SPACES AND NONSTANDARD NORMED SPACES
HSIEN-CHUNG WU
Communicated by the former editorial board
The conventional Hahn-Banach extension theorem based on vector space has been widely used to obtain many important and interesting results in nonlinear analysis, vector optimization and mathematical economics. In this paper, we consider the nonstandard vector space in which the concept of inverse element will not be assumed. In this case, the Hahn-Banach extension theorems based on the concepts of nonstandard vector space and nonstandard normed space can still be created.
AMS 2010 Subject Classification: 46A22.
Key words: nonstandard vector space, nonstandard normed space, nonstandard sublinear functional, null set, Hahn-Banach theorem.
1. INTRODUCTION
It is well-known that the topic of functional analysis is based on the vector space by referring to the monographs [1–5, 9]. However, there are some interesting sets that cannot form the real vector spaces. For example, the set of all closed intervals cannot form a real vector space, which can refer to Wu [6].
Also, the set of all fuzzy elements cannot form a real vector space, which can refer to Wu [7]. The main reason is that there are no additive inverse elements for the above-mentioned spaces. In this case, the conventional Hahn-Banach extension theorem cannot hold true in the space of all closed intervals and the space of all fuzzy elements. Fortunately, under the weaker definition of vector space, which is called the nonstandard vector space, Wu [6, 7] had presented the Hahn-Banach extension theorems for the nonstandard vector spaces consisting of all closed intervals and the nonstandard vector spaces consisting of all fuzzy elements.
Of course, there are other spaces that may not have the inverse elements.
For example, in set-valued analysis, the family of all closed subsets of a topo- logical vector space cannot have the inverse elements, which means that the conventional Hahn-Banach extension theorem cannot hold true in this space.
Inspired by this motivation, we are going to propose the general concept of
MATH. REPORTS16(66),1(2014), 33–59
nonstandard vector space which will be weaker than the conventional vector space in the sense of axioms, and to derive the Hahn-Banach extension theorem based on this general space. Fortunately, this expectation has been done in Wu [8].
Since the concept of inverse elements is not adopted in the nonstandard vector space, every thing will become complicated and non-intuitive. The main contribution of this paper is to establish the different versions of Hahn-Banach extension theorems based on the concepts of nonstandard vector space and nonstandard normed space, which can strengthen the results obtained in Wu [8]. For example, the sufficient conditions in the main Hahn-Banach extension thereom in Wu ([8], Theorem 4.1) are different from the sufficient conditions in Theorem 5.4 in this paper. On the other hand, although the different versions of Hahn-Banach extension theorems based on the interval space have been derived in Wu [6], the sufficient conditions are much neater than that of the sufficient conditions adopted in this paper, since the interval space has owned nice structure which can simplify the sufficient conditions. In other words, the Hahn-Banach extension theorems generalizes the results obtained in Wu [6].
As we shall see in the context of this paper, although the sufficient con- ditions for the Hahn-Banach extension theorems seem complicated, it is not hard to check that if the nonstandard vector space proposed in this paper turns into the conventional vector space, then the sufficient conditions will be satisfied automatically. In other words, the Hahn-Banach extension theorems established in this paper indeed generalizes the conventional Hahn-Banach ex- tension theorem.
In Sections 2 and 3, the concepts of nonstandard vector space and non- standard normed space are proposed. Also many interesting and useful results are also provided in order to derive the Hahn-Banach extension theorems. In Section 4, we also introduce the concept of nonstandard sublinear functional.
Of course, in Sections 5 and 6, we present many versions of Hahn-Banach extension theorems based on the concepts of nonstandard vector space and nonstandard normed space, respectively.
2. NONSTANDARD VECTOR SPACES
Let X be a universal set, and let Fdenote a scalar field, where Fcan be field of real numbers or complex numbers. We assume that the universal set X is endowed with the vector additionx⊕y and scalar multiplication αx for any x, y ∈ X and α ∈ F. We also assume that X is closed under the vector addition and scalar multiplication. In this case, we call X as a universal set overF. In the conventional vector space overF, the additive inverse element of
x is denoted by −x, and it can also be shown that−x=−1x. Here, we shall not consider the concept of inverse element. However, for convenience, we still adopt −x=−1x.
LetY be a subset ofX. We say thatY isclosed under the vector addition if y1, y2 ∈ Y implies y1 ⊕y2 ∈ Y. We say that Y is closed under scalar multiplication ifαy∈Y for any α∈F and y∈Y.
Forx, y∈X, thesubstractionx yis defined byx y=x⊕(−y), where
−y means the scalar multiplication (−1)y. For any x ∈ X and α ∈ F, since
−αxmeans−(αx), we have to mention that (−α)x6=−αxandα(−x)6=−αx in general, unless α(βx) = (αβ)x for any α, β ∈ F. Here, this law will not always be assumed to be true.
• We say that the commutative law for vector addition holds true in X if x⊕y =y⊕x for any x, y∈X.
• We say that the associative law for vector addition holds true in X if (x⊕y)⊕z=x⊕(y⊕z) for anyx, y, z ∈X.
Since the universal set X over F can just own the vector addition and scalar multiplication, in general, the universal set X cannot own the zero el- ement. The set Ω = {x x : x ∈ X} is called the null set of X. Therefore, the null set can be regarded as kind of zero elements of X. Now, we are in a position to define the concept of nonstandard vector space.
Definition 2.1.Let X be a universal set over F. We say that X is a nonstandard vector spaceoverF if the following conditions are satisfied:
(i) 1x=x for any x∈X;
(ii) x=y implies x⊕z=y⊕z and αx=αy for any x, y, z∈X and α∈F. (iii) the commutative and associative laws for vector addition hold true inX.
LetXbe a universal set overF. More laws about the vector addition and scalar multiplication can be defined below.
• We say that the distributive law for vector addition holds true in X if α(x⊕y) =αx⊕αy for any x, y∈X and α∈F.
• We say that thepositively distributive law for vector addition holds true inX ifα(x⊕y) =αx⊕αy for any x, y∈X and α >0.
• We say that theassociative law for scalar multiplicationholds true inX ifα(βx) = (αβ)x for any x∈X and α, β ∈F.
• We say that the associative law for positive scalar multiplication holds true in X ifα(βx) = (αβ)x for any x∈X and α, β >0.
• We say that the distributive law for scalar addition holds true in X if (α+β)x=αx⊕βx for any x∈X and α, β∈F.
• We say that thedistributive law for positive scalar additionholds true in X if (α+β)x=αx⊕βx for any x∈X and α, β >0.
• We say that the distributive law for negative scalar addition holds true inX if (α+β)x=αx⊕βx for any x∈X and α, β <0.
It is easy to see that if the distributive law for positive and negative scalar addition hold true inX, then (α+β)x=αx⊕βx for any x∈X and αβ >0.
It is not hard to see that the distributive law for scalar addition does not hold true in I (the set of all closed intervals). However, the distributive law for positive and negative scalar addition hold true inI.
Let X be a nonstandard vector space over F and Y be a subset of X.
We say that Y is asubspace ofX ifY is closed under the vector addition and scalar multiplication, i.e.,x⊕y ∈Y andαx∈Y for any x, y∈Y and α∈F.
For any x∈ X, since the distributive law (α+β)x =αx⊕βx does not hold true in general, it says thatα1x⊕ · · · ⊕αnx6= (α1+· · ·+αn)xin general.
Therefore, we need to carefully interpret the concept of linear combination.
