When might the Bochner Integral not exist?

116 Views Asked by At

Let $I$ be an integral on a locally compact second countable Hausdorff space $X$ and let $E$ be a Banach space. In this question all functions are taken to be measurable ($E$ and $X$ have the Borel algebras).

Let $f:X\to E$ be such that $\|f\|:X\to\Bbb R$ is an integrable function. One necessary condition on what $I(f)$ should be is:

$$\alpha(I(f))=I(\alpha(f))\quad \forall \alpha\in E^*\tag{1}$$

where $\alpha(f)$ is the function $x\mapsto \alpha(f(x))$. By $|\alpha(f(x))|≤\|\alpha\|\,\|f(x)\|$ the right-hand side always exists. The right-hand side of $(1)$ is linear in $\alpha$ and defines a continuous functional $I(f):E^*\to \Bbb C$, so an element of $E^{**}$.

A function $f:X\to E$ with $\|f\|$ integrable is then called integrable if $I(f)\in E\subset E^{**}$. I think this is the definiton of the Bochner-integral. (Is that the correct term?)

My question is (roughly):

When is it so that $I(f)\in E$?

Two examples:

Let $f\in C_C(X,E)$, then we can approximate $f$ in measure by step-functions with compact support, on which the integral is clearly defined. Since $E$ is Banach the limit of these integrals will again lie in $E$, one checks that this satisfies $(1)$.

For the other case, suppose that $E$ is the dual of some space $E^\times$ (for example $E$ is reflexive or a Hilbert space), then defining for all $\alpha'\in E^\times$: $$\alpha'(I(f)):= I(\alpha'(f)), \quad \alpha'\in E^\times$$ we retrieve an element of $(E^\times)^*=E$. In the reflexive case $E^\times = E^*$ it is automatic that this definition satisfies $(1)$, but it is not clear to me if this holds when $E^\times$ is smaller than $E^*$.

To make my question more precise:

  1. If $E$ has a pre-dual, is $I(f)\in E$ for $\|f\|$ integrable?
  2. Is there another/nicer characterisation (other than is reflexive/has pre-dual) of spaces $E$ that have $I(f)\in E$ for all $\|f\|$ integrable?

Also, is there a nice example of a function $X\to E$ where $E$ has no pre-dual (for example $E=c_{0}(\Bbb N)$) so that $\|f\|$ is integrable but $I(f)\notin E$? I am having a little bit of trouble constructing such a thing on $c_{0}$, but somehow I believe it should exist.