The Sandwich Theorem for Sublinear and Superlinear Functionals
aa r X i v : . [ m a t h . F A ] N ov THE SANDWICH THEOREM FOR SUBLINEAR ANDSUPERLINEAR FUNCTIONALS
A.T. DIAB, S. I. NADA, D. L. FEARNLEY
Abstract.
The Hahn-Banach theorem is an extension theoremfor linear functionals which preserves certain properties. Specifi-cally, if a linear functional is defined on a subspace of a real vectorspace which is dominated by a sublinear functional on the entirespace, then this functional can be extended to a linear functionalon the entire space which is still dominated by a sublinear func-tional. In this paper, we generalize this result to show that a linearfunctional defined on a subspace of a real vector space which isdominated by a sublinear functional and also dominates a super-linear functional on the entire space can be extended to a linearfunctional on the entire space which is also dominated by a sub-linear functional and dominates a superlinear functional. Introduction
Extensions of linear functionals have been studied in locally convexspaces by several authors. In fact, the importance of the Hahn-Banachtheorem arises from its wide variety of applications, including complexand functional analysis and thermodynamics. The Hahn-Banach theo-rem has important implications for convex sets, and is the foundationfor an effective treatment of optimization. It is also used to solve manyproblems in linear algebra, conic duality theory, the minimax theorem,piecewise approximation of convex functionals, extensions of positivelinear functionals, and other results from modern control [1,5,8]. TheHahn-Banach theorem was proved by Hans Hahn (1879- 1934) in 1927and later by Stefan Banach (1892-1945) in 1929. It states that if E is areal vector space, M is a subspace of E and S is a sublinear functionalon E , and f is a linear functional on M such that f ( x ) ≤ S ( x ) forevery x ∈ M (or equivalently, − S ( − x ) ≤ f ( x ) ≤ S ( x )), then thereexists a linear functional L on E such that L | M = f and L ( x ) ≤ S ( x )for every x ∈ E (or equivalently, − S ( − x ) ≤ L ( x ) ≤ S ( x )). We willshow that if P is a superlinear functional on E satisfying the condition Mathematics Subject Classification. A
22 Extension linear functionals,The Hahn - Banach theorem, The classical Hahn - Banach theorem, Sublinearfunctionals and superlinear functionals. that P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M such that f ( x ) ≤ T ( x ) forevery x ∈ M , where T ( x ) = inf y ∈ E { S ( x + y ) − P ( y ) } for every x ∈ E ,then there exists a linear functional L on E such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E . We will refer to this as thesandwich theorem for a sublinear and superlinear functionals. Notethat if we take P ( x ) = − S ( − x ) for every x ∈ E , then we obtain theHahn-Banach theorem. As an application of the Hahn-Banach theoremwe have the following theorem: There exists a linear functional L on E for every sublinear functional S on E such that L ( x ) ≤ S ( x ) for every x ∈ E (we refer to this as the classical Hahn-Banach theorem). We willalso use the sandwich theorem for a sublinear and superlinear function-als to show that, for every a sublinear functional S on E and for everysuperlinear functional P on E , if P ( x ) ≤ S ( x ) for every x ∈ E thenthere exists a linear functional L on E such that P ( x ) ≤ L ( x ) ≤ S ( x )for every x ∈ E . Note that if we take P ( x ) = − S ( − x ) for every x ∈ E ,then this result implies the classical Hahn-Banach theorem.2. Relations between sublinear and superlinearfunctionals defined on a real linear space
In this section we present some basic definitions and results fromfunctional analysis concerning sublinear and superlinear functionals.In the following definitions we suppose that E is a real vector space[6,7,9]. Definition 2.1.
Let E be a real vector space, M be a subspace of E and f be a linear functional on M . Then a linear functional L on E is called an extension functional of f if L ( x ) = f ( x ) for every x ∈ M (denoted L | M = f ). Definition 2.2.
The functional S : E → R is called sublinear if itpossesses the following properties:(1) S ( x + y ) ≤ S ( x ) + S ( y ) for every x, y ∈ E (i.e., S is subadditive),(2) S ( αx ) = αS ( x ) for every x ∈ E and α > S is positivelyhomogeneous). The set of all sublinear functionals on E is denoted by Subl ( E ). Definition 2.3.
