Pettis integral

The Pettis integral is a term named after Billy James Pettis from the mathematical branch of functional analysis . It is an integral for functions on a measure space with values ​​in a Banach space . If the Banach space is equal to the one-dimensional space , then one obtains the usual integral of real-valued functions on the measure space. The Pettis integral not only generalizes the integral of real-valued functions, but also the Bochner integral and the Birkhoff integral , which are also integrals of Banach space-valued functions. ${\displaystyle \mathbb {R} }$

construction

We assume a complete measure space with a finite, positive measure and want to define an integral for functions with values ​​in a Banach space . For the construction described in the following we make use of the fact that for each of the dual space there is a real-valued function and that measure-theoretical terms for such functions are already defined. We call it weakly measurable when there is a measurable function for each . In contrast, as usual, it is called measurable when the archetype of every open set is over. For the relationship between these two notions of measurability, see Pettis' measurability theorem . After all, we call it weakly integrable if there is an integrable function for each . ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ ${\displaystyle \mu }$ ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle X}$ ${\displaystyle \varphi \circ f}$ ${\displaystyle \varphi }$ ${\displaystyle X'}$ ${\displaystyle \Omega \rightarrow \mathbb {R} }$ ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle \varphi \circ f}$ ${\displaystyle \varphi \in X'}$ ${\displaystyle f}$ ${\displaystyle {\mathcal {A}}}$ ${\displaystyle f}$ ${\displaystyle \varphi \circ f}$ ${\displaystyle \varphi \in X'}$

We now consider a weakly integrable function . For each is then , the latter denoting the L ¹ space above the given measurement space, which according to Fischer-Riesz's theorem with regard to the 1-norm is a Banach space. We thus get a linear operator ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle \varphi \in X'}$ ${\displaystyle \varphi \circ f\in L^{1}(\Omega ,{\mathcal {A}},\mu )}$

${\displaystyle S_{f}:X'\rightarrow L^{1}(\Omega ,{\mathcal {A}},\mu ),\,\varphi \mapsto \varphi \circ f}$,

which one can show by means of the theorem of the closed graph that it is even continuous . One can therefore form the adjoint operator . If one identifies the dual space of L ¹ by means of L ^p -duality with , as usual , one obtains an operator ${\displaystyle T_{f}=S_{f}':L^{1}(\Omega ,{\mathcal {A}},\mu )'\rightarrow X''}$ ${\displaystyle L^{\infty }(\Omega ,{\mathcal {A}},\mu )}$

${\displaystyle T_{f}:L^{\infty }(\Omega ,{\mathcal {A}},\mu )\rightarrow X'',\quad (T_{f}(g))(\varphi )=\int _{\Omega }g(t)\varphi (f(t))\,\mathrm {d} \mu (t),\quad \quad g\in L^{\infty }(\Omega ,{\mathcal {A}},\mu ),\varphi \in X'}$.

In particular, one can apply to characteristic functions for measurable quantities . one calls the Dunford integral , after Nelson Dunford , or the Gelfand integral , after Israel Gelfand , and writes ${\displaystyle T_{f}}$ ${\displaystyle \chi _{E}:\Omega \rightarrow \{0,1\}}$ ${\displaystyle E\in {\mathcal {A}}}$ ${\displaystyle T_{f}(\chi _{E})\in X''}$

${\displaystyle \int _{E}f\,\mathrm {d} \mu =\int _{E}f(t)\,\mathrm {d} \mu (t):=T_{f}(\chi _{E})}$.

If one imagines an integral of a function with values ​​in as the mean of the values, one would expect that the integral is again in . Generally this is not the case. But now through the so-called canonical embedding , one defines: ${\displaystyle \textstyle \int _{E}f\,\mathrm {d} \mu }$ ${\displaystyle X}$ ${\displaystyle \mu }$ ${\displaystyle f}$ ${\displaystyle X}$ ${\displaystyle X\subset X''}$

A weakly integrable function is called Pettis integrable , if for all , and one calls the Pettis integral of over . ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle \textstyle \int _{E}f\,\mathrm {d} \mu \in X}$ ${\displaystyle E\in {\mathcal {A}}}$ ${\displaystyle \textstyle \int _{E}f\,\mathrm {d} \mu }$ ${\displaystyle f}$ ${\displaystyle E}$

