Note on the Hahn-Banach Theorem in a Partially Ordered Vector Space

Abstract

Using a fixed point theorem in a partially ordered set, we give a new proof of the Hahn-Banach theorem in the case where the range space is a partially ordered vector space.

1. Introduction

The Hahn-Banach theorem is one of the most fundamental theorems in the functional analysis theory. This theorem is well known in the case where the range space is the real number system as follows.

Let 𝑝 be a sublinear mapping from a vector space 𝑋 into the real number system 𝑅, 𝑌 a subspace of 𝑋, and 𝑞 a linear mapping from 𝑌 into 𝑅 such that 𝑞≤𝑝 on 𝑌. Then there exists a linear mapping 𝑔 from 𝑋 into 𝑅 such that 𝑔=𝑞 on 𝑌 and 𝑔≤𝑝 on 𝑋.

It is known that this theorem is established in the case where the range space is a Dedekind complete Riesz space as follows [1–3].

Let 𝑝 be a sublinear mapping from a vector space 𝑋 into a Dedekind complete Riesz space 𝐸, 𝑌 a subspace of 𝑋 and 𝑞 a linear mapping from 𝑌 into 𝐸 such that 𝑞≤𝑝 on 𝑌. Then there exists a linear mapping 𝑔 from 𝑋 into 𝐸 such that 𝑔=𝑞 on 𝑌 and 𝑔≤𝑝 on 𝑋.

On the other hand, Hirano et al. [4] showed the Hahn-Banach theorem by using the Markov-Kakutani fixed point theorem [5] in the case where the range space is the real number system.

In this paper, motivated by Hirano et al. [4], we give a proof of the Hahn-Banach theorem using a fixed point theorem. We show the Hahn-Banach theorem in the case where the range space is a Dedekind complete partially ordered vector space (Theorem 10). Moreover, we show the Mazur-Orlicz theorem in a Dedekind complete partially ordered vector space (Theorem 11).

2. Preliminaries

Let (𝐸,≤) be a partially ordered set and 𝐹 a subset of 𝐸. The set 𝐹 is called a chain if any two elements are comparable; that is, 𝑥≤𝑦 or 𝑦≤𝑥 for any 𝑥, 𝑦∈𝐹. An element 𝑥∈𝐸 is called a lower bound of 𝐹 if 𝑥≤𝑦 for any 𝑦∈𝐹. An element 𝑥∈𝐸 is called the minimum of 𝐹 if 𝑥 is a lower bound of 𝐹 and 𝑥∈𝐹. If there exists a lower bound of 𝐹, then 𝐹 is said to be bounded from below. An element 𝑥∈𝐸 is called an upper bound of 𝐹 if 𝑦≤𝑥 for any 𝑦∈𝐹. An element 𝑥∈𝐸 is called the maximum of 𝐹 if 𝑥 is an upper bound and 𝑥∈𝐹. If there exists an upper bound of 𝐹, then 𝐹 is said to be bounded from above. If the set of all lower bounds of 𝐹 has the maximum, then the maximum is called an infimum of 𝐹 and denoted by inf 𝐹. If the set of all upper bounds of 𝐹 has the minimum, then the minimum is called a supremum of 𝐹 and denoted by sup 𝐹. An element 𝑥∈𝐹 is called a minimal of 𝐹 if 𝑦≤𝑥 and 𝑦∈𝐹 implies 𝑦=𝑥. A partially ordered set 𝐸 is said to be complete if every nonempty chain of 𝐸 has an infimum; 𝐸 is said to be chain complete if every nonempty chain of 𝐸 which is bounded from below has an infimum; 𝐸 is said to be Dedekind complete if every nonempty subset of 𝐸 which is bounded from below has an infimum. A mapping 𝑓 from 𝐸 into 𝐸 is said to be decreasing if 𝑓(𝑥)≤𝑥 for any 𝑥∈𝐸. For further information of a partially ordered set, see [1, 2, 6–10].

In a complete partially ordered set, the following theorem is obtained; see [11–14].

Theorem 1 (Bourbaki-Kneser). Let 𝐸 be a complete partially ordered set. Let 𝑓 be a decreasing mapping from 𝐸 into 𝐸. Then 𝑓 has a fixed point.