The functional P : E → R is called superlinear if itpossesses the following properties: ANDWICH THEOREM 3 (1) P ( x + y ) ≥ P ( x ) + P ( y ) for every x, y ∈ E (i.e., P is superadditive),(2) P ( αx ) = αP ( x ) for every x ∈ E and α > P is positivelyhomogeneous). The set of all superlinear functionals on E is denotedby Supl ( E ).Note that Lin ( E ) = Subl ( E ) T Supl ( E ), where Lin ( E ) is the setof all linear functionals on E . Also, it is easy to verify that if E ≡ R n , S ( x ) = r n P i =1 x i = k x k and P ( x ) = − s n − P i =1 x i + x n for every x ∈ R n ,then S ∈ Subl ( E ) and P ∈ Supl ( E ). Also, S ∈ Subl ( E ) if and only if − S ∈ Supl ( E ).Throughout the this paper we will use the symbols S and P to denotesublinear and superlinear functionals on the real vector space E . Lemma 2.1. (i) S (0) = 0 and P (0) = 0 , where is the zero vector ofthe real vector space E .(ii) − S ( − x ) ≤ S ( x ) and − P ( − x ) ≥ P ( x ) for every x ∈ E .(iii) A functional f on E is linear if and only if it is additive andpositively homogeneous. Proof. (i) Since sublinear functionals are positively homogeneous, S (0) = S (20) = 2 .S (0 + 0) = S (0) + S (0). Therefore S (0) = 0. Simi-larly, P (0) = 0.(ii) Since 0 = S (0) = S ( − x + x ) ≤ S ( − x ) + S ( x ) for every x ∈ E , then − S ( − x ) ≤ S ( x ). Also, 0 = P (0) = P ( − x + x ) ≥ P ( − x ) + P ( x ) forevery x ∈ E , so − P ( − x ) ≥ P ( x ) for every x ∈ E .(iii) By definition, linear functionals are additive and positively homo-geneous. Let f be an additive positively homogeneous functional on E . It remains to show that f ( αx ) = αf ( x ) for every x ∈ E and α < f is additive, f (0 + 0) = f (0) + f (0). Thus, f (0) = 0, and con-sequently f ( − x ) = − f ( x ) for every x ∈ E . Let x ∈ E and α = − t ,where t >
0. Then f ( αx ) = f ( − tx ) = tf ( − x ) = − tf ( x ) = αf ( x ).Thus f is homogeneous and f is linear. Lemma 2.2. If S ( x ) ≤ P ( x ) for every x ∈ E , then P and S are linearfunctionals on E . Moreover S ≡ P on E . Proof.
Since S is sublinear and P is superlinear, then from(ii)inLemma 2.1, we obtain that − S ( − x ) ≤ S ( x ) and − P ( − x ) ≥ P ( x )for every x ∈ E . So, S ( − x ) ≤ P ( − x ) ≤ − P ( x ) ≤ − S ( x ). Thus, A.T. DIAB, S. I. NADA, D. L. FEARNLEY S ( − x ) = − S ( x ) for every x ∈ E . For all x, y ∈ E , S ( x + y ) = − S ( − x − y ) ≥ − S ( − x ) − S ( − y ). Hence, S is both subadditive andsuperadditive and is therefore additive. Since S is positively homo-geneous and additive, from (iii) in Lemma 2.1 we can conclude that S is linear. By a similar argument, P is linear. Finally, from (ii) inLemma 2.1 we see that − S ( x ) ≤ S ( − x ) ≤ P ( − x ) ≤ − P ( x ). Hence, P ( x ) ≤ S ( x ) for every x ∈ E . Therefore P ( x ) = S ( x ) for every x ∈ E . Lemma 2.3. If f is a homogeneous functional on E and g is anarbitrary functional on E , then the following statements are equivalent:(i) f ( x ) ≤ g ( x ) for every x ∈ E ,(ii) − g ( − x ) ≤ f ( x ) ≤ g ( x ) for every x ∈ E . Proof.
Clearly, (ii) implies (i). Let f ( x ) ≤ g ( x ) for every x ∈ E . Then f ( − x ) ≤ g ( − x ) and consequently − f ( x ) = f ( − x ) ≤ g ( − x ) for every x ∈ E . Therefore − g ( − x ) ≤ f ( x ) and hence − g ( − x ) ≤ f ( x ) ≤ g ( x ). Corollary 2.1. If f is a linear functional on E and f ( x ) ≤ S ( x ) forevery x ∈ E , where S is a sublinear functional on E , then − S ( − x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ E . Proof.
The proof follows directly from Lemma 2.3 and the fact thatevery linear functional is homogeneous.
Lemma 2.4.
