Let $X$ a Banach Space and $(\Omega, \Sigma, \mu)$ a measure space. A function $F:\Omega\rightarrow X$ is Dunford integrable if $x^\ast\circ F$ is $\mu$-integrable for every $x^\ast\in X^\ast$. The space of functions that are Dunford integrable, denoted by $\mathbb{D}(\mu,X)$, is a normed space with $$ \|F\|:=\sup\left\{\int_\Omega|x^\ast\circ F|d\mu:x^\ast\in X^\ast,\,\|x^\ast\|\leq1\right\} $$ Does anybody knows if the space $\mathbb{D}(\mu,X)$ is complete?
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$\begingroup$ Is there any obstacle when trying the usual proof that e.g. $L^1(\Omega,\Sigma,\mu)$ is complete? $\endgroup$– Jochen WengenrothCommented Aug 10, 2015 at 16:50
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$\begingroup$ The usual proof doesn't work in this case, because we consider vectorial functions, and the norm depends on the functionals $x^\ast$. $\endgroup$– Husaí VázquezCommented Aug 11, 2015 at 16:26
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This a revised and expanded version of my post.
No, it is not complete. For simplicity suppose that $X$ is reflexive and separable in which case Dunford and Pettis integrals coincide. Suppose also that $\mu$ is finite and non-atomic. In this case the space of Pettis integrable functions is complete if and only if $X$ is finite-dimensional as proved by Thomas
G.E.F. Thomas, Totally Summable Functions with Values in Locally Convex Spaces, Measure Theory (Oberwolfach 1975), Lecture Notes in Math. Vol. 541 (1976), 117–131.
This is based on a result o his which asserts that there exist an absolutely sequence summable sequence $(x_n)_{n=1}^\infty$ in $X$ (reflexivity is not needed) and a sequence $(f_n)_{n=1}^\infty$ in the unit ball of $L_1(\mu)$ such that the vector measure
$$\nu(A) = \sum_{n=1}^\infty \int\limits_A f_n(t)\,{\rm dt}\cdot x_n$$
does not have a Pettis intregrable density.
G.E.F. Thomas, The Lebesgue-Nikodym Theorem for Vector Valued Radon Measures, Memoirs. of AMS, 139, American Mathematical Society, Providence, 1974.
The sequence $(\sum_{k=1}^n f_k\cdot x_k)$ is Cauchy in $\mathcal{P}$ because $(x_n)_{n=1}^\infty$ is absolutely summable, yet it is not convergent as there is no function $F$ such that
$$\lim_{n\to \infty}\int\limits_A \sum_{k=1}^n f_k(t)\cdot x_k\,{\rm d}t\to \int_A F(t)\,{\rm d}t.$$
It seems to me that Pettis integrable functions form a closed subspace of the space of Dunford integrable functions, hence you may extend the above result as in the case where $X$ is infinite-dimensional, you have an incomplete, closed subspace of a normed space, so the space itself cannot be complete.
Addendum. Let me point out that if you want to do some functional analysis with the space of Pettis integrable functions, even though incomplete, it is barrelled.
answered Aug 11, 2015 at 21:10
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$\begingroup$ Do you know a reference to check that $L_1(\mu)\odot_{\varepsilon}X$ is complete if and only if $X$ is finite-dimensional. Anyways, thanks a lot for your help. I had been thinking for a while if those two spaces were complete. $\endgroup$ Commented Aug 12, 2015 at 4:24
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$\begingroup$ If $X$ and $Y$ are infinite-dimensional Banach spaces and $\alpha$ is any reasonable crossnorm, then $X\odot_\alpha Y$ is incomplete. Choose two sequences of norm-one, linearly independent vectors $(x_n)$ and $(y_n)$ in $X$ and $Y$, respectively. Show that $(\sum_{k=1}^n \tfrac{1}{k^2}x_k\otimes y_k)$ is a Cauchy sequence without a limit in $X\odot_\alpha Y$. $\endgroup$ Commented Aug 12, 2015 at 8:44