Let X be a nonstandard vector space over F and let {x1,· · · , xn} be a finite subset ofX. Alinear combinationof{x1,· · ·, xn}is an expression of the form
y1⊕ · · · ⊕ym,
where yi = αixk for some k ∈ {1,· · ·, n} and the coefficients αi ∈ F for i = 1,· · ·, m. We allow yi = αixk and yj = αjxk for the same xk. In this case, we see that yi ⊕yj = αixk ⊕αjxk 6= (αi +αj)xk in general. For any nonempty subset S of X, the set of all linear combinations of finite subsets of S is called thespanof S, which is also denoted by span(S). Then we see that S ⊆span(S). Wu [8] has shown that span(S) is a subspace of X.
3.NONSTANDARD NORMED SPACES
Let X be a nonstandard vector space over F with the null set Ω. Many kinds of nonstandard normed spaces can be proposed below.
Definition 3.1.Let X be a nonstandard vector space over F. For the nonnegative real-valued function k · k: X → R, we consider the following conditions:
(i) kαxk=|α| kxk for any x∈X and α∈F;
(i’) kαxk=|α| kxk for any x∈X and α∈Fwith α6= 0.
(ii) kx⊕yk≤kxk+kykfor any x, y∈X.
(iii) kxk= 0 impliesx∈Ω.
Different kinds of nonstandard normed spaces are defined below.
• We say that (X,k · k) is a nonstandard pseudo-seminormed spaceif con- ditions (i’) and (ii) are satisfied.
• We say that (X,k · k) is a nonstandard seminormed space if conditions (i) and (ii) are satisfied.
• We say that (X,k · k) is anonstandard pseudo-normed spaceif conditions (i’), (ii) and (iii) are satisfied.
• We say that (X,k · k) is a nonstandard normed space if conditions (i), (ii) and (iii) are satisfied.
We say that the norm k · k satisfies the null condition if condition (iii) is replaced by k x k= 0 if and only if x ∈ Ω. Now we consider the following conditions:
(iv) kx⊕ωk≥kxk for any x∈X and ω∈Ω.
(iv’) kx⊕ωk=kxk for any x∈X and ω∈Ω.
We say that the normk · ksatisfies thenull inequality(resp. null equality) if condition (iv) (resp. (iv’)) is satisfied.
Remark 3.1.We have the following observations.
(i) If the normk · k satisfies the null inequality and null condition, then the norm k · k also satisfies the null equality, since k x⊕ω k≤k x k + k ωk=kxk for anyω ∈Ω.
(ii) If the norm k · k satisfies the null condition , then Ω is closed under the scalar multiplication, since k αω k= |α|· k ω k= 0 by condition (i) in Definition 3.1.
Proposition 3.1 (Wu [8]). The following statements hold true.
(i) Let X be a nonstandard normed space such that the norm k · k satisfies the null inequality. Assume that there existsx¯∈X such thatkω⊕0¯xk=
0 for anyω ∈Ω. Then kxk= 0if and only if x∈Ω, i.e., the norm k · k satisfies the null condition.
(ii) Let X be a nonstandard normed space such that the norm k · k satisfies the null equality. Then the norm k · k satisfies the null condition.
(iii) Let (X,k · k) be a nonstandard pseudo-normed space satisfies the null inequality and null condition. Then it also satisfies the null equality.
(iv) Let (X,k · k) be a nonstandard normed space such that the norm k · k satisfies the null inequality. Then the normk · ksatisfies the null equality if and only if the norm k · k satisfies the null condition.
Definition 3.2. LetX andY be nonstandard vector spaces over the same scalar field F and letT be a function from X intoY.
(i) We say thatT is apseudo-linear operatorif
T(x1⊕x2) =T(x1)⊕T(x2) and T(αx) =αT(x) forx1, x2, x∈X and α∈Fwithα6= 0.
(ii) We say that T is a linear operator if T is a pseudo-linear operator and T(0x) = 0T(x) for anyx∈X.
(iii) If Y is taken as the scalar field F, then the pseudo-linear operator and linear operator are called pseudo-linear functionaland linear functional, respetively.
Proposition 3.2 (Wu [8]). Let X and Y be two nonstandard vector spaces over the same scalar field F and let T be a pseudo-linear operator from X into Y. The following statements hold true.
(i) We haveT(ω)∈ΩY for anyω ∈ΩX, whereΩX andΩY are the null sets of X and Y, respectively.
(ii) If f is a pseudo-linear functional, then f(ω) = 0 for any ω∈ΩX. Definition 3.3.Let (X,k · kX) and (Y,k · kY) be nonstandard pseudo- seminormed spaces. The pseudo-linear operator T : X → Y is said to be bounded if there exists nonnegative real numberc such that
(1) kT(x)kY≤ckxkX
for all x ∈ X. We assume further that the norm k · kX satisfies the null equality. Suppose that X\ΩX 6= ∅. Inspired by part (i) of Proposition 3.1, thenorm of operatorT is denoted bykT kand is defined by
(2) kT k= sup
x∈X\ΩX
kT(x)kY kxkX .
Throughout this paper, when we consider the nonstandard normed space (X,k · kX), we implicitly assume thatX\ΩX 6=∅.
Proposition3.3 (Wu [8]). Let(X,k · kX)and(Y,k · kY)be nonstandard normed spaces such that the norms k · kX and k · kY satisfy the null equality.
Suppose that T :X → Y is a bounded pseudo-linear operator. Then we have kT(x)kY≤kT k · kxkX.
Proposition 3.4 (Wu [8]). Let (X,k · kX) be a nonstandard normed space such that the norm k · ksatisfies the null equality, and let(Y,k · kY)be a nonstandard pseudo-seminormed space. Suppose that T :X→Y is a bounded pseudo-linear operator. The norm of T can be rewritten as
(3) kT k= sup
x∈X\ΩX
kxkX= 1
kT(x)kY= sup
x∈X\ΩX
kxkX≤1
kT(x)kY .
4.NONSTANDARD SUBLINEAR FUNCTIONALS
The Hahn-Banach extension theorem concerns the extension of linear functional dominated by a sublinear functional. Since the concept of non- standard vector space is considered here, we shall propose many concepts of sublinear functionals on the nonstandard vector space over F.
Definition 4.1.Let pbe a real-valued function defined on a nonstandard vector space X overF.
(i) We say thatp ispseudo-sublinear if, for allx, y∈X andλ >0,p(λx) = λp(x) andp(x⊕y)≤p(x) +p(y).
(ii) We say that p is sublinear if, for all x, y ∈X and λ≥0, p(λx) =λp(x) and p(x⊕y)≤p(x) +p(y).
(iii) We say thatp is nonstandard sublinearif p is pseudo-sublinear and sat- isfiesp(ω) = 0 for allω ∈Ω.
We say thatpsatisfies thenull inequalityifp(x⊕ω)≥p(x) for anyx∈X and ω∈Ω.
Remark 4.1.We have the following observations.
(i) Letpbe a sublinear or pseudo-sublinear functional defined on a nonstan- dard vector spaceX overF satisfying ω⊕ω= 2ω for any ω∈Ω.
• If p(ω⊕ω) = p(ω) for any ω ∈ Ω, then we have 2p(ω) = p(2ω) = p(ω⊕ω) =p(ω), which shows thatp(ω) = 0, i.e.,pis a nonstandard sublinear.
• Ifp(ω⊕ω)≤p(ω) for any ω∈Ω, then we have p(ω)≤0.
• Ifp(ω⊕ω)≥p(ω) for any ω∈Ω, then we have p(ω)≥0.