A partially ordered set 𝐸 is called a partially ordered vector space if 𝐸 is a vector space and 𝑥+𝑧≤𝑦+𝑧 and 𝛼𝑥≤𝛼𝑦 hold whenever 𝑥,𝑦,𝑧∈𝐸, 𝑥≤𝑦 and 𝛼 is a nonnegative real number. If a partially ordered vector space 𝐸 is a lattice, that is, any two elements in 𝐸 have a supremum and an infimum, then 𝐸 is called a Riesz space.

Let 𝑋 be a vector space and 𝐸 a partially ordered vector space. A mapping 𝑓 from 𝑋 into 𝐸 is said to be concave if 𝑓(𝑡𝑥+(1−𝑡)𝑦)≥𝑡𝑓(𝑥)+(1−𝑡)𝑓(𝑦) for any 𝑥,𝑦∈𝑋 and 𝑡∈[0,1]. A mapping 𝑝 from 𝑋 into 𝐸 is said to be sublinear if the following conditions are satisfied. (S1) For any 𝑥,𝑦∈𝑋, 𝑝(𝑥+𝑦)≤𝑝(𝑥)+𝑝(𝑦). (S2) For any 𝑥∈𝑋 and nonnegative real number 𝛼, 𝑝(𝛼𝑥)=𝛼𝑝(𝑥).

Let 𝐸^𝑋 be the set of mappings from 𝑋 into 𝐸. Throughout this paper, 𝐸^𝑋 is ordered as follows. For 𝑓,𝑔∈𝐸^𝑋, let 𝑓≤𝑔 mean that 𝑓(𝑥)≤𝑔(𝑥) for any 𝑥∈𝑋. It is easy to check that 𝐸^𝑋 is also a partially ordered vector space.

The following lemmas are useful for the proof of our main results.

Lemma 2. Let 𝑋 be a vector space, 𝐸 a chain complete partially ordered vector space, and 𝑍 a nonempty chain of 𝐸^𝑋 which is bounded from below. Then there exists inf{ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋. Moreover, if 𝑝∈𝐸^𝑋 is defined by 𝑝(𝑥)=inf{ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋, then 𝑝=inf𝑍; that is, 𝐸^𝑋 is chain complete.

Proof. Let 𝑥∈𝑋 be fixed. Since 𝑍 is a nonempty chain, so is {ℎ(𝑥)∣ℎ∈𝑍}. Let 𝑓 be a lower bound of 𝑍. Since 𝑓(𝑥)≤ℎ(𝑥) for any ℎ∈𝑍, {ℎ(𝑥)∣ℎ∈𝑍} is bounded from below. Therefore, since 𝐸 is chain complete, there exists inf{ℎ(𝑥)∣ℎ∈𝑍}.
Define 𝑝∈𝐸^𝑋 by 𝑝(𝑥)=inf{ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋. Then it is clear that 𝑝≤ℎ for any ℎ∈𝑍; that is, 𝑝 is a lower bound of 𝑍. Let 𝑞 be a lower bound of 𝑍. Since 𝑞(𝑥)≤ℎ(𝑥) for any 𝑥∈𝑋 and ℎ∈𝑍, 𝑞(𝑥) is a lower bound of {ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋. Therefore, 𝑞(𝑥)≤inf{ℎ(𝑥)∣ℎ∈𝑍}=𝑝(𝑥) for any 𝑥∈𝑋 and thus 𝑝=inf𝑍.

Lemma 3. Let 𝑋 be a vector space, 𝐸 a Dedekind complete partially ordered vector space, and 𝑍 a nonempty subset in 𝐸^𝑋 which is bounded from below. Then there exists inf{ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋. Moreover, if 𝑝∈𝐸^𝑋 is defined by 𝑝(𝑥)=inf{ℎ(𝑥)∣ℎ∈𝑍} for any 𝑥∈𝑋, then 𝑝=inf𝑍; that is, 𝐸^𝑋 is Dedekind complete.

Proof. The proof is similar to that of Lemma 2.