Let f ∈ Lin ( M ) , and let g, h : M → R , where M is asubspace of E , g (0) = h (0) = 0 , and g ( x ) ≤ h ( x ) for every x ∈ M .Then the following are equivalent:(i) g ( x ) ≤ f ( x ) ≤ h ( x ) for every x ∈ M ,(ii) f ( x ) ≤ h ( x + y ) − g ( y ) for all x, y ∈ M . Proof.
First, we assume that g ( x ) ≤ f ( x ) ≤ h ( x ) for every x ∈ M ,and we want to show that f ( x ) ≤ h ( x + y ) − g ( y ) for every x, y ∈ M .Since g ( x + y ) ≤ f ( x + y ) ≤ h ( x + y ), and − h ( y ) ≤ − f ( y ) ≤ − g ( y ), itfollows that g ( x + y ) − h ( y ) ≤ f ( x + y ) − f ( y ) ≤ h ( x + y ) − g ( y ). Since f ∈ Lin ( M ), then g ( x + y ) − h ( y ) ≤ f ( x ) ≤ h ( x + y ) − g ( y ). Next,assume that f ( x ) ≤ h ( x + y ) − g ( y ) for every x, y ∈ M . Setting y = 0 wesee that f ( x ) ≤ h ( x ). By setting x = − y , we obtain f ( − y ) ≤ − g ( y ),so g ( y ) ≤ f ( y ). Therefore, g ( x ) ≤ f ( x ) ≤ h ( x ) for every x ∈ M . ANDWICH THEOREM 5 The Hahn-Banach theorem and the classicalHahn-Banach theorem
Theorem 3.1. (The Hahn-Banach theorem for sublinear functionals)Let E be a real vector space, M be a subspace of E and S ∈ Subl ( E ) .Furthermore, let f ∈ Lin ( M ) such that f ( x ) ≤ S ( x ) for every x ∈ M (or equivalently, − S ( − x ) ≤ f ( x ) ≤ S ( x ) ), then there exists afunctional L ∈ Lin ( E ) such that L | M = f and L ( x ) ≤ S ( x ) for every x ∈ E (or equivalently, − S ( − x ) ≤ f ( x ) ≤ S ( x ) ) [7 , . The following theorem is the famous application of the Hahn-Banachtheorem and some mathematicians, like Kelly-Namioka[2], Rudin [7],Koing [4], Simons [8,9] and others have proved it. In the following, weintroduce a simple and short proof for it.
Theorem 3.2. (The classical Hahn-Banach theorem for sublinear func-tionals) Let E be a real vector space and S ∈ Subl ( E ) , then there existsa functional L ∈ Lin ( E ) such that L ( x ) ≤ S ( x ) for every x ∈ E . Proof. If S ( x ) = 0 for every x ∈ E , then the result follows immediatelyby taking L ≡ E . Assume that S E . Let x ∈ E , let M = { αx : α ∈ R } , and define f ( αx ) = αS ( x ) for every x ∈ M .We have two cases: Case 1 : M = { } . Since 0 = f (0) = S (0), f ( x ) ≤ S ( x ) for every x ∈ M . By the Hahn-Banach theorem (or Theorem 3.1), there existsa linear functional L on E such that L | M = f and L ( x ) ≤ S ( x ) forevery x ∈ E . Case 2 : M = { } . Let x = αx ∈ M . If α ≥ f ( x ) = f ( αx ) = αf ( x ) ≤ αS ( x ) = S ( x ). If α < α = − t , where t >
0. Then f ( x ) = f ( αx ) = αS ( x ) = − tS ( x ) = − S ( tx ) = − S ( − αx ) = − S ( − x ) ≤ S ( x ) for every x ∈ M . Hence, f ( x ) ≤ S ( x )for every x ∈ M . Thus, from the extension form of the Hahn-Banachtheorem, there exists a linear functional L on E such that L | M = f and L ( x ) ≤ S ( x ) for every x ∈ E .In the following Theorems, we mention a similar fact for superlinearfunctionals and we call it the extension form of the Hahn-Banach theo-rem for superlinear functionals and the classical Hahn-Banach theoremfor superlinear functionals [6]. The proofs of Theorems 3.3 and 3.4 canbe found in [6]. Theorem 3.3. (The extension form of Hahn-Banach theorem for su-perlinear functionals)
A.T. DIAB, S. I. NADA, D. L. FEARNLEY
Suppose that f ∈ Lin ( M ) , where M is a subspace of E such that P ( x ) ≤ f ( x ) for every x ∈ M , where P ∈ Supl ( E ) . Then there existsa functional L ∈ Lin ( E ) such that L | M = f and P ( x ) ≤ L ( x ) forevery x ∈ E . Theorem 3.4. (The classical Hahn-Banach theorem for superlinearfunctionals)Let E be a real vector space and P ∈ Supl ( E ) , then there exists afunctional L ∈ Lin ( E ) such that P ( x ) ≤ L ( x ) for every x ∈ E . The sandwich theorem for sublinear and superlinearfunctionals
In this section we demonstrate three different approaches to provingthat, if S ∈ Subl ( E ), P ∈ Supl ( E ) and f ∈ Lin ( M ), where M is asubspace of E , and P ( x ) ≤ f ( x ) ≤ S ( x ) for each x ∈ M , then thereis an extension functional L ∈ Lin ( E ) so that P ( x ) ≤ L ( x ) ≤ S ( x ) foreach x ∈ E . Throughout the this section, we let T ( x ) = inf y ∈ E { S ( x + y ) − P ( y ) } for every x ∈ E . Lemma 4.1.