Examples

Reflexive spaces

Is reflexive , so is and it is for everyone and every weakly integrable function . This means that every weakly integrable function with values ​​in a reflective space can be Pettis-integrable. ${\displaystyle X}$ ${\displaystyle X=X''}$ ${\displaystyle \textstyle \int _{E}f\,\mathrm {d} \mu \in X}$ ${\displaystyle E\in {\mathcal {A}}}$ ${\displaystyle f}$

Birkhoff integral

Every Birkhoff-integrable function can be Pettis-integrable and the Birkhoff integral corresponds to the Pettis integral. Therefore the Pettis integral is a generalization of the Birkhoff integral. ${\displaystyle f:\Omega \rightarrow X}$

Bochner integral

Every Bochner-integrable function can be Pettis-integrable and the Bochner integral agrees with the Pettis integral. Therefore the Pettis integral is also a generalization of the Bochner integral. It applies ${\displaystyle f:\Omega \rightarrow X}$

Bochner-integrable     Birkhoff-integrable     Pettis-integrable     weakly-integrable. ${\displaystyle \Rightarrow }$ ${\displaystyle \Rightarrow }$ ${\displaystyle \Rightarrow }$

Pettis-integrable but not Bochner-integrable

We consider the unit interval [0,1] with the Lebesgue measure on the σ-algebra of Lebesgue measurable sets as the measure space and the sequence space of the real zero sequences as the Banach space . Let it be the half-open interval and ${\displaystyle \mathrm {d} t}$ ${\displaystyle X=c_{0}}$ ${\displaystyle I_{n}}$ ${\displaystyle \textstyle ({\frac {1}{n+1}},{\frac {1}{n}}]\subset [0,1]}$

${\displaystyle f:[0,1]\rightarrow c_{0},\,f(t):=(n\chi _{I_{n}}(t))_{n\in \mathbb {N} }}$.

Each is actually a null sequence. That is clear for , because it is and for there is exactly one with and therefore is . This function can be integrated by Pettis but not by Bochner. To clarify the above constructions, we carry out the necessary calculations and start with the weak integrability. ${\displaystyle f(t)}$ ${\displaystyle t=0}$ ${\displaystyle f(0)=(0,0,0,\ldots )}$ ${\displaystyle t>0}$ ${\displaystyle n}$ ${\displaystyle t\in I_{n}}$ ${\displaystyle f(t)=(0,0,\ldots ,n,0,0,\ldots )}$

For each is by definition of duality ${\displaystyle \varphi =(v_{n})_{n}\in \ell ^{1}=c_{0}'}$ ${\displaystyle c_{0}'=\ell ^{1}}$

${\displaystyle (\varphi \circ f)(t)=\sum _{n=1}^{\infty }v_{n}\cdot n\chi _{I_{n}}(t)}$

and therefore

${\displaystyle \int _{[0,1]}|(\varphi \circ f)(t)|\,\mathrm {d} t\leq \sum _{n=1}^{\infty }|v_{n}|\cdot n\int _{[0,1]}\chi _{I_{n}}(t)\,\mathrm {d} t=\sum _{n=1}^{\infty }|v_{n}|\cdot n{\frac {1}{n(n+1)}}=\sum _{n=1}^{\infty }|v_{n}|\cdot {\frac {1}{n+1}}<\infty }$,

because the interval has the length . So it is weakly integrable. ${\displaystyle I_{n}}$ ${\displaystyle \textstyle {\frac {1}{n(n+1)}}}$ ${\displaystyle f}$

To determine the Gelfand integrals, consider . If we denote the L ¹ -L ^∞ -duality with angle brackets, then for ${\displaystyle g\in L^{\infty }([0,1])}$ ${\displaystyle \varphi =(v_{n})_{n}\in \ell ^{1}=c_{0}'}$

${\displaystyle \langle T_{f}(g),\varphi \rangle =\langle g,S_{f}(\varphi )\rangle =\langle g,\varphi \circ f\rangle =\int _{[0,1]}g(t)(\varphi \circ f)(t)\,\mathrm {d} t=\int _{[0,1]}g(t)\sum _{n=1}^{\infty }v_{n}\cdot n\chi _{I_{n}}(t)\,\mathrm {d} t}$

${\displaystyle =\sum _{n=1}^{\infty }v_{n}\cdot n\int _{[0,1]}g(t)\chi _{I_{n}}(t)\,\mathrm {d} t=\sum _{n=1}^{\infty }v_{n}\cdot n\int _{I_{n}}g(t)\,\mathrm {d} t=\langle \left(n\cdot \int _{I_{n}}g(t)\,\mathrm {d} t\right)_{n\in \mathbb {N} },(v_{n})_{n\in \mathbb {N} }\rangle =\langle \left(n\cdot \int _{I_{n}}g(t)\,\mathrm {d} t\right)_{n\in \mathbb {N} },\varphi \rangle }$

and you read off

${\displaystyle T_{f}(g)=\left(n\cdot \int _{I_{n}}g(t)\,\mathrm {d} t\right)_{n\in N}\in \ell ^{\infty }=c_{0}''}$.