(ii) Let p be a pseudo-sublinear functional defined on a nonstandard vector space X over Fand satisfy the null inequality. Then, for any ω ∈Ω, we have
p(ω)≤p(ω⊕ω)≤p(ω) +p(ω), which impliesp(ω)≥0.
(iii) Let p be a nonstandard sublinear functional defined on a nonstandard vector space X over Fand satisfy the null inequality. Then we see that p(x⊕ω) =p(x) for anyx∈X andω∈Ω, sincep(x⊕ω)≤p(x) +p(ω) = p(x).
5. HAHN-BANACH EXTENSION THEOREMS IN NONSTANDARD VECTOR SPACES
Let S be a nonempty set which is partially ordered by a reflexive and transitive binary relation “” and letT be a nonempty subset ofS.
• An elementx∗∈S is called anupper boundof T if and only ifxx∗ for all x∈T.
• An element x∗ ∈S is called a lower bound ofT if and only ifx∗ x for all x∈T.
• An elementx∗∈S is called a maximal elementof S if and only if x∈S and x∗ x, thenx x∗. We remark that if the binary relation is also antisymmetric, then an elementx∗∈S is a maximal element ofS if and only if x∈S and x∗ x, thenx=x∗.
• An elementx∗ ∈S is called aminimal element of S if and only if x∈S and x x∗, thenx∗ x. We remark that if the binary relation is also antisymmetric, then an element x∗ ∈S is a minimal element of S if and only if x∈S and xx∗, thenx=x∗.
• The set S is called inductively ordered from above (resp. from below) if and only if each totally ordered subset of S has an upper (resp. lower) bound.
Lemma 5.1 (Zorn’s Lemma). Let S be a nonempty set which is partially ordered by a reflexive and transitive binary relation. If S is inductively ordered from above(resp. from below), thenS has at least one maximal(resp. minimal) element.
Definition 5.1.LetX be a nonstandard vector space overFand let pbe a real-valued function defined on X. We say thatp satisfies theHahn-Banach conditions if the following conditions are satisfied:
(i) p((−α)x) =p(−(αx)) for anyx∈X andα <0;
(ii) p((−α)x⊕z) =p(−(αx)⊕z) for any x, z∈X and α <0;
(iii) p(z⊕(α+β)x)≤p(z⊕αx⊕βx) for any x, z∈X and α, β >0;
(iv) psatisfies the positively distributive law for vector addition, i.e.,p(α(x⊕ y)) =p(αx⊕αy) andp(z⊕α(x⊕y)) =p(z⊕αx⊕αy) for anyx, y, z ∈X and α >0.
In the sequel, we are going to derive many versions of Hahn-Banach the- orems.
Theorem 5.1 (Basic Version of Hahn-Banach Theorem). Let X be a nonstandard vector space over R. For each nonstandard sublinear functional p on X satisfying the null inequality and Hahn-Banach conditions, there ex- ists a pseudo-linear functional f on X such that f satisfies the Hahn-Banach conditions and f(x)≤p(x) for all x∈X.
Proof. We consider the set F of all nonstandard sublinear functionals satisfying the null inequality such that each h inF satisfies the Hahn-Banach
conditions andh(x)≤p(x) for allx∈X. Then we see thatF 6=∅, sincep∈ F. We are going to claim thatF has at least one minimal element with respect to the partial ordering hi hj that is defined to be the pointwise ordering, i.e., hi(x) ≤hj(x) for all x∈X, where hi, hj ∈ F. Suppose that {hi} is a totally ordered subset ofF. We define the functionalh onX by
h(x) = inf
i hi(x).
We want to show that h ∈ F. We firstly need to claim h(x) > −∞
for all x ∈ X. Suppose that there exists x0 ∈ X such that h(x0) = −∞.
Given any real number k, there exists j such that hj(x0) < k, which implies hj(x0) +hj(y)< k+hj(y). Therefore, by the subadditivity of hi, we have
h(x0⊕y) = inf
i hi(x0⊕y)≤inf
i [hi(x0) +hi(y)]
≤hj(x0) +hj(y)< k+hj(y).
Sincekcan be sufficiently small real number, we conclude thath(x0⊕y) =
−∞ for any y∈X. Since x0 x0 =ω ∈Ω andhi∈ F, we have h(z) = inf
i hi(z)≤inf
i hi(z⊕ω) =h(z⊕ω) (4)
=h(z⊕x0 x0) =h(x0⊕(z x0)) =−∞,
which shows h(z) = −∞ for any z ∈ X, i.e., h(ω) = −∞ for any ω ∈ Ω.
Since each hi is also nonstandard sublinear, we have hi(ω) = 0, i.e., h(ω) = infihi(ω) = 0 for any ω ∈Ω. A contradiction occurs. Therefore, we conclude that h(x)>−∞for all x∈X.
It is obvious that h(x)≤p(x), h(λx) =λh(x),h(ω) = 0 andh(x⊕ω)≥ h(x) for anyx∈X,ω∈Ω andλ >0. Since eachhi satisfies the Hahn-Banach conditions, it is also obvious thathsatisfies the Hahn-Banach conditions. Now, we want to show the subadditivity of h. Since h(x) >−∞ and h(y) > −∞, given any >0, there existsj, k such that
infi hi(x) + > hj(x) and inf
i hi(y) + > hk(y), which implies
infi hi(x) + inf
i hi(y) + 2 > hj(x) +hk(y)
≥
hj(x) +hj(y) ifhj hk hk(x) +hk(y) ifhkhj
≥inf
i [hi(x) +hi(y)]≥inf
i hi(x⊕y)
by the subadditivity of hi. Since can be any positive number, we obtain h(x⊕y) ≤ h(x) +h(y), which shows the subadditivity of h. Therefore, we
have h ∈ F and h is also a lower bound of {hi}. This shows that the set F is inductively ordered from below with respect to the pointwise ordering.
By Zorn’s lemma, the set F has at least one minimal element f in F, i.e., f(x)≤p(x) for allx∈X. Next we are going to show thatf is a pseudo-linear functional on X.
For any fixed y∈X, we define the functionalg on X by g(x) = inf
λ>0[f(x⊕λy)−λf(y)]
for all x ∈ X. First of all, we need to claim that the real-valued function g is well-defined, i.e., the infimum exists for each x ∈ X. For each x ∈ X and λ >0, sincex x=ω ∈Ω andf ∈ F, we have
λf(y) =f(λy)≤f(λy⊕ω) =f(λy⊕x x)≤f(x⊕λy) +f(−x).
This says that −∞<−f(−x) ≤f(x⊕λy)−λf(y). By taking infimum forλ >0, we obtain−∞<−f(−x)≤g(x) for allx∈X. This shows that the real-valued functiong is well-defined.
Next, we want to claim that gis a nonstandard sublinear functional sat- isfying the null inequality. For all x ∈ X and all µ > 0, since f satisfies the positively distributive law for vector addition, we have
µg(x) = inf
λ>0[µf(x⊕λy)−µλf(y)] = inf
λ>0[f(µ(x⊕λy))−µλf(y)]
= inf
λ>0[f(µx⊕µλy)−µλf(y)] = inf
ˆλ>0
h
f(µx⊕λy)ˆ −λf(y)ˆ i
=g(µx).
From Remark 4.1, we also have g(ω) = inf
λ>0[f(ω⊕λy)−λf(y)] = inf
λ>0[f(λy)−λf(y)]
= inf
λ>0λ[f(y)−f(y)] = 0.