Let f be a linear functional of a subspace M of E such that: f ( x ) ≤ T ( x ) for every x ∈ M (4.1). Then the followingconditions are satisfied: P ( x ) ≤ S ( x ) for every x ∈ E (4.2), P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M (4.3). Proof.
From the inequality (4.1), we obtain that f ( x ) ≤ S ( x + y ) − P ( y ) for every y ∈ E and x ∈ M . Consequently, by setting x = 0,we see that P ( y ) ≤ S ( y ) for every y ∈ M . By setting y = 0, wesee that f ( x ) ≤ S ( x ) for every x ∈ M , and by setting y = − x , weobtain that f ( − y ) ≤ − P ( y ) or P ( x ) ≤ f ( x ) for every x ∈ M . Hence P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M .The following example shows that the converse of the above lemmais false. Example 4.1.
Let E = R , P ( x ) = −| x | + x , S ( x ) = p x + x ,where x = ( x , x ) and M = { ( x ,
0) : x ∈ R } is an one dimensionalsubspace of E . Suppose that f : M → R is defined as f ( x ) = x for ANDWICH THEOREM 7 every x = ( x , ∈ M . Then inequalities (4.2) and (4.3) are satisfiedbut (4.1) is not. Solution.
It easy to verify that P ∈ Supl ( E ), S ∈ Subl ( E ) and f is a linear functional on M . Let x = ( x , x ) ∈ E . Then S ( x ) = p x + x ≥ x ≥ −| x | + x = P ( x ). Also, if x = ( x , ∈ M then P ( x ) = −| x | ≤ x = f ( x ) ≤ | x | ≤ p x + x = S ( x ), so P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M . Finally, suppose that in-equality (4.1) is true. Then, for each x ∈ M , f ( x ) ≤ S ( x + y ) − P ( y ) for every y ∈ E . Let x = ( x , ∈ M where x >
0. Then x ≤ p ( x + y ) + y + | y | − y for every y = ( y , y ) ∈ E . Inparticular, if y = (0 , y ) and y > x ≤ p x + y − y = ( √ x + y − y )( √ x + y + y )( √ x + y + y ) = x √ x + y + y , which ap-proaches 0 as y → ∞ . But this implies that x ≤
0, a contradiction.
Lemma 4.2. If P ( x ) = − S ( − x ) for every x ∈ E , then T ( x ) = S ( x ) for every x ∈ E . Proof.
From the definition of T , we conclude that T ( x ) ≤ S ( x + y ) − P ( y ) for all x, y ∈ E . Thus T ( x ) ≤ S ( x ). Also, S ( x ) ≤ S ( x + y ) + S ( − y ) = S ( x + y ) − P ( y ) for every y ∈ E , so S ( x ) ≤ T ( x ). Therefore T ( x ) = S ( x ) for every x ∈ E , as required. Corollary 4.1. If P ( x ) = − S ( − x ) for every x ∈ E , then conditions(4.2) and (4.3) are satisfied if and only if condition (4.1) is satisfied. Proof.
Sufficiency follows from Lemma 4.1. Conversely, if conditions(4.2) and (4.3) are satisfied then condition (4.1) follows from Lemma4.2.
Lemma 4.3.
Let P ( x ) ≤ S ( x ) for every x ∈ E . Then P ( x ) ≤ T ( x ) ≤ S ( x ) for every x ∈ E , and T is a sublinear functional. Furthermore, if L is a linear functional on E such that P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E then L ( x ) ≤ T ( x ) for every x ∈ E . Proof.
