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

Authors: Toshiharu Kawasaki; Masashi Toyoda; Toshikazu Watanabe
Journal: Journal of Applied Mathematics
 (2013)

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## 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.


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## Body

## 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.Letp 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].Letp 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).
LetE 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 setE 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.LetX 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 anyx,y∈X, p(x+y)≤p(x)+p(y).(S2)
For anyx∈X and nonnegative real number α, p(αx)=αp(x).LetEX be the set of mappings from X into E. Throughout this paper, EX is ordered as follows. For f,g∈EX, let f≤g mean that f(x)≤g(x) for any x∈X. It is easy to check that EX is also a partially ordered vector space.The following lemmas are useful for the proof of our main results.Lemma 2.
LetX be a vector space, E a chain complete partially ordered vector space, and Z a nonempty chain of EX which is bounded from below. Then there exists inf{h(x)∣h∈Z} for any x∈X. Moreover, if p∈EX is defined by p(x)=inf{h(x)∣h∈Z} for any x∈X, then p=infZ; that is, EX is chain complete.Proof.
Letx∈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}.
Definep∈EX 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.
LetX be a vector space, E a Dedekind complete partially ordered vector space, and Z a nonempty subset in EX which is bounded from below. Then there exists inf{h(x)∣h∈Z} for any x∈X. Moreover, if p∈EX is defined by p(x)=inf{h(x)∣h∈Z} for any x∈X, then p=infZ; that is, EX is Dedekind complete.Proof.
The proof is similar to that of Lemma2.Lemma 4.
LetX, E, EX, Z, and p be the same as in Lemma 2. Suppose that  (1)
for anyh∈Z, x∈X and α>0, there exists h′∈Z such that h(αx)=αh′(x);(2)
p(0)=0;(3)
for anyh1,h2∈Z and x,y∈X, there exists h∈Z such that h(x+y)≤h1(x)+h2(y).
Thenp is sublinear.Proof.
Letx∈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
(1)αh′(x)=αh′(1ααx)=α1αh(αx)=h(αx)
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
(2)p(αx)=inf{h(αx)∣h∈Z}=inf{αh′(x)∣h′∈Z}=αinf{h′(x)∣h′∈Z}=αp(x).
Moreover, (2) shows that p(0x)=p(0)=0=0p(x). Therefore, (S2) holds.
Letx,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
(3)p(x+y)≤h1(x)+h2(y)
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
(4)p(x+y)-h2(y)≤p(x)
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.
LetX, E, EX, Z, and p be the same as in Lemma 3. Suppose that  (1)
for anyh∈Z, x∈X and α>0, there exists h′∈Z such that h(αx)=αh′(x);(2)
p(0)=0;(3)
for anyh1,h2∈Z and x,y∈X, there exists h∈Z such that h(x+y)≤h1(x)+h2(y).
Thenp is sublinear.Proof.
The proof is similar to that of Lemma4.