In fact, this episode is already in , because ${\displaystyle c_{0}}$

${\displaystyle \left|n\cdot \int _{I_{n}}g(t)\,\mathrm {d} t\right|\leq \|g\|_{\infty }n\cdot \int _{I_{n}}\mathrm {d} t=\|g\|_{\infty }\cdot {\frac {1}{n+1}}\rightarrow 0}$.

Therefore Pettis can be integrated. but cannot be integrated into Bochner, because ${\displaystyle f}$ ${\displaystyle f}$

${\displaystyle t\mapsto \|f(t)\|=\|(n\chi _{I_{n}}(t)\|=\sum _{n=1}^{\infty }n\chi _{I_{n}}(t)}$

cannot be integrated.

Difficult to integrate but not Pettis integrable

To construct a weakly integrable function that is not Pettis integrable, we slightly modify the above example. Again we consider the measure space [0,1] with the Lebesgue measure and the Banach space . The function we are looking for is ${\displaystyle c_{0}}$

${\displaystyle f:[0,1]\rightarrow c_{0},\,f(t):=(n(n+1)\chi _{I_{n}}(t))_{n\in \mathbb {N} }}$.

For each is ${\displaystyle \varphi =(v_{n})_{n}\in \ell ^{1}=c_{0}'}$

${\displaystyle (\varphi \circ f)(t)=\sum _{n=1}^{\infty }v_{n}\cdot n(n+1)\chi _{I_{n}}(t)}$

and therefore

${\displaystyle \int _{[0,1]}|(\varphi \circ f)(t)|\,\mathrm {d} t\leq \sum _{n=1}^{\infty }|v_{n}|\cdot n(n+1)\int _{[0,1]}\chi _{I_{n}}(t)\,\mathrm {d} t=\sum _{n=1}^{\infty }|v_{n}|\cdot n(n+1){\frac {1}{n(n+1)}}=\sum _{n=1}^{\infty }|v_{n}|<\infty }$.

So it is weakly integrable. ${\displaystyle f}$

If the constant function is 1, then for each ${\displaystyle e\in L^{\infty }([0,1])}$ ${\displaystyle \varphi =(v_{n})_{n}\in \ell ^{1}=c_{0}'}$

${\displaystyle \langle T_{f}(e),\varphi \rangle =\langle e,S_{f}(\varphi )\rangle =\langle e,\varphi \circ f\rangle =\int _{[0,1]}e(t)(\varphi \circ f)(t)\,\mathrm {d} t=\int _{[0,1]}(\varphi \circ f)(t)\,\mathrm {d} t}$

${\displaystyle =\int _{[0,1]}\sum _{n=1}^{\infty }v_{n}\cdot n(n+1)\chi _{I_{n}}(t)\,\mathrm {d} t=\sum _{n=1}^{\infty }v_{n}\cdot n(n+1)\int _{I_{n}}\,\mathrm {d} t=\sum _{n=1}^{\infty }v_{n}=\langle (1,1,1,\ldots ),\varphi \rangle }$.

So is and that is not over . Therefore Pettis can not be integrated. ${\displaystyle \int _{[0,1]}f(t)\,\mathrm {d} t=T_{f}(e)=(1,1,1,\ldots )\in \ell ^{\infty }=c_{0}''}$ ${\displaystyle c_{0}}$ ${\displaystyle f}$

properties

Weak compactness

If Pettis can be integrated with the above notation , the associated operator is weakly compact . ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle T_{f}:L^{\infty }(\Omega ,{\mathcal {A}},\mu )\rightarrow X}$

Operators

Let there be a finite, complete measure space, a Banach space and Pettis integrable. If there is a continuous, linear operator between Banach spaces, then Pettis is also integrable and it holds ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ ${\displaystyle X}$ ${\displaystyle f:\Omega \rightarrow X}$ ${\displaystyle A:X\rightarrow Y}$ ${\displaystyle A\circ f:\Omega \rightarrow Y}$

${\displaystyle \int _{E}A\circ f\,\mathrm {d} \mu =A\left(\int _{E}f\,\mathrm {d} \mu \right)}$for any measurable amount . ${\displaystyle E\subset \Omega }$