Lemma 4. Let 𝑋, 𝐸, 𝐸^𝑋, 𝑍, and 𝑝 be the same as in Lemma 2. Suppose that (1)for any ℎ∈𝑍, 𝑥∈𝑋 and 𝛼>0, there exists ℎ^′∈𝑍 such that ℎ(𝛼𝑥)=𝛼ℎ^′(𝑥); (2)𝑝(0)=0; (3)for any ℎ₁,ℎ₂∈𝑍 and 𝑥,𝑦∈𝑋, there exists ℎ∈𝑍 such that ℎ(𝑥+𝑦)≤ℎ₁(𝑥)+ℎ₂(𝑦).
Then 𝑝 is sublinear.

Proof. Let 𝑥∈𝑋 and 𝛼>0 be fixed. It is clear from (1) that {ℎ(𝛼𝑥)∣ℎ∈𝑍}⊂{𝛼ℎ^′(𝑥)∣ℎ^′∈𝑍}. Since 𝛼𝑥∈𝑋 and 1/𝛼>0, by (1), for any ℎ^′∈𝑍 there exists ℎ∈𝑍 such that 𝛼ℎ^′(𝑥)=𝛼ℎ^′1𝛼1𝛼𝑥=𝛼𝛼ℎ(𝛼𝑥)=ℎ(𝛼𝑥) (1) and hence {𝛼ℎ^′(𝑥)∣ℎ^′∈𝑍}⊂{ℎ(𝛼𝑥)∣ℎ∈𝑍}. Therefore, we conclude that {ℎ(𝛼𝑥)∣ℎ∈𝑍}={𝛼ℎ^′(𝑥)∣ℎ^′∈𝑍}. Thus we obtain that 𝑝(𝛼𝑥)=inf{ℎ(𝛼𝑥)∣ℎ∈𝑍}=inf𝛼ℎ^′(𝑥)∣ℎ^′ℎ∈𝑍=𝛼inf^′(𝑥)∣ℎ^′∈𝑍=𝛼𝑝(𝑥). (2) Moreover, (2) shows that 𝑝(0𝑥)=𝑝(0)=0=0𝑝(𝑥). Therefore, (S2) holds.
Let 𝑥,𝑦∈𝑋 be fixed. By (3), for any ℎ₁,ℎ₂∈𝑍, there exists ℎ∈𝑍 such that ℎ(𝑥+𝑦)≤ℎ₁(𝑥)+ℎ₂(𝑦). Thus we have 𝑝𝑥+𝑦≤ℎ₁(𝑥)+ℎ₂𝑦 (3) for any ℎ₁,ℎ₂∈𝑍. This shows that 𝑝(𝑥+𝑦)−ℎ₂(𝑦) is a lower bound of {ℎ(𝑥)∣ℎ∈𝑍} for any ℎ₂∈𝑍 and hence we have 𝑝𝑥+𝑦−ℎ₂𝑦≤𝑝(𝑥) (4) for any ℎ₂∈𝑍. This shows that 𝑝(𝑥+𝑦)−𝑝(𝑥) is a lower bound of {ℎ(𝑦)∣ℎ∈𝑍} and hence we have 𝑝(𝑥+𝑦)−𝑝(𝑥)≤𝑝(𝑦). Therefore, (S1) holds. This completes the proof.

Lemma 5. Let 𝑋, 𝐸, 𝐸^𝑋, 𝑍, and 𝑝 be the same as in Lemma 3. Suppose that (1)for any ℎ∈𝑍, 𝑥∈𝑋 and 𝛼>0, there exists ℎ^′∈𝑍 such that ℎ(𝛼𝑥)=𝛼ℎ^′(𝑥); (2)𝑝(0)=0; (3)for any ℎ₁,ℎ₂∈𝑍 and 𝑥,𝑦∈𝑋, there exists ℎ∈𝑍 such that ℎ(𝑥+𝑦)≤ℎ₁(𝑥)+ℎ₂(𝑦).
Then 𝑝 is sublinear.

Proof. The proof is similar to that of Lemma 4.

3. Main Results

To obtain our main results, we need the following.