Now, we have g(x⊕z) = inf
λ>0[f(x⊕z⊕λy)−λf(y)]
≤f(x⊕z⊕(λ+µ)y)−(λ+µ)f(y)
≤f(x⊕λy⊕z⊕µy)−λf(y)−µf(y)
(by referring to the Hahn-Banach conditions)
≤(f(x⊕λy)−λf(y)) + (f(z⊕µy)−µf(y)) (by the sublinearity of f).
By taking infimum on both sides, we have g(x⊕z)≤ inf
λ>0inf
µ>0[(f(x⊕λy)−λf(y)) + (f(z⊕µy)−µf(y))]
= inf
λ>0(f(x⊕λy)−λf(y)) + inf
µ>0(f(z⊕µy)−µf(y)), which concludes g(x⊕z)≤g(x) +g(z). Since f ∈ F, we have
g(x⊕ω) = inf
λ>0[f(x⊕ω⊕λy)−λf(y)]≥ inf
λ>0[f(x⊕λy)−λf(y)] =g(x).
This shows that g is a nonstandard sublinear functional satisfying the null inequality.
Next, we want to show that g satiesfies the Hahn-Banach conditions.
Since f satisfies the Hahn-Banach conditions, we also have g((−α)x) = inf
λ>0[f((−α)x⊕λy)−λf(y)]
= inf
λ>0[f(−(αx)⊕λy)−λf(y)] =g(−(αx)) (whereα <0) g((−α)x⊕z) = inf
λ>0[f((−α)x⊕z⊕λy)−λf(y)]
= inf
λ>0[f(−(αx)⊕z⊕λy)−λf(y)] =g(−(αx)⊕z) (where α <0) g(z⊕(α+β)x) = inf
λ>0[f(z⊕(α+β)x⊕λy)−λf(y)]
≤ inf
λ>0[f(z⊕αx⊕βx⊕λy)−λf(y)] =g(z⊕αx⊕βx) (whereα, β >0) g(α(x⊕y)) = inf
λ>0[f(α(x⊕y)⊕λy)−λf(y)]
= inf
λ>0[f(αx⊕αy⊕λy)−λf(y)] =g(αx⊕αy) (where α >0) g(z⊕α(x⊕y)) = inf
λ>0[f(z⊕α(x⊕y)⊕λy)−λf(y)]
= inf
λ>0[f(z⊕αx⊕αy⊕λy)−λf(y)] =g(z⊕αx⊕αy) (whereα >0).
This shows that g satisfies the Hahn-Banach conditions. Since (5) f(x⊕λy)−λf(y)≤f(x) +λf(y)−λf(y) =f(x)≤p(x),
i.e.,g(x)≤p(x) for allx∈X (by taking infimum on both sides), we conclude that g∈ F.
From (5), we also have g(x) ≤f(x) for all x∈X by taking infimum on both sides. Sincef is the minimal element ofF andg∈ F, we havef(x)≤g(x) for all x∈X. Therefore, we obtain g=f. Furthermore, we have
f(x) =g(x) = inf
λ>0[f(x⊕λy)−λf(y)]≤f(x⊕y)−f(y),
which implies f(x) +f(y)≤f(x⊕y). Sincef is also subadditive, we conclude that f is additive. Since g =f, i.e., f(λx) = λf(x) and f(ω) = 0 for λ > 0
and ω∈Ω. It remains to show thatf(λx) =λf(x) for λ <0. For anyx∈X, we let ω=x x∈Ω. Then, by the additivity of f, we have
0 =f(ω) =f(x x) =f(x⊕(−x)) =f(x) +f(−x),
which says that f(−x) = −f(x). Therefore, for λ < 0, since f satisfies the Hahn-Banach conditions, we have
−λf(x) =f((−λ)x) =f(−(λx)) =−f(λx),
which impliesf(λx) =λf(x). This shows thatfis a pseudo-linear functional.
Corollary 5.1 (Basic Version of Hahn-Banach Theorem). Let X be a nonstandard vector space over Rsuch thatα(x⊕y) =αx⊕αyand(α+β)x= αx⊕βx for any x, y ∈X and α, β > 0, and (−λ)x =−(λx) for any λ < 0.
For each nonstandard sublinear functionalponX satisfying the null inequality, there exists a pseudo-linear functional f on X such that f satisfies the Hahn- Banach conditions and f(x)≤p(x) for all x∈X.
Proof. Under the assumptions, the Hahn-Banach conditions will be staif- sied automatically. Therefore, the results follows from Theorem 5.1 immedi- ately.
It is not hard to see that if X is taken as the setI of all closed intervals, then we automatically have α(x⊕y) =αx⊕αy and (α+β)x=αx⊕βx for any x, y ∈ X and α, β > 0, and (−λ)x = −(λx) for any λ < 0. Therefore, Corollary 5.1 is applicable for the space I.
Definition 5.2. Let X be a nonstandard vector space over R and let S be a subset of X. We say that S is nonstandard convex if, for any a, b ∈ S, λa⊕(1−λ)b∈Sforλ∈(0,1). Letf :S→Rbe a real-valued function defined on a nonstandard convex subset S. We say that f is nonstandard concave if f(λx⊕(1−λ)y)≥λf(x) + (1−λ)f(y) for any x, y∈X and λ∈(0,1).
For the nonstandard convexity, it is not necessary to consider the cases of λ= 0 and λ= 1.
Definition 5.3.LetX be a nonstandard vector space overFand let pbe a real-valued function defined onX.
(i) We say thatp satisfies the positively distributive law for vector addition if p(α(x⊕y)) = p(αx⊕αy) and p(z⊕α(x⊕y)) = p(z⊕αx⊕αy) for any x, y, z∈X and α >0.
(ii) We say that psatisfies the associative law for positive scalar multiplica- tion if p(α(βx)) = p((αβ)x) and p(z⊕α(βx)) = p(z⊕(αβ)x) for any x, z ∈X and α, β >0.
(iii) We say thatp satisfies the positively sub-distributive law for vector ad- dition ifp(α(x⊕y))≤p(αx⊕αy) andp(z⊕α(x⊕y))≤p(z⊕αx⊕αy) for any x, y, z∈X and α >0.
(iv) We say thatp satisfies the sub-associative law for positive scalar multi- plication if p(α(βx)) ≤ p((αβ)x) and p(z⊕α(βx)) ≤ p(z⊕(αβ)x) for any x, z∈X and α, β >0.
Remark 5.1.Let X be a nonstandard vector space overF and let p be a real-valued function defined on X. We have the following observations.
• Ifp is a pseudo-sublinear functional, then, for anyα, β >0, we have p(α(βx)) =α·p(βx) =αβ·p(x) =p((αβ)x).
• If the positively distributive law for vector addition and associative law for positive scalar multiplication hold true in X, then the laws in Defini- tion 5.3 are automatically satisfied. We also remark that the positively distributive law for vector addition and associative law for positive scalar multiplication hold true automatically in the space I of all closed inter- vals.
Lemma 5.2. Let S be a nonempty subset of a nonstandard vector space X over R. Let p be a pseudo-sublinear functional defined on X satisfying the null inequality, and let h be a given real-valued function defined on S with h(x)≤p(x) for allx∈S. Letf be a real-valued function defined on X by
f(x) = inf
y∈S,λ>0[p(x⊕λy)−λh(y)]. Then the following statements hold true.
(i) The real-valued function f satisfies the inequality f(x) ≤ p(x) for all x∈X.
(ii) If p(ω)≤0 for any ω ∈Ω, then, for a pseudo-linear functiona l defined onX, we have l(x)≤p(x) for all x∈X andh(x)≤l(x) for all x∈S if and only if l(x)≤f(x) for all x∈X.
