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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?

Tesla
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C-star-W-star
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2 Answers2

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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.

C-star-W-star
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Disintegrating By Parts
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  • 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. – Disintegrating By Parts 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 01:56
  • 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 02:08
  • @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. – Disintegrating By Parts Nov 26 '14 at 02:16
  • By convention yes but one is not restricted to confine oneself to only compact intervals. In fact every finite measure space serves for a generalized Riemann integral. – C-star-W-star Nov 26 '14 at 02:19
  • Do you think $\sin(1/x)$ is Riemann integrable on $[0,1]$ if you set the value of the function to--say--0 at $x=0$? A bounded function $f$ is Riemann integrable on $[0,1]$ iff its set of discontinuities is of Lebesgue measure $0$. – Disintegrating By Parts Nov 26 '14 at 02:25
  • No of course it is not Riemann integrable only improperly but one can swap it around and construct a continuous and bounded function over the whole real line that oscillates more and more too say $\sin(x^2)$ together with a finite Borel measure say $\int_Ae^{-x^2}\mathrm{d}x$. – C-star-W-star Nov 26 '14 at 02:28
  • If you let $f(x)=\sin(1/x)$ on $(0,1]$ and $f(0)=0$, then $f$ is bounded and its set of discontinuities has Lebesgue measure $0$. So $f$ is Riemann integrable on $[0,1]$. – Disintegrating By Parts Nov 26 '14 at 02:30
  • A big theorem proved by Riemann-Stieltjes: If $f$ is of bounded variation on $[0,2\pi]$, then the Fourier series for $f$ converges everywhere to the mean of the left- and right-hand limit of $f$. If $f$ is continuous on $[a,b]\subseteq [0,2\pi]$, then the convergence is uniform on any $[a',b']\subset(a,b)$. Furthermore, this convergence is the same for all the orthogonal expansions associated with regular Sturm-Liouville problems on $[0,2\pi]$. Do you deal with a lot of functions which are not of bounded variation? – Disintegrating By Parts Nov 26 '14 at 02:35
  • Besides the improperness of $\int_0^1\sin(\frac{1}{x})\mathrm{d}x$ can perfectly be cured by allowing measurable partitions instead of intervals only. Then as $\sin(\frac{1}{x})$ is bounded plus measurable it is uniform limit of simple functions and thus Riemann integrable in the generalized sense. – C-star-W-star Nov 26 '14 at 02:37
  • @Freeze_S : Nothing improper about the Riemann integral of the $f$ as described using $\sin(1/x)$. Give $f(0)$ a value and $f$ is Riemann integrable on $[0,1]$. – Disintegrating By Parts Nov 26 '14 at 02:38
  • Ups, right it is properly Riemann integrable. 'dough' – C-star-W-star Nov 26 '14 at 02:43
  • Hmm I'm not as versatile with PDE. – C-star-W-star Nov 26 '14 at 02:45
  • @Freeze_S : And you can have such oscillatory singularities at $1/2,1/3,1/4,\cdots$ and it is still Riemann integrable. – Disintegrating By Parts Nov 26 '14 at 02:46
  • @Freeze_S : Legendre polynomial expansions convergence everywhere in $(a,b)$ to $\frac{1}{2}(f(x+0)+f(x-0))$ if $f$ is of bounded variation on $[a,b]$. Same of Bessel functions, Hermite, Laguerre ... you name it. Where do you need more than bounded variation in Physics? And the convergence is uniform in $[a',b']\subset (a,b)$ if $f$ is continuous in $[a,b]$ as well. That's all Riemann-Stieltjes. Physics limits the path just because of energy, speed and time, at least normally. – Disintegrating By Parts Nov 26 '14 at 02:49
  • In fact, I like physics and my program is about physics but I love abstract math. So when I encounter such an expression as above I'm curious and would like to understand it in a more abstract setting. That is: Finite measure space and Banach-space-valued functions (or Hausdorff TVS). – C-star-W-star Nov 26 '14 at 02:57
  • Well yes Laguere Hermite and Legendre :) and maybe Bessel yes but for curiosity - I hope you understand – C-star-W-star Nov 26 '14 at 02:59
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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!

C-star-W-star
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