Since P ( u ) ≤ L ( u ), it follows that L ( u ) ≤ − P ( − u ). Hence,from the linearity of L we see that L ( u + v ) ≤ S ( v ) − P ( − u ) forevery u, v ∈ E . Setting u + v = x and u = − y , we obtain L ( x ) ≤ S ( x + y ) − P ( y ). Taking the infimum over all y ∈ E , we see that L ( x ) ≤ T ( x ) for every x ∈ E . We next prove that T : E → R is A.T. DIAB, S. I. NADA, D. L. FEARNLEY sublinear. Since P ( y ) ≤ S ( y ) ≤ S ( x + y ) + S ( − x ), it follows that − S ( − x ) ≤ S ( x + y ) − P ( y ). Taking the infimum over all y ∈ E, wesee that T ( x ) ≥ − S ( − x ), so T : E → R . It follows that T is positivelyhomogeneous since S and P are positively homogeneous. Thus, toprove that T is sublinear it remains to show that T is subadditive. Let u, v ∈ E . Then T ( u + v ) = inf y + z ∈ E { S ( u + v + y + z ) − P ( y + z ) } ≤ inf y ∈ E { S ( u + z ) − P ( y ) } + inf z ∈ E { S ( v + z ) − P ( z ) } = T ( u ) + T ( v ). Thus T issubadditive and, consequently, sublinear. Since T ( x ) ≤ S ( x + y ) − P ( y ),it follows that T ( x ) ≤ S ( x + 0) − P (0), so T ( x ) ≤ S ( x ) for every x ∈ E . Also, T ( − y ) ≤ S (0) − P ( y ), so T ( − y ) ≤ − P ( y ) and − T ( y ) ≤ T ( − y ) ≤ − P ( y ), and hence P ( y ) ≤ T ( y ) for every y ∈ E . Therefore P ( x ) ≤ T ( x ) ≤ S ( x ) for every x ∈ E , and the proof is complete.In the following theorem we introduce the first method for provingthe statement above, which we call the sandwich theorem for sublinearand superlinear functionals. Throughout this section, we assume that f is a linear functional on M , where M is a subspace of E . As before, S ∈ Subl ( E ) and P ∈ Supl ( E ). Theorem 4.1.
There exists an extension linear functional L on E ofthe linear functional f on M such that P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E if and only if f ( x ) ≤ T ( x ) for every x ∈ M . Proof.
From Lemma 4.3 we know that T ∈ Subl ( E ) and P ( x ) ≤ T ( x ) ≤ S ( x ) for every x ∈ E . Also by lemma 4.3, if P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M then f ( x ) ≤ T ( x ) for every x ∈ M since f is alinear functional on M . Next, assume that f ( x ) ≤ T ( x ) for every x ∈ M . By lemma 4.1, P ( x ) ≤ S ( x ) for all x ∈ E and P ( x ) ≤ f ( x ) ≤ S ( x )for all x ∈ M . By Theorem 3.3 there is an extension linear functional L on E such that f ( x ) = L ( x ) for every x ∈ M and P ( x ) ≤ L ( x ) forevery x ∈ E . By lemma 4.3 L ( x ) ≤ T ( x ) ≤ S ( x ) for every x ∈ E .Hence, it follows that P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E .The following corollary shows that if we let P ( x ) = − S ( − x ) for every x ∈ E , then we obtain the Hahn-Banach theorem. Corollary 4.2. If P ( x ) = − S ( − x ) for every x ∈ E and f is a linearfunctional on M such that f ( x ) ≤ S ( x ) for every x ∈ M . Then thereexists a linear functional L on E such that L | M = f and L ( x ) ≤ S ( x ) for every x ∈ E . ANDWICH THEOREM 9
Proof.
The proof follows directly from theorem 4.1 and corollary 4.1,as required.We now introduce the second proof of the Sandwich Theorem forSuperlinear and Sublinear Functionals.
Lemma 4.4.
Let x / ∈ M, a = sup y ∈ M { f ( y ) − S ( y − x ) } , b = inf y ∈ M {− f ( y )+ S ( y + x ) } , c = inf y ∈ M { f ( y ) − P ( y − x ) } and d = sup y ∈ M {− f ( y )+ P ( y + x ) } ,and assume that a ≤ b and d ≤ c . Then the following statements areequivalent:(i) f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and y ∈ E .(ii) a ≤ c and d ≤ b . Proof.