Lemma 6. Let 𝑔 be a sublinear mapping from a vector space 𝑋 into a chain complete partially ordered vector space 𝐸 and 𝑦∈𝑋. Let 𝜙 be a mapping from 𝑋 into 𝐸 defined by 𝜙(𝑥)=inf𝑔𝑥+𝑡𝑦−𝑔𝑡𝑦∣𝑡≥0 (5) for any 𝑥∈𝑋. Then 𝜙 is sublinear and 𝑔^∗≤𝜙≤𝑔 on 𝑋, where 𝑔^∗ is a mapping from 𝑋 into 𝐸 defined by 𝑔^∗(𝑥)=−𝑔(−𝑥) for 𝑥∈𝑋.

Proof. For any 𝑥∈𝑋 and 𝑡≥0, put 𝜏_𝑡(𝑥)=𝑔(𝑥+𝑡𝑦)−𝑔(𝑡𝑦). Then 𝑍={𝜏_𝑡∣𝑡≥0} is a nonempty chain and bounded from below in 𝐸^𝑋. Indeed, since 𝑔=𝜏₀∈𝑍, 𝑍 is nonempty. If 𝑠≤𝑡, then 𝜏_𝑠(𝑥)−𝜏_𝑡(𝑥)−𝑔+𝑔+()𝑔𝑦()𝑦()𝑦=𝑔𝑥+𝑠𝑦−𝑔𝑠𝑦𝑥+𝑡𝑦−𝑔𝑡𝑦=𝑔𝑥+𝑠𝑦𝑡𝑦−𝑔𝑠𝑦−𝑔𝑥+𝑡𝑦=𝑔𝑥+𝑠𝑦𝑡−𝑠−𝑔𝑥+𝑡𝑦=𝑔𝑥+𝑠𝑦+𝑔𝑡−𝑠−𝑔𝑥+𝑡𝑦≥𝑔𝑥+𝑠𝑦+𝑡−𝑠−𝑔𝑥+𝑡𝑦=0 (6) for any 𝑥∈𝑋. Thus 𝑍 is a chain in 𝐸^𝑋. Since 𝜏_𝑡(𝑥)()()=𝑔𝑥+𝑡𝑦−𝑔𝑡𝑦≥𝑔𝑡𝑦−𝑔−𝑥−𝑔𝑡𝑦=−𝑔−𝑥=𝑔^∗(𝑥) (7) for any 𝑥∈𝑋 and 𝑡≥0, 𝑔^∗ is a lower bound of 𝑍. Hence 𝑍 is bounded from below in 𝐸^𝑋. Lemma 2 shows that 𝜙(𝑥)=inf𝑍 is well defined.
We next check (1), (2), and (3) in Lemma 4. Let 𝑡≥0, 𝑥∈𝑋, and 𝛼>0. We have 𝜏_𝑡()𝑔𝑡𝛼𝑥=𝑔𝛼𝑥+𝑡𝑦−𝑔𝑡𝑦=𝛼𝑥+𝛼𝑦𝑡−𝑔𝛼𝑦=𝛼𝜏_𝑡/𝛼(𝑥). (8) Clearly, 𝜏_𝑡/𝛼∈𝑍 and hence (1) in Lemma 4 holds. Since 𝜙(0)=inf{0∣𝑡≥0}=0, (2) in Lemma 4 holds. Let 𝑡₁,𝑡₂≥0 and 𝑥₁,𝑥₂∈𝑋. Since we have 𝜏_𝑡1+𝑡2𝑥₁+𝑥₂=𝑔𝑥₁+𝑥₂+𝑡₁+𝑡₂𝑦−𝑔𝑡₁+𝑡₂𝑦≤𝑔𝑥₁+𝑡₁𝑦+𝑔𝑥₂+𝑡₂𝑦−𝑡₁+𝑡₂𝑔𝑦=𝑔𝑥₁+𝑡₁𝑦−𝑔𝑡₁𝑦+𝑔𝑥₂+𝑡₂𝑦−𝑔𝑡₂𝑦=𝜏_𝑡1𝑥₁+𝜏_𝑡2𝑥₂, (9) (3) in Lemma 4 holds. Therefore, Lemma 4 implies that 𝜙 is sublinear.
Finally, it is clear that 𝜙≤𝑔. This inequality and (7) imply that 𝑔^∗≤𝜙≤𝑔 on 𝑋.