Vector space of the Pettis integrable functions

It is easy to show that sums and scalar multiples of Pettis-integrable functions are again Pettis-integrable and that the integral behaves linearly, that is

${\displaystyle \int _{\Omega }(\alpha f+g)\,\mathrm {d} \mu =\alpha \int _{\Omega }f\,\mathrm {d} \mu +\int _{\Omega }g\,\mathrm {d} \mu }$

for Pettis-integrable functions and . ${\displaystyle f,g:\Omega \rightarrow X}$ ${\displaystyle \alpha \in \mathbb {R} }$

The measurable, Pettis-integrable functions therefore form a vector space . The set of functions that take on the value μ-almost-everywhere form a sub-vector space, and the quotient space after this sub-space is denoted by . In the conventional view of measurement theory, this is the space of measurable, Pettis-integrable functions, whereby μ-almost-everywhere identical functions are identified. ${\displaystyle \Omega \rightarrow X}$ ${\displaystyle P_{(1)}(\Omega ,{\mathcal {A}},\mu ,X)}$ ${\displaystyle 0\in X}$ ${\displaystyle P_{1}(\Omega ,{\mathcal {A}},\mu ,X)}$

The 1 standard for Pettis-integrable functions

Is measurable with the above designations and Pettis integrable, then is ${\displaystyle f:\Omega \rightarrow X}$

${\displaystyle |f|_{1}:=\sup \left\{\int _{\Omega }|\varphi \circ f|\,\mathrm {d} \mu ;\,\varphi \in X',\|\varphi \|\leq 1\right\}=\sup\{T_{f}(g);\,g\in L^{\infty }(\Omega ,{\mathcal {A}},\mu ),\|g\|_{\infty }\leq 1\}}$

at last. is a semi-norm up and a norm up . This normalized space is usually not complete ; let it be completed . ${\displaystyle |\cdot |_{1}}$ ${\displaystyle P_{(1)}(\Omega ,{\mathcal {A}},\mu ,X)}$ ${\displaystyle P_{1}(\Omega ,{\mathcal {A}},\mu ,X)}$ ${\displaystyle {\overline {P_{1}}}(\Omega ,{\mathcal {A}},\mu ,X)}$

Injective tensor product

Let it again be a finite, complete measure space and a Banach space. Then ${\displaystyle (\Omega ,{\mathcal {A}},\mu )}$ ${\displaystyle X}$

${\displaystyle L^{1}(\Omega ,{\mathcal {A}},\mu )\times X\rightarrow {\overline {P_{1}}}(\Omega ,{\mathcal {A}},\mu ,X),\,(f,x)\mapsto f(\cdot )x}$

is a bilinear map , and it holds

${\displaystyle \int _{\Omega }f(\cdot )x\,\mathrm {d} \mu =\left(\int _{\Omega }f\,\mathrm {d} \mu \right)x\in X}$.

This bilinear mapping defines a linear mapping on the tensor product

${\displaystyle L^{1}(\Omega ,{\mathcal {A}},\mu )\otimes X\rightarrow {\overline {P_{1}}}(\Omega ,{\mathcal {A}},\mu ,X)}$.

If one completes this tensor product to the injective tensor product , one obtains an isometric isomorphism

${\displaystyle L^{1}(\Omega ,{\mathcal {A}},\mu ){\hat {\otimes }}_{\varepsilon }X\rightarrow {\overline {P_{1}}}(\Omega ,{\mathcal {A}},\mu ,X)}$.

Individual evidence

1. ^ Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces , Springer-Verlag 2002, ISBN 1-85233-437-1 , Chapter 3.3: The Dual Space of and the Pettis Integral. ${\displaystyle L_{1}(\mu ){\hat {\otimes }}_{\varepsilon }X}$
2. ^ Joseph Diestel: Geometry of Banach Spaces - Selected Topics , Lecture Notes in Mathematics 485, Springer-Verlag (1975), ISBN 3-540-07402-3 , Chapter 6, §1, Theorem 3
3. ^ Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces , Springer-Verlag 2002, ISBN 1-85233-437-1 , page 53
4. ^ Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces , Springer-Verlag 2002, ISBN 1-85233-437-1 , page 52
5. ^ Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces , Springer-Verlag 2002, ISBN 1-85233-437-1 , sentence 3.7
6. ^ Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces , Springer-Verlag 2002, ISBN 1-85233-437-1 , sentence 3.13