Suppose that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and y ∈ E . Then f ( w − z ) ≤ S ( w − x ) − P ( z − x ) for every w, z ∈ M ,where x = w − z and y = z − x . Consequently, from the linearity of f , we obtain that f ( w ) − S ( w − x ) ≤ f ( z ) − P ( z − x ) for every w, z ∈ M . Hence, { f ( w ) − S ( w − x ) : w ∈ M } is bounded aboveand has a least upper bound, a . Similarly, we can see that there is a c = inf z ∈ M { f ( z ) − P ( z − x ) } . Since f ( w − z ) ≤ S ( w − x ) − P ( z − x )for every w, z ∈ M , it follows that a ≤ c . If we let x = w − z and y = z + x in the inequality f ( x ) ≤ S ( x + y ) − P ( y ), we obtain that − f ( z ) + P ( z − x ) ≤ − f ( w ) + S ( w + x ) for every w, z ∈ M andhence d ≤ b . Conversely, assume that a ≤ c and d ≤ b . Since a ≤ c ,it follows that sup w ∈ M { f ( w ) − S ( w − x ) } ≤ inf z ∈ M { f ( z ) − P ( z − x ) } andhence f ( w ) − S ( w − x ) ≤ f ( z ) − P ( z − x ) for every w, z ∈ M ,and consequently f ( w − z ) ≤ S ( w − x ) − P ( z − x ). Thus, f ( x ) ≤ S ( x + y ) − P ( y ) for every x = w − z ∈ M and y = z − x ∈ E . Remark 4.1.
Condition (i) in the Lemma 4.4 is equivalent to stat-ing that f ( x ) ≤ T ( x ) = inf y ∈ E { S ( x + y ) − P ( y ) } for every x ∈ M .Also, condition (ii) in Lemma 4.4 is equivalent to the condition that [ a, b ] T [ d, c ] = ∅ . Lemma 4.5.
Let x / ∈ M and let M be the the smallest linear subspaceof E containing M and x . Let f be a linear functional on M suchthat f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and for every y ∈ M .Then there exists a linear functional L on M such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ M . Proof.
From Lemma 4.1, we have f ( x ) ≤ S ( x ) for every x ∈ M .Therefore for arbitrary y , y ∈ M we have f ( y + y ) ≤ S ( y + y ). So,from the linearity of f and the sublinearity of S we get f ( y )+ f ( y ) = f ( y + y ) ≤ S ( y + y ) ≤ S ( y − x ) + S ( y + x ). Thus, we obtain: f ( y ) − S ( y − x ) ≤ − f ( y ) + S ( y + x ) (4.4).By fixing y and letting y vary over M , we see that { f ( y ) − S ( y − x ) : y ∈ M } is bounded above. Let a = sup y ∈ M { f ( y ) − S ( y − x ) } . By asimilar argument, we can find b = inf y ∈ M {− f ( y ) + S ( y + x ) } . It isclear from (4.4) that a ≤ b . Similarly, from Lemma 4.1, we know that P ( x ) ≤ f ( x ) for every x ∈ M . Therefore, for arbitrary y , y ∈ M wehave P ( y + y ) ≤ f ( y + y ). So, from the linearity of f and fromthe super linearity of P , we get P ( y + x + P ( y − x ) ≤ P ( y + y ) ≤ f ( y ) + f ( y ). Thus, we have: − f ( y ) + P ( y + x ) ≤ f ( y ) − P ( y − x ) (4.5).By fixing y and letting y vary over M , we see that {− f ( y ) + P ( y + x ) : y ∈ M } is bounded above. Let d = sup y ∈ M {− f ( y ) + P ( y + x ) } .By a similar argument, there exists a c = inf y ∈ M { f ( y ) − P ( y − x ) } , andfrom inequality (4.5) we see that d ≤ c . By Remark 4.1, we note that[ a, b ] ∩ [ d, c ] = ∅ . Let ξ ∈ [ a, b ] ∩ [ d, c ]. Note that, for every x ∈ M , thereis a unique y ∈ M and α ∈ R such that x = y + αx . We define a realvalued functional L on M by setting L ( x ) = L ( y + αx ) = f ( y ) + αξ .We wish to verify that L is linear such that L ( x ) = f ( x ) for every x ∈ M , and that P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ M . First, weshow that L is linear. Let z , z ∈ M . Then there exist y , y ∈ M and α , α ∈ R such that z = y + α x and z = y + α x . Hence, forevery α, β ∈ R we get L ( αz + βz ) = L (( αy + βy ) + ( αα + βα ) x ) = f ( αy + βy ) + ( αα + βα ) ξ = α ( f ( y ) + α ξ ) + β ( f ( y ) + α ξ ) = L ( z ) + L ( z ). Next, we wish to show that L ( x ) = f ( x ) for every x ∈ M . If x ∈ M , then since x = y + αx for a unique choice of y ∈ M and α ∈ R , it follows that that x = y and α = 0. Thus, L ( x ) = f ( x )for every x ∈ M , so L | M = f . To show that P ( x ) ≤ L ( x ) ≤ S ( x ) forevery x = y + αx ∈ M , we consider three cases. Case 1. α = 0. We have already shown that L ( x ) = f ( x ), andtherefore P ( x ) ≤ L ( x ) ≤ S ( x ), for every x ∈ M . Case 2. α >