(iii) Suppose that p satisfies the positively distributive law for vector addition and sub-associative law for positive scalar multiplication. If the set S is nonstandard convex and h is nonstandard concave, then f is a pseudo- sublinear functional satisfying the null inequality. If we assume further that p is a nonstandard sublinear functional satisfying the null inequal- ity, then f is also a nonstandard sublinear functional satisfying the null inequality.
Proof. First of all, we need to claim that the real-valued function f is well-defined, i.e., the infimum exists for each x ∈X. For each x ∈ X,y ∈S
and λ >0, we havex x=ω∈Ω and
λh(y)≤λp(y) =p(λy)≤p(λy⊕ω)
=p(λy⊕x x)≤p(x⊕λy) +p(−x).
This says that −∞ < −p(−x) ≤ p(x⊕λy)−λh(y), i.e., the infimum exists. In other words, the real-valued function f is well-defined.
To prove part (i), by the sublinearity of p, we have
p(x⊕λy)−λh(y)≤p(x) +λp(y)−λh(y) =p(x) +λ(p(y)−h(y)) for all x∈X,y∈S and λ >0. By taking infimum on both sides, we have
y∈S,λ>0inf [p(x⊕λy)−λh(y)]≤p(x) + inf
y∈S,λ>0λ(p(y)−h(y)) =p(x), since p(y)−h(y) ≥ 0 for all y ∈S and λ > 0, which shows thatf(x) ≤p(x) for all x∈X.
To prove part (ii), let l be a pseudo-linear functional defined on X such thatl(x)≤p(x) for all x∈X and h(x)≤l(x) for allx∈S. Forx∈X,y∈S and λ >0, we let ω =λy λy∈Ω. Then, by Proposition 3.2, we have
l(x) =l(x)+l(ω) =l(x⊕ω) =l(x⊕λy λy) =l(x⊕λy)−λl(y)≤p(x⊕λy)−λh(y).
By taking infimum on both sides, we obtain l(x) ≤ f(x) for all x ∈ X.
Conversely, using (i), we immediately have l(x)≤p(x) for all x∈X. On the other hand, for someλ0 >0 and for allx∈S, we have
−λ0l(x) =l(−λ0x)≤f(−λ0x) = inf
y∈S,λ>0[p(−λ0x⊕λy)−λh(y)]
≤p(−λ0x⊕λ0x)−λ0h(x) =p(ω)−λ0h(x)≤ −λ0h(x), sincep(ω)≤0, which says thath(x)≤l(x).
To prove part (iii), for all x ∈ X and all µ > 0, since p satisfies the positively distributive law for vector addition, we have
µf(x) = inf
y∈S,λ>0[µp(x⊕λy)−µλh(y)] = inf
y∈S,λ>0[p(µ(x⊕λy))−µλh(y)]
= inf
y∈S,λ>0[p(µx⊕µλy)−µλh(y)]
= inf
y∈S,λ>0ˆ
h
p(µx⊕λy)ˆ −ˆλh(y) i
=f(µx).
Now, we want to show that f is subadditive. For any u, v ∈ S and λ, µ >0, sinceS is nonstandard convex, we have
w= λ
λ+µu⊕ µ
λ+µv∈S.
Since p satisfies the positively distributive law for vector addition and sub-associative law for positive scalar multiplication, we obtain
f(x⊕y) = inf
z∈S,η>0[p((x⊕y)⊕ηz)−ηh(z)]
≤p(x⊕y⊕(λ+µ)w)−(λ+µ)h(w) (by taking z=wand η=λ+µ)
≤p(x⊕λu⊕y⊕µv)−λh(u)−µh(v) (sinceh is nonstandard concave)
≤(p(x⊕λu)−λh(u)) + (p(y⊕µv)−µh(v)). By taking infimum on both sides, we have
f(x⊕y)≤ inf
u∈S,λ>0 inf
v∈S,µ>0[(p(x⊕λu)−λh(u)) + (p(y⊕µv)−µh(v))]
= inf
u∈S,λ>0
(p(x⊕λu)−λh(u)) + inf
v∈S,µ>0(p(y⊕µv)−µh(v))
= inf
u∈S,λ>0(p(x⊕λu)−λh(u)) + inf
v∈S,µ>0(p(y⊕µv)−µh(v)), which concludes f(x⊕y)≤f(x) +f(y). Finally, we have
f(x⊕ω) = inf
y∈S,λ>0[p(x⊕ω⊕λy)−λh(y)]
≥ inf
y∈S,λ>0[p(x⊕λy)−λh(y)] =f(x).
Now, we assume that p is a nonstandard sublinear functional satisfying the null inequality. From Remark 4.1, we also have
f(ω) = inf
y∈S,λ>0[p(ω⊕λy)−λh(y)] = inf
y∈S,λ>0[p(λy)−λh(y)]
= inf
y∈S,λ>0λ[p(y)−h(y)] = 0,
since p(y)−h(y) ≥0 for all y ∈S. This shows that f is also a nonstandard sublinear functional satisfying the null inequality. We complete the proof.
Theorem 5.2 (Sandwich Version of Hahn-Banach Theorem). Let X be a nonstandard vector space over R. Let p be a nonstandard sublinear func- tional on X satisfying the null inequality, Hahn-Banach conditions and sub- associative law for positive scalar multiplication. If S is a nonstandard convex subset of X andh is a nonstandard concave functional onS with h(x)≤p(x) for all x ∈ S, then there exists a pseudo-linear functional f on X such that f satisfies the Hahn-Banach conditions with f(x) ≤ p(x) for all x ∈ X and h(x)≤f(x) for all x∈S.
Proof. Letq be a functional defined on X by
(6) q(x) = inf
y∈S,λ>0[p(x⊕λy)−λh(y)].
Then Proposition 5.2 (iii) shows that q is a nonstandard sublinear func- tional satisfying the null inequality. Since p satisfies the Hahn-Banach condi- tions, it is not hard to see that q also satisfies the Hahn-Banach conditions.
Now applying Theorem 5.1, there exists a pseudo-linear functionalf onXsuch that f satisfies the Hahn-Banach conditions and f(x) ≤ q(x) for all x ∈ X.
Finally, sincep(ω) = 0 for anyω ∈Ω, using Proposition 5.2 (ii), we obtain the desired sandwich result. We complete the proof.
Corollary 5.2 (Sandwich Version of Hahn-Banach Theorem). Let X be a nonstandard vector space over R such that α(x ⊕y) = αx ⊕αy and (α+β)x = αx⊕βx for any x, y ∈ X and α, β > 0, and (−λ)x = −(λx) for any λ < 0. Let p be a nonstandard sublinear functional on X satisfying the null inequality. If S is a nonstandard convex subset of X and h is a nonstandard concave functional on S with h(x) ≤ p(x) for all x ∈ S, then there exists a pseudo-linear functional f on X such that f satisfies the Hahn- Banach conditions with f(x) ≤ p(x) for all x ∈ X and h(x) ≤ f(x) for all x∈S.
Proof. Under the assumptions, the Hahn-Banach conditions will be satisfied automatically. Therefore, the result follows from Theorem 5.2 im- mediately.
Theorem 5.3 (Convex Version of Hahn-Banach Theorem). Let X be a nonstandard vector space over R. Let p be a nonstandard sublinear func- tional on X satisfying the null inequality, Hahn-Banach conditions and sub- associative law for positive scalar multiplication. If S is a nonstandard convex subset ofX and letk= infx∈Sp(x), then there exists a pseudo-linear functional f onX such that f satisfies the Hahn-Banach conditions withf(x)≤p(x) for all x∈X and
x∈Sinf f(x) = inf
x∈Sp(x) =k.
