# Stone's Theorem Integral: Avanced Integral

Reference

This problem grew out from: Stone's Theorem Integral: Basic Integral

Problem

Given the real line as measure space $$\mathbb{R}$$ and a Hilbert space $$\mathcal{H}$$.

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

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

This time the integral is taken over an infinite measure: $$\int_0^\infty e^{-\lambda s}\varphi(s) \, \mathrm ds$$ What interpretations are available and how do they agree?

The Riemann integral for vector functions with values in a Banach space $X$ is essentially the same as for scalar functions. $$\int_{a}^{b} F(t)\,dt = \lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{\mathcal{P}} F(t_{j}^{\star})\Delta_{j}t,$$ where $\mathcal{P}$ is a partition with partition points $$a = t_{0} < t_{1} < t_{2} < \cdots < t_{n} =b$$ and augmented with evaluation points $t_{j}^{\star}\in [t_{j-1},t_{j}]$. If you're used to the Darboux-Riemann integral using upper and lower integrals, then forget that one. Stick with the classical definition stated above. For scalar functions, these definitions are equivalent. Of course that's not the case here because The Darboux-Riemann integral makes no sense.

For scalar functions $F$, the integral exists iff the set of discontinuities of $F$ is of Lebesgue measure $0$. I believe the 'if' part is still true for Banach-space-valued functions. However, if you're concerned, piecewise continuous is always safe. Your integral over the infinite interval has to be an improper Riemann integral, just as it is for scalar functions, and the integral makes sense if the integrand is absolutely Riemann integrable on the infinite interval. So the following makes sense as an improper Riemann integral: $$\int_{0}^{\infty}e^{-\lambda t}U(t)x\,dt,\;\;\; \mathcal{R}\lambda > 0,\;\; x\in X.$$ This Laplace transform integral is fundamental to $C^0$ semigroup theory because it gives the resolvent of the generator.

• Aha, ok so as an improper Riemann integral continuity plus absolutely improper integrable suffice? – C-Star-W-Star Nov 25 '14 at 22:23
• @Freeze_S : Yes, continuity plus absolute integrability suffice. – DisintegratingByParts Nov 25 '14 at 22:35
• So formally this is saying that the net of integrals over (compact) measurable sets of finite mass converges. On compact sets continuity then bounds the function and makes the partial sums converge. Abstracting this to general measure space one has to drop compactness as the concept may not be available anymore. But the net of integrals over measurable sets of finite mass is still available. On the other hand even for finite Borel measures plus continuity this may be bad even when the functions is additionally bounded. – C-Star-W-Star Nov 26 '14 at 1:56
• Consider something rather bad as $\sin(\frac{1}{x})$ with Lebesgue measure on $[0,1]$. This one is not continuous but it can be turned a bounded and continuous example over a finite measure space oscillating faster and faster $\sin(x^2)$ with Borel measure having weight far outside. – C-Star-W-Star Nov 26 '14 at 2:08
• @Freeze_S : Riemann integrals require considering only bounded functions on $[a,b]$. Indeed, if $\int_{a}^{b}F(t)\,dt$ exists, then, for every $\epsilon > 0$, there exists $\delta > 0$ such that $\|\sum_{\mathcal{P}}F(t_{j}^{\star})\Delta_{j}t-\sum_{\mathcal{P'}}F(t_{k}^{\star})\Delta_{k}t\| < \epsilon$ whenever $\mathcal{P}$ and $\mathcal{P'}$ are partitions whose norms are bounded by $\delta$. Obviously that cannot happen if $F$ is unbounded because the intermediate points are allowed to vary arbitrarily. Nets are not needed for convergence; just a basic Cauchy test. – DisintegratingByParts Nov 26 '14 at 2:16

Bochner

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

Also it is absolutely integrable: $$\int\|\varphi(s)\|\alpha(s)\mathrm{d}s=\frac{1}{\lambda}\|\varphi\|<\infty$$

So the Bochner integral exists!

Improper Bochner

This one coincides with the former by dominated convergence.

Improper Riemann

Especially, it is bounded: $$\lambda(A)<\infty:\quad\|\alpha\varphi\|_A\leq1\|\varphi\|<\infty$$ so for subspaces of finite measure: $$\lambda(A)<\infty:\quad\alpha\varphi\in\mathcal{L}_\mathfrak{R}(A)\cap\mathcal{L}_\mathfrak{B}(A)$$ But the real line is $\sigma$-finite so one has: $$\int_A\alpha(s)\varphi(s)\mathrm{d}s\to\int_0^\infty\alpha(s)\varphi(s)\mathrm{d}s$$

So the improper Riemann integral exists and agrees with others!

Induced Bochner

Consider the induced Borel measure: $$\mu(A):=\int_A\alpha(s)\mathrm{d}s:\quad\mu(\mathbb{R})=\frac{1}{\lambda}<\infty$$

Then one has absolute integrability as: $$\int\|\varphi(s)\|\mathrm{d}\mu(s)=\int\|\varphi(s)\|\alpha(s)\mathrm{d}s<\infty$$ and again measurability by continuity.

So it is Bochner integrable.

As the function is measurable and bounded one can construct: $$\|\sigma_n\|_\infty\leq\|\varphi\|_\infty+1:\quad\sigma_n\to\varphi$$ Thus one obtains by dominated convergence: $$\int\varphi(s)\mathrm{d}\mu(s)\leftarrow\int\sigma_n(s)\mathrm{d}\mu(s)=\int\sigma_n(s)\alpha(s)\mathrm{d}s\to\int\varphi(s)\alpha(s)\mathrm{d}s$$

So the induced Bochner integral exists and agrees with others!

Induced Riemann

One last time by boundedness: $$\|\varphi\|_\infty<\infty:\quad\varphi\in\mathcal{L}_\mathfrak{R}(\mu)\cap\mathcal{L}_\mathfrak{B}(\mu)$$

So the induced Riemann integral exists and agrees with others!