Let h, h′ ∈K. Note that for all h, e(h) is in K. The indifference h∼e(h) implies by additivity that h + h′ ∼e(h) + h_. Additivity and h_ ∼e(h_) imply that e(h) + h_ ∼e(h) + e(h′). Transitivity gives h + h′ ∼e(h) + e(h′). Hence, e(h + h′) = e(h) + e(h′).
This and monotonicity imply by a similar line of proof as in Lemma 3 and Theorem 4 in Weymark (1981) that e(h) is linear: e(h)