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I am interested in $\textbf{Integration in Banach spaces}$. Here is a little motivation for my question:

Let $\left(X,\|\cdot\| \right)$ be a Banach space, $a,b \in \mathbb{R}$ with $a<b$ and $f \colon [a,b] \longrightarrow X$ a function. How can we integrate such a function?

I could already find an answer with the $\textbf{Riemann Integral for Banach space-valued functions}$ (which is quite similar to the comon Riemann Integral) and the $\textbf{Bochner Integral}$ (which is similar to the Lebesgue Integral).

But so far I only know some theoretical results about those integrals (only the basical ones) and I have not yet seen or calculated a practical example.

Now I wonder if anybody could present me different examples of such a integral. (I am looking for such nice and epical integrals we know from Complex analysis or we could calculate using an $d$-dimensional Spherical coordinate system or something similar.)

I am also looking for any kind of (nice) calculations involving Integration in Banach Spaces. If anybody knows a rewarding (not too hard) theorem/proof involving Integration in Banach Spaces this would also interest me.

I hope you understand what I am searching for...

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    There's nothing fundamentally new in terms of the calculations. Any reasonable definition of a Banach space-valued integral should have the property that if $v^{\ast} : B \to \mathbb{R}$ is a linear functional, then $v^{\ast} (\int f) = \int v^{\ast} f$, where the latter is an ordinary one-dimensional integral. – Qiaochu Yuan Apr 06 '14 at 05:02
  • Is this a soft version of Hille's Theorem? If so, where can I find a proof for it? I already studied the proof of Hille's Theorem, but I need some more time because you need a lot of measure theory for it. –  Apr 06 '14 at 13:24

3 Answers3

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One useful example is the holomorphic functional calculus. It allows us to generalize Cauchy's integral formula from complex analysis in one variable to evaluate functions of operators.

Let $V$ be a Banach space and let $T$ be a bounded linear operator on $V$. If $\Gamma$ is a positively oriented rectifiable Jordan curve such that the spectrum of $T$ is contained in the interior of $\Gamma$, then for each function $f$ holomorphic on and inside $\Gamma$,

$$ f(T) = \frac{1}{2 \pi i} \oint_{\Gamma} f(\zeta) (\zeta I - T)^{-1} \, dz $$

The integrand is a function whose arguments are in $\mathbb{C}$ and that takes values in $V$, and hence it requires Bochner integration to make well-defined. The above formula is the proper generalization of the Cauchy integral formula

$$ f(z) = \frac{1}{2 \pi i} \oint_{\Gamma} f(\zeta) (\zeta - z)^{-1} \, dz,$$

where $\Gamma$ encloses $z$ (the value $z$ being the only element in the spectrum of the map $x \mapsto zx$).

This formula allows you to derive Bochner integral formulations for expressions like $\exp(T)$ or $\log(T)$ for certain linear operators $T$. In the case that $V = \mathbb{C}^{n \times n}$, then $T$ is a matrix and the Cauchy integral formulation for $\exp(T)$ matches the regular definition of the matrix exponential.

Christopher A. Wong
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  • Very nice useful example! I can need this one. Can I use your statement also for "toy contours" as they are defined in "COMPLEX ANALYSIS - Stein & Shakarchi"? –  Apr 04 '14 at 23:46
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    This material is used, among other places, in some of the foundational results leading up to the proof of the Gelfand-Naimark theorem; see http://qchu.wordpress.com/2012/07/17/banach-algebras-the-gelfand-representation-and-the-commutative-gelfand-naimark-theorem/ for some indications of how this works (I black-boxed the complex analysis though). – Qiaochu Yuan Apr 06 '14 at 05:05
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First of all, in order to "calculate" something explicitly you need a reasonably nice Banach space $B$ as your target. For instance, suppose your Banach space is $B=C([a,b])$. Then a continuous function $f: [0,T]\to B$ is nothing but a continuos function of two variables $F(x,t)$, $x\in [a,b], t\in [0,T]$: $$ f(t)(x)= F(x,t). $$ Now, computing the Bochner integral $$ \int_{0}^T f(t)dt $$ simply amounts to (if you follow the definition) computing the integral $$ \int_{0}^T F(x,t)dt $$ which is something you surely saw in a calculus of several variables class.

You can use a similar computation if your target is, say, $L^p([a,b])$ and so on.

Moishe Kohan
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    How would one in general compute the last integral? If the function were “simple”, a linear combination of continuous functions on $[a,b]$, then this can be done easily? Is the integral then computed/defined by approximating f by such simple functions? If so, is this something that can always be done, just like measurable functions can be compute by simple functions on a measure space if the target space is R or C. – user82261 Apr 22 '19 at 10:17
  • Ah, I see. You have the measure space [a,b] as your domain so we are in the Calc-II case. I was considering the case where the domain could be a measure space, for example a compact metric space with a Radon measure defined on it. I suppose computations in that case would be messier. – user82261 Apr 22 '19 at 10:29
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Elaborating a bit on @QiaochuYuan's comment: for almost every purpose, we certainly want the property that $T\int f=\int Tf$, where $f$ is $V$-valued, and $T$ is a continuous linear map $V\to W$. Under very mild hypotheses on the spaces $V,W$ (locally convex, quasi-complete suffices), by Hahn-Banach the desired property already follows from requiring $\lambda\int f=\int \lambda\circ f$ for continuous linear functionals $\lambda$. (We would also want estimates on $\int f$, but/and that does/will come out of one of several constructions.) This characterization is often called a Gelfand-Pettis integral, or weak integral (not because the conclusion is weak, but because the hypothesis is weak).

The condition with linear functionals already implies (by Hahn-Banach) that if there exists any vector $v=\int f\in V$ at all, it is unique. There's no ambiguity in what it is... if it exists.

Not all uniquely characterized things exist, as it turns out, so proof of existence (under various hypotheses) is important. Constructions suffice, and may prove useful incidental properties.

This set-up already allows the Schwartz-Grothendieck vector-valued extension of Cauchy-Goursat theory of complex functions in one variable.

Many basic purposes are fulfilled by looking at Banach-space-valued functions, perhaps continuous with compact support. But in practice we often want the weaker "strong" topology on operators. And we may want test-function-valued functions (not Frechet, but strict-colimit-of-Frechet), or distribution-valued. A small amount of work does verify that all spaces $V$ are "quasi-complete" (and locally convex), so that compactly-supported, continuous, $V$-valued functions (as a useful example) have integrals. In a very positive sense, we can prove that the situation is fairly "idiot-proof", although understanding that point is considerably more complicated than simply using the ideas in blissful naivete. :)

paul garrett
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