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Wikipedia http://en.wikipedia.org/wiki/Pettis_integral defines the Pettis Integral for Banach space valued functions wrt to some measure space by duality.

It calls the Pettis & Bochner integral the weak & strong integral respectively, which implies some kind of relationship; also, apparently there is a Dunford integral which specialises to the Pettis integral.

My question is: Why are these weak integrals useful, what is the definition of the Dunford Integral and how does it specialise to the Pettis one, and what is the relationship between the Pettis Integral & the Bochner Integral?.

I've just noted: Weak and Strong Integration of vector-valued functions, which answers the part of my question about the Pettis Integral.

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Let $f \colon \Omega \to E$ be your function. $\mu$ is a measure on $\Omega$. Assume, for every $x^* \in E^*$, the composition $x^* \circ f$ is $\mu$-integrable. The Dunford integral in general lies in $E^{**}$, namely $\int_A f d\mu$ is the element $u^{**} \in E^{**}$ defined by $u^{**}(x^*) = \int_A x^{*}\circ f\,d\mu$ for all $x^* \in E^*$. And then, of course, if $\int_A f\,d\mu \in E$ for all $A$, we say $f$ is Pettis integrable.

If $f$ is Bochner integrable, then it is also Pettis integrable, and the two integrals agree.

If $f$ has almost all its values in a separable space, then $f$ is Bochner integrable if and only if it is Pettis integrable, and the scalar integral $\int \|f\| d\mu$ converges.

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great chareacterisation! –  Mozibur Ullah Dec 2 '12 at 21:17
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Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription $$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$ I first learnt this in Talagrand's AMS memoir:


but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute $$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$ and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that $$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$ as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals $F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$ do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).

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Well, it seems others beat me to it. I should learn to write faster! –  Ollie Margetts Apr 3 '12 at 12:55
@margetts:I like the examples, it does a good job of highlighting the differences between the integrals. –  Mozibur Ullah Apr 4 '12 at 1:35
A nice example, Ollie! –  Tomek Kania Apr 5 '12 at 12:01
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The Dunford integral is a version of the weak integral, still weaker than the more widely used Pettis integral; for a $X$-valued function, it takes value in $X^{**}$. See the paper http://www.ams.org/journals/proc/1986-096-03/S0002-9939-1986-0822428-0/S0002-9939-1986-0822428-0.pdf for a detailed study of relations between the Dunford and Pettis integrals.

This notion should not be confused with the Riesz-Dunford integral, a Cauchy type integral in the complex domain defining a holomorphic functional calculus for some classes of operators.

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