By Theorem 1 and Lemma 6, we obtain the following. For the case that 𝐸 is a Dedekind complete Riesz space, see [2].

Theorem 7. Let 𝑓 be a sublinear mapping from a vector space 𝑋 into a chain complete partially ordered vector space 𝐸. Then there exists a linear mapping 𝑔 from 𝑋 into 𝐸 such that 𝑔≤𝑓 on 𝑋.

Proof. Let 𝑌 be a subset of 𝐸^𝑋 defined by 𝑌=ℎ∈𝐸^𝑋∣ℎissublinear,𝑓^∗≤ℎ≤𝑓, (10) where 𝑓^∗ is defined by 𝑓^∗(𝑥)=−𝑓(−𝑥) for any 𝑥∈𝑋. Then it is clear that 𝑓∈𝑌 and hence 𝑌 is nonempty. Moreover 𝑌 is complete. In fact, let 𝑍⊂𝑌 be a nonempty chain. Since for any ℎ∈𝑍, 𝑓^∗≤ℎ, 𝑍 is bounded from below. It follows from Lemma 2 that there exists inf𝑍∈𝐸^𝑋. By Lemma 4, inf𝑍 is sublinear. Since 𝑓^∗≤ℎ≤𝑓 for any ℎ∈𝑍, we have 𝑓^∗≤inf𝑍≤𝑓. Thus inf𝑍∈𝑌 and hence 𝑌 is complete. Furthermore 𝑌 has a minimal. In fact, we suppose that 𝑌 does not have a minimal element. Then, for any ℎ∈𝑌, there exists ℎ∈𝑌 such that ℎ≤ℎ and ℎ≠ℎ. We define a mapping 𝑇 from 𝑌 into 𝑌 by 𝑇ℎ=ℎ. Since the mapping 𝑇 is decreasing, there exists ℎ₀∈𝑌 satisfying 𝑇ℎ₀=ℎ₀ by Theorem 1. This is a contradiction.
Let 𝑔 be a minimal in 𝑌. Let 𝑥∈𝑋. Let 𝜙 be a mapping from 𝑋 into 𝐸 defined by 𝜙(𝑧)=inf𝑔(𝑧+𝑡𝑥)−𝑔(𝑡𝑥)∣𝑡≥0 (11) for any 𝑧∈𝑋, then 𝜙 is sublinear and 𝑔^∗≤𝜙≤𝑔 on 𝑋 by Lemma 6. Moreover 𝜙∈𝑌. In fact, since 𝑔≤𝑓 and 𝑓^∗≤𝑔^∗, we have 𝑓^∗≤𝑔^∗≤𝜙≤𝑔≤𝑓 for any 𝑓∈𝑍. This shows that 𝜙∈𝑌. Since 𝑔 is minimal, 𝜙=𝑔. Then we have 𝑔()()𝑔()()()(𝑥)(0)(𝑥)(𝑥).−𝑥=𝜙−𝑥=inf−𝑥+𝑡𝑥−𝑔𝑡𝑥∣𝑡≥0≤𝑔−𝑥+𝑥−𝑔=𝑔−𝑔=−𝑔 (12) Since 𝑔 is sublinear and 0=𝑔(0)≤𝑔(𝑥+𝑧)+𝑔(−𝑥−𝑧), we have ()()()()(𝑥)(𝑧).−𝑔𝑥+𝑧≤𝑔−𝑥−𝑧≤𝑔−𝑥+𝑔−𝑧≤−𝑔−𝑔 (13) Thus 𝑔(𝑥)+𝑔(𝑧)≤𝑔(𝑥+𝑧). Since 𝑔 is sublinear, we also have 𝑔(𝑥+𝑧)≤𝑔(𝑥)+𝑔(𝑧) for any 𝑥,𝑧∈𝑋. Then we obtain that for any 𝑥,𝑧∈𝑋, 𝑔(𝑥+𝑧)=𝑔(𝑥)+𝑔(𝑧). Let 𝑥∈𝑋 and 𝛼>0. Since 0=𝑔(𝛼𝑥−𝛼𝑥)=𝛼𝑔(𝑥)+𝑔(−𝛼𝑥), (14) we have 𝑔(−𝛼𝑥)=−𝛼𝑔(𝑥). Then for any real number 𝛼, we have 𝑔(𝛼𝑥)=𝛼𝑔(𝑥). Thus 𝑔 is linear. Therefore, 𝑔 is a linear mapping from 𝑋 into 𝐸 such that 𝑔≤𝑓 on 𝑋.

