Stone's Theorem Integral: Basic Integral

Question

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?

Answer 1

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!

Answer 2

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.