x,y)$, and the weak equalities don't need proof because they define the operator. Furthermore, weak measurability is all you need.

freishahiri
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Disintegrating By Parts
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  • Yepp, certainly the best way to do so!! ;) ...Besides I just thought how much is behind the sayings in literature that they can take it as Riemann integral and how this relates to some sort of uniformly bounded integral and - now - what really surprised me is that the latter two really can be different?! – freishahiri Nov 04 '14 at 05:32
  • @Freeze_S : These techniques work for reflexive Banach spaces, too in pretty much the same way. And they can work in non-reflexive spaces very often with a little more work. – Disintegrating By Parts Nov 04 '14 at 05:35
  • Btw isn't that one precisely Pettis? – freishahiri Nov 04 '14 at 05:36
  • @Freeze_S : I'm not sure. I never studied the Pettis integral. – Disintegrating By Parts Nov 04 '14 at 05:40
  • Can you explain that with reflexive spaces? – freishahiri Nov 04 '14 at 05:50
  • I mean according to the definition it's just the Pettis integral not knowing to much about it either... – freishahiri Nov 04 '14 at 05:51
  • For fixed $x$, you can define $x^{\star}\mapsto \int x^{\star}(F(t)x),d\mu(t)$, which becomes an element of $X^{\star\star}$. If reflexive, then $\int x^{\star}(F(t)x),d\mu(t)=x^{\star}(y)$ for a unique $y\in X$, which is defined to be $\int F(t)x,d\mu(t)$. – Disintegrating By Parts Nov 04 '14 at 05:55
  • Ok sorry for that long break (it took me a while to go through the details of double dual again) but shouldn't there be any assumption on $F$ in order for the expression to become an element of the double dual? – freishahiri Nov 04 '14 at 06:54
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    Fix $x$. Assume $|F(t)x|$ is measurable and $x^{\star}(F(t)x)$ is measurable for all $x^{\star}$, with $\int |F(t)x|d\mu(t) < \infty$. Then $|\int x^{\star}(F(t)x),d\mu(t)| \le |x^{\star}|\int |F(t)x|,d\mu(t)$. That's enough for reflexive spaces. You can get work up to spaces such as $L^{1}$: Look at the first lemma of http://www.ams.org/journals/proc/1986-097-02/S0002-9939-1986-0835903-X/S0002-9939-1986-0835903-X.pdf . – Disintegrating By Parts Nov 04 '14 at 07:00
  • Do they talk about the algebraic dual $X^*$ or the topological dual $X'$? – freishahiri Nov 04 '14 at 07:35
  • That is really Pettis ;): The existence of an element representing the Pettis integral is guaranteed by reflexivity plus assumptions here. – freishahiri Nov 04 '14 at 07:57
  • They're talking about convergence in the AMS link that I posted. This is about the topological dual $X^{\star}$. Look at the first lemma. – Disintegrating By Parts Nov 04 '14 at 10:44
  • Alright I just wasn't sure since their always mentioning 'linear functional' so linear continuous indeed. – freishahiri Nov 04 '14 at 10:47
  • @Freeze_S : You really should look at that first lemma. If I'm not wrong, it implies: If (a) $|f(t)|$ is measurable and integrable (b) $x^{\star}(f)$ is measurable for all $x^{\star}$ and (c) the range of $f$ lies in separable closed subspace of a Banach space $X$, then you have an integral. The case of reflexive $X$ is even easier, of course, because no assumption about the range of $f$ is necessary in that case. Krein-Smulian is very powerful, even in the results leading up to it. – Disintegrating By Parts Nov 04 '14 at 12:25
  • Yep, so for the separable case knowing that the weak expression being measurable and the norm expression being integrable already implies that there exists an integral. Nice. :) Unfortunately there are non-weakly-integrable functions precisely because of nonseparable range. That was also one of my driving forces to try to set up another notion of integral (Reinventing the wheel part 2...) – freishahiri Nov 04 '14 at 12:41
  • A nice example of a Riemann integrable not Bochner (nor Pettis afaik) is given by $F:[0,1]\to\ell_2([0,1]):t\mapsto\chi_t$. That even takes values in a Hilbert space but a nonseparable one... – freishahiri Nov 04 '14 at 12:46
  • @Freeze_S : If it's Riemann integrable, then $x^{\star}\circ F$ should be Riemann integrable, too. What are you claiming about this example? – Disintegrating By Parts Nov 04 '14 at 12:58
  • Yep right, I claim that it is an example of a Riemann integrable function having nonseparable image and therefore being at least not Bochner integrable. But you're right as it is Riemann integrable (in any kind of sense of sums or whatsoever) it must be weakly integrable as well. Moreover as its norm is integrable and since Hilbert spaces are reflexive it is integrable in the sense described above or equivalently in the sense of Pettis. So very good! ^^ – freishahiri Nov 04 '14 at 15:03
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Uniform Integral

Note that this is a potential nonexample: Stone's Theorem: Bad Example!

Bochner Integral

Since it is separable valued: $$\varphi\in\mathcal{C}(\mathbb{R},E):\quad\mathbb{R}\text{ separable}\implies\varphi(\mathbb{R})\text{ separable}$$ and weakly measurable: $$l\in E':\quad\varphi\text{ continuous}\implies l\circ\varphi\text{ measurable}$$ so by Pettis' criterion strongly measurable: $$\varphi\text{ Bochner measurable}$$

Also it is absolutely integrable: $$\int\|\varphi(s)\|\mathrm{d}s=\|\varphi\|\mu(\mathbb{R})<\infty$$

So the Bochner integral exists!

Riemann

One has lipschitz continuity: $$\|\varphi(t)-\varphi(s)\|\leq L_\varphi|t-s|$$ and a uniform bound: $$\|\varphi(t)-\varphi(s)\|\leq2\|\varphi\|$$ Thus choose a partition: $$\mathcal{P}_\varepsilon:=\{(-\infty,-R_\varepsilon],\ldots,(R\frac{k-1}{K_\varepsilon(R)},R\frac{k}{K_\varepsilon(R)}],\ldots,(R_\varepsilon,\infty)\}$$ in order to obtain: $$\|\sum_{A\in\mathcal{P}}\varphi(a)\mu(A)-\sum_{A'\in\mathcal{P}'}\varphi(a')\mu(A')\|\\\leq2\|\varphi\|\mu(-\infty,-R]+L_\varphi\frac{R}{K_\varepsilon(R)}\mu(-R,R]+2\|\varphi\|_\infty\mu(R,\infty)<3\left(\frac{\varepsilon}{3}\right)\quad(\mathcal{P},\mathcal{P}'\geq\mathcal{P}_\varepsilon)$$ So the Riemann integral exists!

Bochner vs. Riemann

This was no surprise as for finite measures: $$\|\varphi\|_\infty<\infty:\quad\varphi\in\mathcal{L}_\mathfrak{B}\implies\varphi\in\mathcal{L}_\mathfrak{R}$$ but it is not evident from the calculations that: $$\int_\mathfrak{B}\varphi(s)\mathrm{d}s=\int_\mathfrak{R}\varphi(s)\mathrm{d}s$$ So the integrals agree!

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freishahiri
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