Since Dedekind completeness implies chain completeness, we obtain the following.

Corollary 8. Let 𝑓 be a sublinear mapping from a vector space 𝑋 into a Dedekind complete partially ordered vector space 𝐸. Then there exists a linear mapping 𝑔 from 𝑋 into 𝐸 such that 𝑔≤𝑓 on 𝑋.

To give the Hahn-Banach Theorem in the case where the range space is a Dedekind complete partially ordered vector space, we need the following.

Lemma 9. Let 𝑝 be a sublinear mapping from a vector space 𝑋 into a Dedekind complete partially ordered vector space 𝐸, 𝐾 a nonempty convex subset of 𝑋, and 𝑞 a concave mapping from 𝐾 into 𝐸 such that 𝑞≤𝑝 on 𝐾. For any 𝑥∈𝑋, let 𝜙(𝑥)=inf𝑝𝑥+𝑡𝑦−𝑡𝑞𝑦𝑡≥0,𝑦∈𝐾. (15) Then 𝜙 is a sublinear mapping such that 𝜙≤𝑝 on 𝑋. Moreover, if 𝑔 is a linear mapping from 𝑋 into 𝐸, then 𝑔≤𝜙 on 𝑋 is equivalent to 𝑔≤𝑝 on 𝑋 and 𝑞≤𝑔 on 𝐾.