Proof. Suppose that k = −∞. Theorem 5.1 says that there exists a pseudo-linear functional f on X such thatf satisfies the Hahn-Banach condi- tions and f(x)≤p(x) for allx∈X. This shows that
x∈Sinf f(x)≤ inf
x∈Sp(x) =−∞; that is, inf
x∈Sf(x) = inf
x∈Sp(x) =−∞.
Now, we assume that k >−∞. Then we define the functionalh on S by h(x) =k for allx∈S. Thenh(x)≤p(x) for allx∈S andh is a nonstandard concave functional on S, since αk+ (1−α)k = k for any α ∈ (0,1). From
Theorem 5.2, there exists a pseudo-linear functionalf onXsuch thatfsatisfies the Hahn-Banach conditions and f(x) ≤p(x) for all x ∈X and h(x) ≤f(x) for all x ∈ S. Therefore, for x ∈ S, we have k = h(x) ≤ f(x) ≤ p(x). By taking infimum on the inequalities, we obtain
k≤ inf
x∈Sf(x)≤ inf
x∈Sp(x) =k.
We complete the proof.
Corollary 5.3 (Convex Version of Hahn-Banach Theorem). LetX be a nonstandard vector space over Rsuch thatα(x⊕y) =αx⊕αyand(α+β)x= αx⊕βx for any x, y ∈X and α, β > 0, and (−λ)x =−(λx) for any λ < 0.
Let pbe a nonstandard sublinear functional onX satisfying the null inequality.
If S is a nonstandard convex subset of X and let k= infx∈Sp(x), then there exists a pseudo-linear functionalf onX such thatf satisfies the Hahn-Banach conditions with f(x)≤p(x) for allx∈X and
x∈Sinf f(x) = inf
x∈Sp(x) =k.
Proof. Under the assumptions, the Hahn-Banach conditions will be staif- sied automatically. Therefore, the result follows from Theorem 5.3 immedi- ately.
Theorem 5.4 (Hahn-Banach Extension Theorem). Let X be a nonstan- dard vector space over R. Let p be a nonstandard sublinear functional on X satisfying the null inequality, Hahn-Banach conditions and sub-associative law for positive scalar multiplication. Let Z be a subspace of X. If f is a pseudo- linear functional defined on Z withf(x)≤p(x) for allx∈Z, then there exists a pseudo-linear functional fˆon X such thatfˆsatisfies the Hahn-Banach con- ditions with f(x)ˆ ≤p(x) for all x∈X and fˆ(x) =f(x) for all x∈Z.
Proof. Since Z is a subspace of X, it says that Z is also a nonstandard convex subset ofX. Sincef is a pseudo-linear functional defined onZ, we also see thatf is a nonstandard concave functional defined onZ. Theorem 5.2 says that there exists a pseudo-linear functional ˆf on X such that ˆf satisfies the Hahn-Banach conditions and ˆf(x) ≤p(x) for all x ∈X and f(x) ≤fˆ(x) for all x∈Z. For anyx∈Z, since−x∈Z, we have
f(x)≤f(x) =ˆ −fˆ(−x)≤ −f(−x) =−(−f(x)) =f(x)
for all x ∈ Z. This shows that ˆf(x) = f(x) for all x ∈ Z. We complete the proof.
We also remark that ifXis a real vector space, then all the assumptions of Theorem 5.4 will be satisfied (it is not hard to check). In this case, Theorem 5.4 is reduced to the conventional Hahn-Banach theorem.
Corollary 5.4 (Hahn-Banach Extension Theorem). Let X be a non- standard vector space over R such that α(x⊕y) = αx⊕αy and (α+β)x = αx⊕βx for any x, y ∈X and α, β > 0, and (−λ)x =−(λx) for any λ < 0.
Let pbe a nonstandard sublinear functional onX satisfying the null inequality.
Let Z be a subspace ofX. Iff is a pseudo-linear functional defined on Z with f(x) ≤ p(x) for all x ∈ Z, then there exists a pseudo-linear functional fˆ on X such that fˆsatisfies the Hahn-Banach conditions with fˆ(x) ≤ p(x) for all x∈X and fˆ(x) =f(x) for all x∈Z.
Proof. Under the assumptions, the Hahn-Banach conditions will be staif- sied automatically. Therefore, the results follows from Theorem 5.4 immedi- ately.
The following result is useful for investigating the generalized Hahn- Banach extension theorem.
Lemma 5.3. Let X be a nonstandard vector space over F and let p be a real-valued function defined on X satisfying the following conditions:
(i) p(x⊕y)≤p(x) +p(y) for anyx, y∈X;
(ii) p(αx) =|α|p(x) for any x∈X and α∈R with α6= 0;
(iii) p(x⊕ω)≥p(x) for any x∈X andω ∈Ω.
Then p(x)≥0 for all x∈X.
Proof. For anyω∈Ω, by conditions (i) and (iii), we have 2p(ω) =p(ω) +p(ω)≥p(ω⊕ω)≥p(ω), which implies p(ω)≥0. By conditions (i) and (ii), we also have
0≤p(ω) =p(x x) =p(x⊕(−x))≤p(x) +p(−x) = 2p(x), which implies p(x)≥0 for anyx∈X. We complete the proof.
Theorem5.5 (Generalized Hahn-Banach Extension Theorem). LetX be a nonstandard vector space overFsuch that the following condition is satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x=α(ix); if Fis taken as R, then this condition is not needed;
Let p be a real-valued function defined on X such that the following con- ditions are satisfied:
(ii) p satisfies the sub-associative law for positive scalar multiplication;
(iii) p satisfies the Hahn-Banach conditions;
(iv) p(x⊕y)≤p(x) +p(y) for anyx, y∈X;
(v) p(αx) =|α|p(x) for any x∈X and α∈F with α6= 0;
(vi) p(x⊕ω)≥p(x) for any x∈X andω ∈Ω;
(vii) p(ω) = 0 for anyω∈Ω.
Suppose that f is a pseudo-linear functional defined on a subspace Z of X such that
(7) |f(z)| ≤p(z) for all z∈Z.
Then there exists a pseudo-linear functional fˆonX such thatfˆsatisfies the Hahn-Banach conditions with |fˆ(x)| ≤p(x) for allx∈X and f(z) =ˆ f(z) for all z∈Z.
Proof. Suppose that F is taken asR. From (7), we have f(z)≤ |f(z)| ≤ p(z) for all z∈Z. By Theorem 5.4, there is a linear extension ˆf from Z toX such that ˆf satisfies the Hahn-Banach conditions and
(8) fˆ(x)≤p(x) for allx∈X,
where ˆf is a pseudo-linear functional. Using (8) and condition (v), we obtain
−fˆ(x) = ˆf(−x)≤p(−x) =| −1|p(x) =p(x),
that is, ˆf(x) ≥ −p(x). Combining it with (8), we obtain |f(x)| ≤ˆ p(x) for all x∈X.
