Let $(\Omega,\mu)$ be a measure space, say $\sigma$-finite for the sake of simplicity, and let $L^1 := L^1(\Omega,\mu)$ denote the real-valued $L^1$-space over $(\Omega,\mu)$.

For all $f,h \in L^1$ we call the set \begin{align*} [f,h] := \{g \in L^1: \; f \le g \le h\} \end{align*} the order interval between $f$ and $h$. Order intervals have the following property:

Proposition 1. Every order interval in $L^1$ is weakly compact.

However, the only proof of this fact that I know of makes heavy use of the theory of Banach lattices. It runs as follows:

Proof of Proposition 1. Let $(L^1)_+ := \{f \in L^1: \; f \ge 0\}$. If $(f_j)_{j \in J}$ is an increasing and norm bounded net in $(L^1)_+$, then it follows from the additivity of the norm on $(L_1)_+$ (i.e. $\|f+g\| = \|f\| + \|g\|$ for all $f,g \in (L^1)_+$) that $(f_j)_{j \in J}$ is a Cauchy net and thus convergent. Banach lattices with this property are called KB-spaces, and it is known that every KB-spaces is a Banach lattice with order continuous norm (which means that $\|g_j\| \to 0$ for every decreasing net (g_j) in the positive cone which has infimum $0$); see for instance [MN91, the first paragraph after Definition 2.4.11]. Hence, $L^1$ is a Banach lattice with order continuous norm, and for those spaces it is known that every order interval is weakly compact, see e.g. [MN91, Theorem 2.4.2(i) and (vi)]. Hence, the assertion follows.

Reference: [MN91] Peter Meyer-Nieberg, Banach lattices, Springer (1991).

Unfortunately, this proof is far from being elementary (and certainly completely unsuited for an introductionary course on functional analysis). Indeed, the fact that every KB-space has order continuous norm is not difficult to prove, but the fact that every order interval is weakly compact in a Banach lattice with order continuous norm heavily relies on non-trivial Banach lattice theory.

[In case that anybody is interested, here is the argument in a nutshell: it uses that a Banach latice $E$ with order continuous norm is a so-called ideal in its bi-dual $E''$, which implies that every order interval in $[f,g] \subseteq E$ coincides with the order interval between $f$ und $g$ in $E''$. Since the latter set is weak${}^*$-compact in $E''$ by the Banach-Alaoglu theorem, it follows that it is weakly compact in the space $E$.

Of course, the essence of this argument is that a Banach lattice with order continuous norm is an ideal in its bi-dual, which is again non-trivial. Even the fact that the dual of a Banach lattice is again a (Banach) lattice is not completely obvious (although it is essential and well-known in Banach lattice theory, of course).]

When giving a course on functional analysis, a colleague and I discussed whether there is a simpler proof for Proposition 1 which we could include in the course. It seems very likely that Proposition 1 was known before Banach lattice theory was developed, so there should probably be a simpler (or more elementary) proof. Yet, we weren't able to come up with one. So my question is:

Question. Is there an elemetary proof of Proposition 1?

The course where we wanted to include this is already over, but in order to make precise what I mean by "elementary", let us defined it as "well-suited to be tought in a first course on functional analysis" (let us also assume that the Banach-Alaoglu theorem and the Eberlein-Smulian theorem are already known to the course, so that we have the most important tools concerning weak compactness and weak${}^*$-compactness at hands).

  • $\begingroup$ I think that an easy arguement with the classical James' compactness theorem works. $\endgroup$ Feb 14, 2018 at 10:56
  • $\begingroup$ Using that the dual Banach space of $L^1$ is $L^\infty$. The integral representation of any element of $(L^1)^\ast$ can be easily maximized on the interval. $\endgroup$ Feb 14, 2018 at 10:58
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    $\begingroup$ But James' theorem is far beyond a first course in functional analysis. $\endgroup$ Feb 14, 2018 at 11:20
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    $\begingroup$ @Littlefield Thank you, this is an interesting observation. Yet, I probably would not call James' theorem "elementary". $\endgroup$ Feb 14, 2018 at 11:28
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    $\begingroup$ Is one allowed to use the characterization of (relative) weak compactness by uniform integrability? $\endgroup$ Feb 14, 2018 at 11:42

1 Answer 1


As Jochen Wengenroth mentioned in the comments, the weak compactness of order intervals follows from their uniform integrability. I think the proof of the Dunford-Pettis theorem is reasonably elementary; anyway, here is a sketch tailored for this special situation.

Let $\mathcal{F}$ be the weak-$\ast$ closure of $[f,h]$ in $(L^\infty)^\prime$. We already know that $\mathcal{F}$ is weak-$\ast$ compact, so we only have to show that $\mathcal{F}\subset L^1$.

Let $G\in\mathcal{F}$ and let $(g_\alpha)_{\alpha\in I}$ be a net in $[f,h]$ such that $g_\alpha\to G$ in the weak-$\ast$ topology. If we can show that $A\mapsto G(\chi_A)$ is a measure, then it follows from the Radon-Nikodym theorem that there exists $g\in L^1$ such that $G(\chi_A)=\int_A g\,d\mu$. From that it is not hard to see that $G(\phi)=\int g\phi\,d\mu$ for all $\phi\in L^\infty$.

So let's get to the $\sigma$-additivity of the map $A\mapsto G(\chi_A)$. Let $(A_k)$ be a sequence of disjoint measurable sets and $A=\bigcup_k A_k$. We have $$ \lvert G(\chi_{A\setminus\bigcup_{k=1}^nA_k})\rvert\leq\limsup_\alpha \int_{A\setminus\bigcup_{k=1}^n A_k}\lvert g_\alpha\rvert\,d\mu\leq\int_{A\setminus\bigcup_{k=1}^n A_k}(\lvert f\rvert+\lvert h\rvert)\,d\mu\to 0 $$ as $n\to\infty$. Thus $G(\chi_A)=\lim_{n\to\infty}G(\chi_{\bigcup_{k=1}^n A_k})=\sum_{k=1}^\infty G(\chi_{A_k})$, and we are done.


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