Assume that either X is compact or Y is compact. Then: Proof. The inequality α ≤ β is always true and α ≥ β follows from Remark 18. We end this section with a short proof of the Markov-Kakutani fixed point theorem for abelian families of continuous affine mappings in linear topological spaces having separating duals $^{1}$. Remark 20 If both X and Y are compact, the infsup equality in Corollary 19 is a minmax equality and is equivalent to the existence of a saddle point ( x 0 , y 0 ) for the function f ( x, y ) , i.e., f ( x, y 0 ) ≤ f ( x 0 , y ) , ∀ ( x, y ) ∈ X × Y. Recall that a mapping φ from a convex set X into a vector space is said to be affine if and only if φ ( ∑ n i $_{=1}$λ$_{i}$ x$_{i}$ ) = ∑ n i =1 λ$_{i}$φ ( x$_{i}$ ) for any convex combination ∑ n i =1 λ$_{i}$x$_{i}$, λ$_{i}$ ≥ 0 , ∑ n i =1 λ$_{i}$ = 1 , in X. The key ingredient is the following fixed point property for continuous affine transformations of a compact convex set. Corollary 21 Let X be a non-empty compact convex subset of a t.v.s. E with separating dual E ′ and let φ : X −→ X be a continuous affine mapping. Then φ has a fixed point. Proof. The proof is a simplification of the treatment in [9]. Define f : X × E ′ −→ R by f ( x, ℓ ) = ℓ ( φ ( x ) − x ) , ( x, ℓ ) ∈ X × E $^{′}$. It suffices to prove the existence of x 0 ∈ X such that f ( x 0 , ℓ ) ≤ 0 , ∀ ℓ ∈ E $^{′}$, for this would imply ℓ ( φ ( x 0 ) − x 0 ) = 0 , ∀ ℓ ∈ E $^{′}$, i.e., φ ( x 0 ) − x 0 = 0 and the proof is complete. $^{1}$A t.v.s. E has separating dual if for each x ∈ E, x = 0 , there exists a bounded linear form ℓ ∈ E $^{′}$, the topological dual of E, such that ℓ ( x ) = 0 . Locally convex topological vector spaces have separating duals. Sequence spaces ℓ p , 0 < p < 1 , and Hardy spaces H p , 0 < p < 1 , are instances of non-locally convex spaces with separating duals. 18 α = sup X inf Y f ( x, y ) = inf Y sup X f ( x, y ) = β