Suppose that F is taken as C. The functional f is a complex-valued function, which can be written asf(x) =f1(x) +if2(x), wheref1(x) = Ref(x) and f2(x) = Imf(x) are real-valued functions. Now we define
X¯ ={ix:x∈X}, Z¯ ={iz :z∈Z}, Ω =¯ {iω:ω∈Ω}
and consider the new sets
Xˆ =X∪X¯ and ˆZ =Z∪Z.¯
Then, under condition (i), we see that the nonstandard vector space X over C is equivalent to the nonstandard vector space ˆX over R in the sense that each element inX overCcan be rewritten as an element in ˆX overRand vice versa. By definition, the null set ˆΩ of ˆX is given by
Ω =ˆ n
ˆ
x xˆ: ˆx∈Xˆ o
={x x:x∈X} ∪ {ix ix:x∈X}. By condition (i), we also have
Ω =ˆ {x x:x∈X} ∪ {i(x x) :x∈X}= Ω∪Ω.¯
Since Z is a complex-valued subspace of X, we see that ˆZ is a subspace of ˆX. Furthermore, the complex-valued subspace Z is equivalent to the real- valued subspace ˆZ in the sense that each element inZ overCcan be rewritten as an element in ˆZ overRand vice versa. Under these settings, conditions (ii)–
(vii) of this theorem can be regarded as the corresponding conditions based on
the nonstandard vector space ˆX overR. Since f is a pseudo-linear functional on the complex-valued subspace Z, we are going to show that f1 and f2 are also pseudo-linear functionals on the real-valued subspace ˆZ. Indeed, for any x, y∈Zˆ=Z∪Z, we have¯
[f1(x) +f1(y)] +i[f2(x) +f2(y)] =f1(x) +if2(x) +f1(y) +if2(y)
=f(x) +f(y) =f(x⊕y) =f1(x⊕y) +if2(x⊕y) and, for α∈Rwith α6= 0 andx∈Z, we haveˆ
αf1(x) +iαf2(x) =αf(x) =f(αx) =f1(αx) +if2(αx).
By comparing the real and imaginary parts, we see that f1 and f2 are the pseudo-linear functionals on the real-valued subspace ˆZ. Since the real part of a complex number cannot exceed the absolute value, from (7), we have f1(x)≤ |f(x)| ≤p(x) for allx∈Zˆ. By Theorem 5.4, there is a linear extension fˆ1 of f1 from ˆZ to ˆX such that ˆf1 satisfies the Hahn-Banach conditions and fˆ1(x) ≤p(x) for all x ∈X, where ˆˆ f1 is also a pseudo-linear functional on ˆX.
Now, for everyx∈Z, we have
i[f1(x) +if2(x)] =if(x) =f(ix) =f1(ix) +if2(ix).
Comparing the real parts on both sides, we obtain (9) f2(x) =−f1(ix) for all x∈Z.
For x∈X, we set
(10) f(x) = ˆˆ f1(x)−ifˆ1(ix).
Therefore, if x ∈ Z ⊆ Z,ˆ i.e., ix ∈ Z¯ ⊆ Zˆ, then ˆf1(x) = f1(x) and fˆ1(ix) = f1(ix), since ˆf1 is a linear extension of f1 from ˆZ to ˆX. Therefore, from (9) and (10), we see that
fˆ(x) =f1(x) +if2(x) =f(x)
forx∈Z,i.e., ˆf is an extension off from Z toX. Now we are going to show that ˆf is a pseudo-linear functional on the nonstandard vector space X over C. From (10), we have
fˆ(x⊕y) = ˆf1(x⊕y)−ifˆ1(i(x⊕y))
= ˆf1(x⊕y)−ifˆ1(ix⊕iy) (by condition (i))
= ˆf1(x) + ˆf1(y)−ifˆ1(ix)−ifˆ1(iy) (by the linearity of ˆf1 on ˆX)
= ˆf(x) + ˆf(y)
and, for any complex numberα+iβ forα, β∈R, we have f((αˆ +iβ)x) = ˆf(αx⊕iβx) (by condition (i))
= ˆf1(αx⊕iβx)−ifˆ1(α(ix)⊕(−β)x) (by condition (i))
=αfˆ1(x) +βfˆ1(ix)−i[αfˆ1(ix)−βfˆ1(x)]
(by the linearity of ˆf1 on ˆX)
= (α+iβ)[ ˆf1(x)−ifˆ1(ix)]
= (α+iβ) ˆf(x).
This shows that ˆf is indeed a pseudo-linear functional on the nonstandard vector space X over C. We remain to prove |fˆ(x)| ≤ p(x) for all x ∈ X.
Now, for x ∈ X with ˆf(x) = 0, it is easy to see that |f(x)|ˆ = 0 ≤ p(x) by Lemma 5.3. Therefore, we consider x∈X with ˆf(x) 6= 0. Then we can write fˆ(x) =|fˆ(x)|eiγ,i.e.,
(11) |fˆ(x)|= ˆf(x)e−iγ= ˆf(e−iγx).
Since |fˆ(x)|is real, using (11), we obtain ˆf(e−iγx) = Re ˆf(e−iγx). There- fore, using (10) and condition (v), we have
|fˆ(x)|= ˆf(e−iγx) = Re ˆf(e−iγx) = ˆf1(e−iγx)≤p(e−iγx) =|e−iγ|p(x) =p(x).
Therefore, we obtain the desired result. Finally, since ˆf1 satisfies the Hahn-Banach conditions, using (10) and condition (i), we see that ˆf also sat- isfies the Hahn-Banach conditions. We complete the proof.
Corollary 5.5 (Generalized Hahn-Banach Extension Theorem). Let X be a nonstandard vector space over F such that the following conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x=α(ix); if Fis taken as R, then this condition is not needed;
(ii) (−λ)x=−(λx) for any x∈X and λ <0;
(iii) α(x⊕y) = αx⊕αy and (α+β)x = αx⊕βx for any x, y ∈ X and α, β >0.
Let p be a real-valued function defined on X satisfying the following con- ditions:
(iv) p(x⊕y)≤p(x) +p(y) for anyx, y∈X;
(v) p(αx) =|α|p(x) for any x∈X and α∈F with α6= 0.
(vi) p(x⊕ω)≥p(x) for any x∈X andω ∈Ω.
(vii) p(ω) = 0 for anyω∈Ω.
Suppose that f is a pseudo-linear functional defined on a subspace Z of X such that |f(z)| ≤ p(z) for all z ∈ Z. Then there exists a pseudo- linear functional fˆ on X such that fˆ satisfies the Hahn-Banach conditions with |fˆ(x)| ≤p(x) for allx∈X and fˆ(z) =f(z) for all z∈Z.
Proof. Under conditions (ii) and (iii), we see that p satisfies the Hahn-Banach conditions. Therefore, the conditions in Theorem 5.5 are all satisfied.
We also remark that conditions (ii) and (iii) in Corollary 5.5 are auto- matically satisfies in the space I of all closed intervals.
6. HAHN-BANACH EXTENSION THEOREMS IN NONSTANDARD NORMED SPACES
In the sequel, we are going to derive the Hahn-Banach theorems in non- standard normed spaces. In order for the convenient discussion, we introduce many terminologies.
Definition 6.1.Let (X,k · k) be a nonstandard pseudo-seminormed space.
(i) We say that the norm k · k satisfies the positively distributive law for vector addition if k α(x⊕y) k=k αx⊕αy k and k z⊕α(x⊕y) k=k z⊕αx⊕αyk for any x, y, z∈X andα >0.
(ii) We say that the normk · ksatisfies thedistributive law for positive scalar additionifk(α+β)xk=kαx⊕βxkandkz⊕(α+β)xk=kz⊕αx⊕βxk for any x, z ∈X and α, β >0.
(iii) We say that the normk · ksatisfiesthe associative law for positive scalar multiplication if kα(βx) k=k (αβ)xk and kz⊕α(βx)k=kz⊕(αβ)xk for any x, z ∈X and α, β >0.