0. Since ξ ∈ [ a, b ] ∩ [ d, c ], it follows that d = sup y ∈ M {− f ( y )+ P ( y + x ) } ≤ ξ ≤ b = inf y ∈ M {− f ( y ) + S ( y + x ) } . Thus, − f ( y ) + P ( y + x ) ≤ ξ ≤ − f ( y ) + S ( y + x ) for every y ∈ M . Replacing y by yα , andmultiplying the inequality by α , we get P ( y + αx ) ≤ f ( y ) + αξ ≤ ANDWICH THEOREM 11 S ( y + αx ) for every y ∈ M . Therefore P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ M , where α > Case 3. α <
0. Let α = − t , where t >
0. We know that a = sup y ∈ M { f ( y ) − S ( y − x ) } ≤ ξ ≤ c = inf y ∈ M { f ( y ) − P ( y − x ) } so f ( y ) − S ( y − x ) ≤ ξ ≤ f ( y ) − P ( y − x ) for every y ∈ M . Thus, − S ( y − x ) ≤ − f ( y )+ ξ ≤ − P ( y − x ) for every y ∈ M . Replacing y by yt , and multiplying both sides by t , we get − S ( y − tx ) ≤ − f ( y ) + tξ ≤− P ( y − tx ), so − S ( y + αx ) ≤ − f ( y ) − αξ ≤ − P ( y + αx ). Thismeans that P ( y + αx ) ≤ f ( y ) + αξ ≤ S ( y + αx ) for every y ∈ M .Therefore, P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ M . This completes theproof. Theorem 4.2.
Let M and E be two linear subspaces of E such that M ⊆ E ⊆ E . Suppose that f is a linear functional on M such that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and y ∈ E . Then there existsa linear functional L on E such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E . Proof.
The proof follows directly from Lemma 4.5 and Zorn’s Lemma(See [3], p. 134 for this argument).It is worth pointing out that such an extension of a linear functional f on M does not always exist on the whole space E . If (4.1) is notvalid on the whole space E but is valid on some linear subspace E containing M , then by using the above theorem we can extend f to alinear functional L on E so that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x )for every x ∈ E , but it does not follow that we can extend f to E . Example 4.2.
Let S ( x ) = s P i =1 x i and P ( x ) = − s P i =1 x i + x forevery x ∈ R . Let M = { ( x , , ,
0) : x ∈ R } be a subspace of R , and define a linear functional f on M as f ( x ) = x for every x = ( x , , , ∈ M . Then we show that there exists a linear functional L on a subspace E = R such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ R , but there does not exist a linear functional L on R such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ R . Solution.
It is easy to verify that S ∈ Subl ( R ) and P ∈ Supl ( R ).First, we show that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and forevery y ∈ M = { ( x , x , x ,
0) : x , x , x ∈ R } . Let x = ( x , , , ∈ M and y = ( y , y , y , ∈ M . Then f ( x ) = x , S ( x + y ) = p ( x + y ) + y + y and P ( y ) = − p y + y + y . Therefore f ( x ) = x ≤ | x | = | x | − | y | + | y | ≤ | x + y | + | y | ≤ p ( x + y ) + y + y + p y + y + y ≤ S ( x + y ) − P ( y ). Hence, by Theorem 4.2, there ex-ists a linear functional L on M such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ M . Suppose there is a linear func-tional L on R such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) forevery x ∈ R . Then by Theorem 4.1 and Remark 4.1, it must fol-low that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and for every y ∈ R . But if we let x = (10 , , , ∈ M and y = (0 , , , ∈ R , then f ( x ) = 10, S ( x + y ) = √ , P ( y ) = 1 and consequently S ( x + y ) − P ( y ) ∼ = 9 . <
10 = f ( x ), which contradicts the assump-tion that f ( x ) ≤ S ( x + y ) − P ( y ). Theorem 4.3.
Let f be a linear functional on M such that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and x ∈ E . Then there exists a linearfunctional L on E such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) forevery x ∈ E . Proof.