Proof. First, we show that 𝜙 is well defined and 𝜙(𝑥)≥−𝑝(−𝑥) for any 𝑥∈𝑋. Let 𝑍={𝜏_𝑡,𝑦∣𝑡≥0and𝑦∈𝐾}, where 𝜏_𝑡,𝑦(𝑥)=𝑝𝑥+𝑡𝑦−𝑡𝑞𝑦 (16) for any 𝑥∈𝑋 and 𝑡≥0. For any 𝜏_𝑡,𝑦∈𝑍 and 𝑥∈𝑋, 𝜏_𝑡,𝑦(𝑥)𝑦()𝑦(),=𝑝𝑥+𝑡𝑦−𝑡𝑞≥𝑝𝑡𝑦−𝑝−𝑥−𝑡𝑞≥−𝑝−𝑥 (17) and thus 𝜙(𝑥)≥−𝑝(−𝑥) and 𝑍 is bounded from below in 𝐸^𝑋. Since 𝐸 is Dedekind complete, 𝜙 is well defined by Lemma 3.
We next check (1), (2), and (3) in Lemma 5. (1)Let 𝜏_𝑡,𝑦∈𝑍. For any 𝑥∈𝑋 and 𝛼>0, we have 𝜏_𝑡,𝑦()𝑦𝑝𝑡𝛼𝑥=𝑝𝛼𝑥+𝑡𝑦−𝑡𝑞=𝛼𝑥+𝛼𝑦−𝑡𝛼𝑞𝑦=𝛼𝜏_{𝑡/𝛼,𝑦}(𝑥). (18)(2)By the definition of 𝜙, 𝜙(𝑥)≤𝑝(𝑥) for any 𝑥∈𝑋. Therefore 𝜙(0)≤𝑝(0)=0. Since 𝑝≥𝑞 on 𝐾, we have 𝜙(0)𝑝𝑦𝑦𝑦.=inf𝑡𝑦−𝑡𝑞𝑡≥0,𝑦∈𝐾=inf𝑡𝑝−𝑡𝑞𝑡≥0,𝑦∈𝐾≥0 (19)
Hence we have 𝜙(0)=0. (3)Let 𝜏_𝑡1,𝑦1,𝜏_𝑡2,𝑦2∈𝑍 satisfying 𝑡₁+𝑡₂≠0. Let 𝑥₁,𝑥₂∈𝑋. Since 𝐾 is convex and 𝑞 is concave, we have 𝜏_𝑡1,𝑦1𝑥₁+𝜏_𝑡2,𝑦2𝑥₂=𝑝𝑥₁+𝑡₁𝑦₁−𝑡₁𝑞𝑦₁+𝑝𝑥₂+𝑡₂𝑦₂−𝑡₂𝑞𝑦₂≥𝑝𝑥₁+𝑥₂+𝑡₁+𝑡₂𝑤−𝑡₁+𝑡₂𝑞(𝑤)=𝜏_{𝑡1+𝑡2,𝑤}𝑥₁+𝑥₂, (20) where 𝑤=(1/(𝑡₁+𝑡₂))(𝑡₁𝑦₁+𝑡₂𝑦₂)∈𝐾. Since 𝑝 is sublinear, we have 𝜏_0,𝑤𝑥₁+𝑥₂=𝑝𝑥₁+𝑥₂≤𝑝𝑥₁+𝑝𝑥₂=𝜏_0,𝑦1𝑥₁+𝜏_0,𝑦2𝑥₂. (21) Therefore, for any 𝑥₁,𝑥₂∈𝑋 and 𝑡₁,𝑡₂≥0, we have 𝜏_𝑡1,𝑦1(𝑥₁)+𝜏_𝑡2,𝑦2(𝑥₂)≥𝜏_{𝑡1+𝑡2,𝑤}(𝑥₁+𝑥₂).
Thus by Lemma 5, 𝜙 is sublinear. Moreover, by the definition of 𝜙, we have 𝜙≤𝑝 on 𝑋.
Let 𝑔 be a linear mapping from 𝑋 into 𝐸. Suppose that 𝑔≤𝜙 on 𝑋. Since 𝜙≤𝑝 on 𝑋, we have 𝑔≤𝑝 on 𝑋. Moreover, since for any 𝑦∈𝐾, 𝑦𝑦𝑦,−𝑔=𝑔−𝑦≤𝜙−𝑦≤𝑝−𝑦+𝑦−𝑞=−𝑞 (22) we have 𝑔≥𝑞 on 𝐾. To prove the converse, suppose that 𝑔≤𝑝 on 𝑋 and 𝑞≤𝑔 on 𝐾. For any 𝑥∈𝑋, 𝑦∈𝐾 and 𝑡≥0, we have 𝑔(𝑥)=𝑔𝑥+𝑡𝑦−𝑡𝑔𝑦≤𝑝𝑥+𝑡𝑦−𝑡𝑞𝑦. (23) This implies that 𝑔≤𝜙 on 𝑋.

By Corollary 8 and Lemma 9, we have the Hahn-Banach theorem in the case where the range space is a Dedekind complete partially ordered vector space. For the case that 𝐸 is a Dedekind complete Riesz space, see [2].

Theorem 10. Let 𝑝 be a sublinear mapping from a vector space 𝑋 into a Dedekind complete partially ordered vector space 𝐸, 𝑌 a subspace of 𝑋, and 𝑞 a linear mapping from 𝑌 into 𝐸 such that 𝑞≤𝑝 on 𝑌. Then there exists a linear mapping 𝑔 from 𝑋 into 𝑌 such that 𝑔=𝑞 on 𝑌 and 𝑔≤𝑝 on 𝑋.

Proof. Let 𝜙 be a mapping from 𝑋 into 𝐸 defined by 𝜙(𝑥)=inf𝑝𝑥+𝑡𝑦−𝑡𝑞𝑦𝑡≥0,𝑦∈𝐾 (24) for any 𝑥∈𝑋. By Lemma 9, 𝜙 is a sublinear mapping such that 𝜙≤𝑝 on 𝑋. By Corollary 8, there exists a linear mapping 𝑔 such that 𝑔≤𝜙 on 𝑋. Then putting 𝐾=𝑌 in Lemma 9, we have 𝑔≤𝑝 on 𝑋 and 𝑞≤𝑔 on 𝑌. Since 𝑌 is a subspace, for any 𝑦∈𝑌, we have −𝑦∈𝑌. Then 𝑞(−𝑦)≤𝑔(−𝑦). Since 𝑞 and 𝑔 are linear, we have −𝑞(𝑦)≤−𝑔(𝑦). Then 𝑔≤𝑞 on 𝑌. Thus 𝑔=𝑞 on 𝑌.

