This amounts to showing that ⋂ ℓ ∈ E ′ A ( ℓ ) =  for the relation A := { ( ℓ, x ) ∈ E ′ × X : f ( x, ℓ ) ≤ 0 } . Since for each fixed ℓ ∈ E $^{′}$, the function f ( x, ℓ ) is l.s.c. in x, then each A ( ℓ ) is a closed, hence compact, subset of X. It suffices, therefore, to show that the collection { A ( ℓ ) : ℓ ∈ E $^{′}$} has the finite intersection property. Consider, to this aim, a finite collection of bounded linear functionals L := { ℓ 1 , . . . , ℓ$_{n}$ } ⊂ E $^{′}$, and let Y = conv ( L ) , a convex compact subset of E $^{′}$. The restriction of f ( x, ℓ ) to X × Y is obviously u.s.c. and quasiconcave in x and l.s.c. and quasiconvex in ℓ. Since both X and Y are compact and convex, it follows from Remark 20 that there exists ( x 0 , ℓ 0 ) ∈ X × Y with f ( x 0 , ℓ ) ≤ f ( x, ℓ 0 ) for all ( x, ℓ ) ∈ X × Y, i.e., ℓ ( φ ( x 0 ) − x 0 ) ≤ ℓ 0 ( φ ( x ) − x ) for all ( x, ℓ ) ∈ X × Y. Let ˆ x be such that ℓ 0 (ˆ x ) = max$_{x}$$_{∈}$$_{X}$ ℓ 0 ( x ) . Since φ (ˆ x ) ∈ X, it follows that ℓ 0 ( φ (ˆ x ) − ˆ x ) ≤ 0 and, consequently, ℓ ( φ ( x 0 ) − x 0 ) ≤ 0 , for all ℓ ∈ Y, in particular, ℓ$_{i}$ ( φ ( x 0 ) − x 0 ) = f ( x 0 , ℓ$_{i}$ ) ≤ 0 for all ℓ$_{i}$ ∈ L, and the proof is complete. The Markov-Kakutani follows by a standard compactness argument. Recall that a family F = { φ } of mappings is said to be abelian if φ$_{1}$φ$_{2}$ = φ$_{2}$φ$_{1}$ for all φ$_{1}$, φ$_{2}$ ∈ F. Corollary 22 (Theorem of Markov-Kakutani) Let X be a non-empty compact convex subset of a t.v.s. E with separating dual E ′ and let F be an abelian family of continuous affine transformations from X into itself . Then, there exists x 0 ∈ X such that φ ( x 0 ) = x 0 for every φ ∈ F. Proof. For any given φ ∈ F, let F ix ( φ ) be the set of its fixed points. We show that ⋂ φ ∈ F F ix ( φ ) =  . Clearly, for each φ ∈ F, F ix ( φ ) is non-empty (by Corollary 21), convex (as φ is affine), and closed hence compact in X. It suffices to show that the family { F ix ( φ ) : φ ∈ F } has the finite intersection property, i.e., ⋂ n i =1 F ix ( φ$_{i}$ ) =  for any { φ$_{1}$, . . . , φ$_{n}$ } ⊂ F . The proof is by induction on n. For n = 1 , clearly F ix ( φ$_{1}$ ) =  (Corollary 21). Assume that the statement is 19