The theorem follows directly from Theorem 4.2, setting E = E .We now move to our third proof of the Sandwich Theorem for Su-perlinear and Sublinear Functionals. Throughout the remainder of thispaper, let Γ = { L ∈ Lin ( E ) : L | M = f and L ( x ) ≤ S ( x ) for every x ∈ E } and let Γ = { L ∈ Lin ( E ) : L | M = f and P ( x ) ≤ L ( x ) forevery x ∈ E } . Remark 4.2.
Let f ∈ Lin ( M ) , P ∈ Supl ( E ) , and S ∈ Subl ( E ) , sothat P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M , and P ( x ) ≤ S ( x ) for every x ∈ E . Then by Theorem 3.1 and see Theorem 3.3, it follows that Γ and Γ are non-empty. Theorem 4.4.
A necessary and sufficient condition for Γ ∩ Γ = φ isthat f ( x ) ≤ T ( x ) for every x ∈ M , where T ( x ) = inf y ∈ E { S ( x + y ) − P ( y ) } . Proof.
Suppose there is a linear functional L ∈ Γ ∩ Γ . Then L | M = f , L ( x ) ≤ S ( x ) and P ( x ) ≤ L ( x ) for every x ∈ E . Since P ( u ) ≤ L ( u ), P ( − u ) ≤ L ( − u ) = − L ( u ), and L ( u ) ≤ − P ( − u ) for every u ∈ E .Therefore, from the linearity of L , we have L ( u + v ) = L ( u ) + L ( v ) ≤ S ( v ) − P ( − u ). This means that L ( x ) ≤ S ( x − u ) − P ( − u ), so L ( x ) ≤ ANDWICH THEOREM 13 S ( x + y ) − P ( y ), where x = u + v ∈ E and y = − u ∈ E . Since L | M = f , it follows that f ( x ) ≤ S ( x + y ) − P ( y ) for every x ∈ M and y ∈ E . Taking the infimum over all y ∈ E , f ( x ) ≤ T ( x ) for every x ∈ M . Conversely, assume that f ( x ) ≤ T ( x ) for every x ∈ M . Itis easy to verify that T ∈ Subl ( E ) (see Lemma 4.3). By lemma 4.1, P ( x ) ≤ S ( x ) for all x ∈ E and P ( x ) ≤ f ( x ) ≤ S ( x ) for all x ∈ M . ByTheorem 3.3 there is an extension linear functional L on E such that f ( x ) = L ( x ) for every x ∈ M and P ( x ) ≤ L ( x ) for every x ∈ E . Bylemma 4.3 L ( x ) ≤ T ( x ) ≤ S ( x ) for every x ∈ E . Hence, it follows that P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E . Hence, L ∈ Γ ∩ Γ . Theorem 4.5.
Let f be a linear functional on M , where M is asubspace of E such that P ( x ) ≤ f ( x ) ≤ S ( x ) for every x ∈ M , where S ∈ Subl ( E ) and P ∈ Supl ( E ) . Then the necessary and sufficientcondition that there exists a linear functional L on E such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x ) for every x ∈ E is that f ( x ) ≤ T ( x ) forevery x ∈ M . Proof.
Since f is a linear functional on M and P ( x ) ≤ f ( x ) ≤ S ( x )for every x ∈ M , then P ( x ) ≤ f ( x ) and f ( x ) ≤ S ( x ), then fromRemark 4.2, we conclude that Γ = ∅ and Γ = ∅ . Assume that L isa linear functional on E such that L | M = f and P ( x ) ≤ L ( x ) ≤ S ( x )for every x ∈ E . Then Γ ∩ Γ = φ . Hence from Theorem 4.4, wehave f ( x ) ≤ T ( x ) for every x ∈ M . Conversely, let f ( x ) ≤ T ( x ) forevery x ∈ M . Then T ∈ Subl ( E ), and by Theorem 4.4 we obtain thatΓ ∩ Γ = φ , and consequently there exists a linear functional L on E such that L | M = f and P ( x ) ≤ L ( x ) for every x ∈ E and L ( x ) ≤ S ( x )for every x ∈ E . REFERENCES1. A.T.Diab, N. Faried and Afaf Abou Elmatty, Extending LinearContinuous Functionals with Preservation of Positivity , Proc. Math.Phys. Soc. Egypt, 84, 2, p.p. 125-132, (2006).2. J. L. Kelley, I. Namioka, and co-authors,
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Dept. of Math., Faculty of Science, Ain Shams University, Cairo,EgyptDept. of Math., Faculty of Science, Qatar University, Doha, QatarDept. of Math., Utah Valley University, Orem, Utah, USA
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