Moreover, by Corollary 8, we obtain the Mazur-Orlicz theorem in a Dedekind complete partially ordered vector space. For the case that 𝐸 is a Dedekind complete Riesz space, see [1, 15].

Theorem 11. Let 𝑝 be a sublinear mapping from a vector space 𝑋 into a Dedekind complete partially ordered vector space 𝐸. Let {𝑥_𝑗∣𝑗∈𝐽} be a family of elements of 𝑋 and {𝑦_𝑗∣𝑗∈𝐽} a family of elements of 𝐸. Then the following (1) and (2) are equivalent. (1)There exists a linear mapping 𝑔 from 𝑋 into 𝐸 such that 𝑔≤𝑝 on 𝑋 and 𝑦_𝑗≤𝑔(𝑥_𝑗) for any 𝑗∈𝐽. (2)For any natural number 𝑛, nonnegative real numbers 𝛼₁,𝛼₂,…,𝛼_𝑛≥0 and 𝑗₁,𝑗₂,…,𝑗_𝑛∈𝐽, one has ^𝑛_𝑖=1𝛼_𝑖𝑦_𝑗𝑖≤𝑝^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖. (25)

Proof. Let 𝛼₁,𝛼₂,…,𝛼_𝑛≥0 and 𝑗₁,𝑗₂,…,𝑗_𝑛∈𝐽 for a natural number 𝑛. By (1), we have ^𝑛_𝑖=1𝛼_𝑖𝑦_𝑗𝑖≤^𝑛_𝑖=1𝛼_𝑖𝑔𝑥_𝑗𝑖=𝑔^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖≤𝑝^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖. (26) Thus (2) is established.
Next by (2), for any 𝑥∈𝑋, we have ^𝑛_𝑖=1𝛼_𝑖𝑦_𝑗𝑖≤𝑝^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖=𝑝𝑥+^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖−𝑥≤𝑝𝑥+^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖(),()+𝑝−𝑥−𝑝−𝑥≤𝑝𝑥+^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖−^𝑛_𝑖=1𝛼_𝑖𝑦_𝑗𝑖 (27) for any natural number 𝑛, 𝛼₁,𝛼₂,…,𝛼_𝑛≥0 and 𝑗₁,𝑗₂,…,𝑗_𝑛∈𝐽. Put 𝑝₀(𝑥)=inf𝑝𝑥+^𝑛_𝑖=1𝛼_𝑖𝑥_𝑗𝑖−^𝑛_𝑖=1𝛼_𝑖𝑦_𝑗𝑖𝛼∣𝑛∈𝑁,_𝑖≥0,𝑗_𝑖∈𝐽,𝑖=1,2,…,𝑛 (28) for 𝑥∈𝑋, where 𝑁 is the set of all natural numbers. By Lemma 3, 𝑝₀ is well defined. Since 𝑝 is sublinear, 𝑝₀ is also sublinear. Thus by Corollary 8, there exists a linear mapping 𝑔 from 𝑋 into 𝐸 such that 𝑔≤𝑝₀ on 𝑋. Since 𝑝₀(−𝑥_𝑗)≤𝑝(−𝑥_𝑗+𝑥_𝑗)−𝑦_𝑗=−𝑦_𝑗, we have 𝑦_𝑗≤−𝑝₀−𝑥_𝑗≤−𝑔−𝑥_𝑗=𝑔𝑥_𝑗. (29)
Since 𝑝₀(𝑥)≤𝑝(𝑥) for any 𝑥∈𝑋, we have 𝑔(𝑥)≤𝑝(𝑥) for any 𝑥∈𝑋. Thus (1) is established.

Acknowledgment

The authors would like to express their hearty thanks to Professor Koji Aoyama for careful reading and many valuable suggestions.