Product of two weakly compact endomorphisms is compact I've seen the statement that if a Banach lattice $E$ satisfies the property that 
$$||x + y|| = ||x|| + ||y||, \: \forall x,y \in E_+$$
then if $S,T:E \to E$ are weakly compact endomorphisms, then $ST$ is compact.
I have yet to see a proof of this statement though. Any help or hints would be appreciated.
 A: Such Banach lattices $E$, where the norm is additive
on the positive cone, are called AL-spaces (which is an abbreviation for "Abstract $L$-space"). 
There is a well-known structural result that for any AL-space $E$ there exists a measure space $X = (X, \Sigma, \mu)$ and a Banach lattice isomorphism $\pi : E \to L^1(X)$ (c.f. Theorem 8.5 in "Banach Lattices and Positive Operators" by Schaefer e.g.). 
Now there is another result I would like to mention that comes from the classical Dunford-Pettis theory: Let $X = (X, \Sigma, \mu)$ be a measure space. Then the product of two weakly compact operators on $L^1(X)$ is compact. (c.f. Corollary VI.8.13 in "Linear Operators Part I: General Theory" by Dunford and Schwartz e.g.). 
Finally, keep in mind that any composition of a (weakly) compact operator with a bounded operator is again (weakly) compact.
Using the two theorems above one can deduce your statement as follows: Let $S, T$ be weakly compact operators on an AL-space $E$. Since both $\pi$ and $\pi^{-1}$ are bounded, it follows that $\pi S \pi^{-1}$ and $\pi T \pi^{-1}$ are weakly compact on $L^1(X)$. Hence by the theorem above $\pi S T \pi^{-1} = (\pi S \pi^{-1}) (\pi T \pi^{-1})$ is compact on $L^1(X)$. So it follows that $ST = \pi^{-1} (\pi S T \pi^{-1}) \pi$ is compact on $E$.
I hope things got a little bit clearer :)
