# Banach space integral via defining it in $X^{**}$ and then proving it's in $X$

Vector-valued integration is something I generally try not to think about very much. I have the impression that it can be a sort of "rabbit hole" of a subtlety if one allows it to be. So, I tend to treat it as a black box. Eventually, though, adopting this attitude is bound to make a person feel guilty -- and then it's time to firm things up a bit. I know there are various ways of defining the integral of a Banach space-valued function. What I'm asking about here is an approach that I've always had in the back of my mind as something that "should work", but, when I tried to write down the details, I became less confident. Can anybody help me put this on rigorous footing? Or gently inform me that this approach is not good?

Here is the sort of setting I have in mind. Let $X$ be a Banach space, say over $\mathbb{C}$, and let $f : \mathbb{R} \to X$ be continuous and such that $M : = \int_\mathbb{R} \|f(t)\| \ dt < \infty$. Now, for any $\varphi \in X^*$, we have $$| \int \varphi \circ f | \leq \int |\varphi(f(t))| \ dt \leq \int \| \varphi \| \| f(t) \| \ dt = M \|\varphi\|$$ so it is not difficult to see that $\varphi \mapsto \int \varphi \circ f : X^* \to \mathbb{C}$ is a bounded linear functional, that is, an element of $X^{**}$.

What I want to do now show is show that this element of the double dual is actually in $X \subset X^{**}$ so that there is an element $I \in X$ satisfying $\varphi(I) = \int \varphi \circ f$ for all $\varphi \in X^*$. One then defines $\int f := I$. The way I always envisioned doing this was by an application of the following fact:

An element $\psi \in X^{**}$ is actually in $X \subset X^{**}$ if (and only if) it is continuous $X^* \to \mathbb{C}$ when $X^*$ has the weak-star topology.

So, let's try to see if $\varphi \mapsto \int \varphi \circ f$ has this kind of continuity. Suppose that $\varphi_i$ is any net in $X^*$, convergent in the weak-start topology to some $\varphi \in X^*$. That is $\varphi_i(x) \to \varphi(x)$ for all $x \in X$. We want to show that $$\int_\mathbb{R} \varphi_i(f(t)) \ dt \to \int_\mathbb{R} \varphi(f(t)) \ dt.$$ In our favour, we have that $\varphi_i \circ f \to \varphi \circ f$ pointwise over $\mathbb{R}$ (from the weak-star continuity). So, one thinks that Lebesgue's dominated convergence function may apply. But the problems with this are two-fold.

1. We need a dominating function for the family of functions $t \mapsto \varphi_i(f(t))$. I think maybe the uniform boundedness principle can help here?
2. The Lebesgue dominated convergence theorem applies to sequences not nets. Unless there is some way to finesse around this issue, I think this may be a more critical error.

To reiterate, the slightly flabby question I'm asking is:

Can the approach to defining $\int f$ I just outlined be "rigourized"?

• I'm not sure how one would best prove that $\Lambda\colon\varphi \mapsto \int \varphi\circ f$ is in (comes from an element of) $X$, but that's basically the Pettis integral. – Daniel Fischer Aug 4 '13 at 21:54

Consider first the integral over a compact interval $[a,b]$. Then $\frac{1}{b-a} \int_a^b f(t) dt$ is in the weak$^*$ closure in $X^{**}$ of the convex hull $\mathrm{Conv}(f([a,b]))$ of $f([a,b])$ (morally, this is clear since the integral with respect to a probability measure is a generalized convex combination, the proof uses this property for real-valued functions). By the continuity of $f$ the range $f([a,b])$ is compact in $X$ and hence its convex hull is precompact and thus relatively compact in $X$. Since compact subsets of $X$ are weak$^*$ closed in $X^{**}$ you get that the integral $\int_a^b f(t) dt$ belongs indeed to $X$ (and not only to the bidual).

For the general case you can approximate $f$ by $f \chi_n$ where $\chi_n$ is $1$ on $[-n,n]$, $0$ in $\mathbb R \setminus [-(n+1),n+1]$ and afin-linear on the remaining two intervals. Then you show $\int_{\mathbb R} f_n(t)dt = \int _{-(n+1)}^{n+1} f_n(t) dt \to \int_{\mathbb R} f(t)dt$ in the norm of $X^{**}$ so that the integral is in $X$ (because $X$ is closed in its bidual).

• Do the terms "precompact" and "relatively compact" mean different things to you? – Mike F Aug 6 '13 at 6:25
• In Banach (or complete metric) spaces they are equivalent. By definition, a set is precompact, if for every $\varepsilon >0$ the set is covered by finitely many balls of that radius. Relatively compact means that the closure is compact. – Jochen Aug 6 '13 at 7:56
• Hmm OK so your "precompact" is my "totally bounded". – Mike F Aug 6 '13 at 16:44

From this answer we know that $\int_\mathbb{R}f(t)dt$ exists as some element $x_f\in X$ if

1) $f(\mathbb{R}\setminus A)$ is separable for some null set $A$

2) $\varphi \circ f$ is continuous for every $\varphi\in X^*$

We are going to prove that 1) and 2) holds in our case. For every $n\in\mathbb{N}$ the set $f([-n,n])$ is compact as continuous image of compact set, and hence separable. Thus $f(\mathbb{R})=\bigcup_{n\in\mathbb{N}} f([-n,n])$ is separable as union of separable sets. The second condition holds trivially because $f$ is continuous.

Hence for any $\varphi\in X^*$ we get $$\int_\mathbb{R}(\varphi\circ f)(t)dt=\varphi\left(\int_\mathbb{R} f(t)dt\right)=\varphi(x_f)$$ So the map $\varphi\mapsto \int_\mathbb{R}(\varphi\circ f)(t)dt$ is just the evaluation of $\varphi$ at $x_f$.

• It occurs to me, since $f$ vanishes at $\infty$ (being continuous and integrable), it extends ctsly to the 1-pt compactification $\mathbb{R} \cup \{\infty\}$ by $\infty \mapsto 0$. So, $f(\mathbb{R}) \cup \{0\}$ is already compact. Just another argument for separability of $f(\mathbb{R})$. – Mike F Dec 16 '13 at 4:48