## 3. Main Results

To obtain our main results, we need the following.Lemma 6.
Letg 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
(5)ϕ(x)=inf{g(x+ty)-g(ty)∣t≥0}
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 anyx∈X and t≥0, put τt(x)=g(x+ty)-g(ty). Then Z={τt∣t≥0} is a nonempty chain and bounded from below in EX. Indeed, since g=τ0∈Z, Z is nonempty. If s≤t, then
(6)τs(x)-τt(x)=g(x+sy)-g(sy)-(g(x+ty)-g(ty))=g(x+sy)+(g(ty)-g(sy))-g(x+ty)=g(x+sy)+(t-s)g(y)-g(x+ty)=g(x+sy)+g((t-s)y)-g(x+ty)≥g(x+sy+(t-s)y)-g(x+ty)=0
for any x∈X. Thus Z is a chain in EX. Since
(7)τt(x)=g(x+ty)-g(ty)≥g(ty)-g(-x)-g(ty)=-g(-x)=g*(x)
for any x∈X and t≥0, g* is a lower bound of Z. Hence Z is bounded from below in EX. Lemma 2 shows that ϕ(x)=infZ is well defined.
We next check (1), (2), and (3) in Lemma4. Let t≥0, x∈X, and α>0. We have
(8)τt(αx)=g(αx+ty)-g(ty)=α(g(x+tαy)-g(tαy))=ατt/α(x).
Clearly, τt/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
(9)τt1+t2(x1+x2)=g(x1+x2+(t1+t2)y)-g((t1+t2)y)≤g(x1+t1y)+g(x2+t2y)-(t1+t2)g(y)=g(x1+t1y)-g(t1y)+g(x2+t2y)-g(t2y)=τt1(x1)+τt2(x2),
(3) in Lemma 4 holds. Therefore, Lemma 4 implies that ϕ is sublinear.
Finally, it is clear thatϕ≤g. This inequality and (7) imply that g*≤ϕ≤g on X.By Theorem1 and Lemma 6, we obtain the following. For the case that E is a Dedekind complete Riesz space, see [2].Theorem 7.
Letf 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.
LetY be a subset of EX defined by
(10)Y={h∈EX∣hissublinear,f*≤h≤f},
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∈EX. 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.
Letg be a minimal in Y. Let x∈X. Let ϕ be a mapping from X into E defined by
(11)ϕ(z)=inf{g(z+tx)-g(tx)∣t≥0}
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
(12)g(-x)=ϕ(-x)=inf{g(-x+tx)-g(tx)∣t≥0}≤g(-x+x)-g(x)=g(0)-g(x)=-g(x).
Since g is sublinear and 0=g(0)≤g(x+z)+g(-x-z), we have
(13)-g(x+z)≤g(-x-z)≤g(-x)+g(-z)≤-g(x)-g(z).
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
(14)0=g(αx-αx)=αg(x)+g(-αx),
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.
Letf 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.
Letp 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
(15)ϕ(x)=inf{p(x+ty)-tq(y)t≥0,y∈K}.
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 thatϕ is well defined and ϕ(x)≥-p(-x) for any x∈X. Let Z={τt,y∣t≥0andy∈K}, where
(16)τt,y(x)=p(x+ty)-tq(y)
for any x∈X and t≥0. For any τt,y∈Z and x∈X,
(17)τt,y(x)=p(x+ty)-tq(y)≥p(ty)-p(-x)-tq(y)≥-p(-x),
and thus ϕ(x)≥-p(-x) and Z is bounded from below in EX. Since E is Dedekind complete, ϕ is well defined by Lemma 3.
We next check (1), (2), and (3) in Lemma5. (1)
Letτt,y∈Z. For any x∈X and α>0, we have
(18)τt,y(αx)=p(αx+ty)-tq(y)=α(p(x+tαy)-tαq(y))=ατt/α,y(x).(2)
By the definition ofϕ, ϕ(x)≤p(x) for any x∈X. Therefore ϕ(0)≤p(0)=0. Since p≥q on K, we have
(19)ϕ(0)=inf{p(ty)-tq(y)t≥0,y∈K}=inf{tp(y)-tq(y)t≥0,y∈K≥0}.
Hence we haveϕ(0)=0. (3)
Letτt1,y1,τt2,y2∈Z satisfying t1+t2≠0. Let x1,x2∈X. Since K is convex and q is concave, we have
(20)τt1,y1(x1)+τt2,y2(x2)=p(x1+t1y1)-t1q(y1)+p(x2+t2y2)-t2q(y2)≥p(x1+x2+(t1+t2)w)-(t1+t2)q(w)=τt1+t2,w(x1+x2), where w=(1/(t1+t2))(t1y1+t2y2)∈K. Since p is sublinear, we have
(21)τ0,w(x1+x2)=p(x1+x2)≤p(x1)+p(x2)=τ0,y1(x1)+τ0,y2(x2).
Therefore, for any x1,x2∈X and t1,t2≥0, we have τt1,y1(x1)+τt2,y2(x2)≥τt1+t2,w(x1+x2).
Thus by Lemma5, ϕ is sublinear. Moreover, by the definition of ϕ, we have ϕ≤p on X.
Letg 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,
(22)-g(y)=g(-y)≤ϕ(-y)≤p(-y+y)-q(y)=-q(y),
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
(23)g(x)=g(x+ty)-tg(y)≤p(x+ty)-tq(y).
This implies that g≤ϕ on X.By Corollary8 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.
Letp 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.
Letϕ be a mapping from X into E defined by
(24)ϕ(x)=inf{p(x+ty)-tq(y)t≥0,y∈K}
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 Corollary8, 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.
Letp be a sublinear mapping from a vector space X into a Dedekind complete partially ordered vector space E. Let {xj∣j∈J} be a family of elements of X and {yj∣j∈J} a family of elements of E. Then the following (1) and (2) are equivalent.  (1)
There exists a linear mappingg from X into E such that g≤p on X and yj≤g(xj) for any j∈J.(2)
For any natural numbern, nonnegative real numbers α1,α2,…,αn≥0 and j1,j2,…,jn∈J, one has
(25)∑i=1n‍αiyji≤p(∑i=1n‍αixji).Proof.
Letα1,α2,…,αn≥0 and j1,j2,…,jn∈J for a natural number n. By (1), we have
(26)∑i=1n‍αiyji≤∑i=1n‍αig(xji)=g(∑i=1n‍αixji)≤p(∑i=1n‍αixji).
Thus (2) is established.
Next by (2), for anyx∈X, we have
(27)∑i=1n‍αiyji≤p(∑i=1n‍αixji)=p(x+∑i=1n‍αixji-x)≤p(x+∑i=1n‍αixji)+p(-x),-p(-x)≤p(x+∑i=1n‍αixji)-∑i=1n‍αiyji
for any natural number n, α1,α2,…,αn≥0 and j1,j2,…,jn∈J. Put
(28)p0(x)=inf{p(x+∑i=1n‍αixji)-∑i=1n‍αiyji∣n∈N,mmmmmmmαi≥0,ji∈J,i=1,2,…,n(x+∑i=1n‍αixji)}
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(-xj)≤p(-xj+xj)-yj=-yj, we have
(29)yj≤-p0(-xj)≤-g(-xj)=g(xj).
Sincep0(x)≤p(x) for any x∈X, we have g(x)≤p(x) for any x∈X. Thus (1) is established.



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