(iv) We say that the norm k · k satisfies the positively sub-distributive law for vector addition ifkα(x⊕y)k≤kαx⊕αy kand kz⊕α(x⊕y)k≤k z⊕αx⊕αyk for any x, y, z∈X andα >0.
(v) We say that the norm k · k satisfies the sub-associative law for positive scalar multiplication if k α(βx) k≤k (αβ)x k and k z⊕α(βx) k≤k z⊕ (αβ)xk for any x, z∈X and α, β >0.
Let (X,k · k) be a nonstandard pseudo-seminormed space. Ifk · ksatisfies the positively distributive law for vector addition, then, by induction onn, we have kα(x1⊕ · · · ⊕xn)k=kαx1⊕ · · · ⊕αxnk andkz⊕α(x1⊕ · · · ⊕xn)k=k z⊕αx1 ⊕ · · · ⊕αxn k for any α ∈F and z, x1,· · ·, xn ∈ X. However, if the equality k z⊕α(x⊕y) k=k z⊕αx⊕αy k is omitted in the definition, then we cannot have the equality k α(x1⊕ · · · ⊕ xn) k=k αx1 ⊕ · · · ⊕αxn k by
induction on n. The similar situation also applies to the case of distributive law for positive scalar addition.
Definition 6.2.Let (X,k · k) be a nonstandard pseudo-seminormed space.
We say that k · k satisfies the Hahn-Banach conditionsif the following condi- tions are satisfied:
(i) k(−α)xk=k −(αx)k for any x∈X and α <0;
(ii) k(−α)x⊕zk=k −(αx)⊕zkfor any x, z ∈X andα <0;
(iii) kz⊕(α+β)xk≤kz⊕αx⊕βxk for any x, z∈X and α, β >0.
(iv) k · k satisfies the positively distributive law for vector addition.
Theorem6.1 (Hahn-Banach Extension Theorem in Nonstandard Normed Space – I). Let(X,k · k) be a nonstandard pseudo-normed space. Suppose that the following conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x = α(ix); if F is taken as the field of real numbers, then this condition is not needed;
(ii) the norm k · k satisfies the sub-associative law for positive scalar multi- plication;
(iii) the norm k · k satisfies the Hahn-Banach conditions.
(iv) the norm k · k satisfies the null condition;
(v) the norm k · k satisfies the null inequality.
Suppose that f is a pseudo-linear functional defined on a subspace Z of X such that |f(z)| ≤k z k for all z ∈ Z. Then there exists a pseudo-linear functional fˆonX such that |fˆ(x)| ≤kx k for all x∈X and f(z) =ˆ f(z) for all z∈Z.
Proof. We define the function p(x) =k x k≥ 0 for all x ∈ X. Then p satisfies conditions (ii)–(vii) of Theorem 5.5. Therefore, the result follows from Theorem 5.5 immediately.
Corollary 6.1 (Hahn-Banach Extension Theorem in Nonstandard Normed Space – II). Let (X,k · k) be a nonstandard pseudo-normed space.
Suppose that the following conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x = α(ix); if F is taken as the field of real numbers, then this condition is not needed;
(ii) (−λ)x=−(λx) for any x∈X and λ <0;
(iii) α(x⊕y) = αx⊕αy and (α+β)x = αx⊕βx for any x, y ∈ X and α, β >0.
(iv) the norm k · k satisfies the null condition;
(v) the norm k · k satisfies the null inequality.
Suppose that f is a pseudo-linear functional defined on a subspace Z of X such that |f(z)| ≤k z k for all z ∈ Z. Then there exists a pseudo-linear functional fˆon X satisfying |fˆ(x)| ≤kx k for all x ∈X and fˆ(z) =f(z) for all z∈Z.
Proof. We define the function p(x) =k x k≥0 for all x ∈ X. Then the result follows from Corollary 5.5 immediately.
Theorem6.2 (Hahn-Banach Extension Theorem in Nonstandard Normed Space – III). Let (X,k · k) be a nonstandard normed space such that the fol- lowing conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x = α(ix); if F is taken as the field of real numbers, then this condition is not needed;
(ii) the norm k · k satisfies the null condition;
(iii) the norm k · k satisfies the null inequality;
(iv) the norm k · k satisfies the sub-associative law for positive scalar multi- plication;
(v) the norm k · k satisfies the Hahn-Banach conditions.
Suppose thatf is a bounded pseudo-linear functional defined on a subspace Z of X, where Z 6= Ω. Then there exists a bounded pseudo-linear functionalfˆ on X such that fˆ(z) =f(z) for all z∈Z and kfˆkX=kf kZ, where
kfˆkX= sup
x∈X\Ω kxk= 1
|fˆ(x)|and kf kZ= sup
x∈Z\Ω kxk= 1
|f(x)|.
If Z = Ω then f(z) = 0 for all z∈Z and its linear extension isfˆ(x) = 0 for all x∈X. In this case, we have kf kZ= 0 =kfˆkX.
Proof. By Proposition 3.1, we see that the norm k · k also satisfies the null equality. ForZ 6= Ω, from Proposition 3.3, we have |f(x)| ≤kf kZ · kxk for all x ∈ Z. Let p(x) =k f kZ · kx k. Then we see that p can be defined on all of X. Since we are going to apply Theorem 5.5, we need to check that p(x) satisfies conditions (ii)–(vii). Conditions (ii) and (iii) in Theorem 5.5 are obviously satisfied by conditions (iv) and (v) of this theorem.
For condition (iv), we have
p(x⊕y) =kf kZ· kx⊕y k≤kf kZ (kxk+kyk) =p(x) +p(y).
For condition (v), we have, α∈Fwith α6= 0,
p(αx) =kf kZ · kαxk=|α| kf kZ · kxk=|α|p(x).
For condition (vi), we have
p(x⊕ω) =kf kZ · kx⊕ω k≥kf kZ· kxk=p(x).
For condition (vii), we have p(ω) =k f kZ · k ω k= 0 for any ω ∈ Ω by Proposition 3.1 (ii). Therefore, Theorem 5.5 says that there exists a pseudo- linear functional ˆf on X such that ˆf(z) = f(z) for all z ∈ Z and |fˆ(x)| ≤ p(x) =kf kZ · kxk for anyx ∈X. Applying Proposition 3.4 and taking the supremum over all x∈X with norm 1, we obtain the inequality
(12) kfˆkX= sup
x∈X\Ω kxk= 1
|fˆ(x)| ≤kf kZ .
Since ˆf is an extension of f, the norm should increase, i.e., k fˆkX≥k f kZ. Combining it with (12), we obtain kfˆkX=kf kZ. Finally, forZ = Ω, since the norm k · k satisfies the null condition, we have |f(z)| ≤k z k= 0, which shows that f(z) = 0 for all z ∈ Z. Therefore, the extension is ˆf = 0.
We complete the proof.
Corollary 6.2 (Hahn-Banach Extension Theorem in Nonstandard Normed Space – IV). Let (X,k · k) be a nonstandard normed space such that the following conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x = α(ix); if F is taken as the field of real numbers, then this condition is not needed;
(ii) the norm k · k satisfies the and null condition;
(iii) the norm k · k satisfies the null inequality;
(iv) (−λ)x=−(λx) for any x∈X and λ <0;
(v) α(x⊕y) = αx⊕αy and (α+β)x = αx⊕βx for any x, y ∈ X and α, β >0.
Suppose thatf is a bounded pseudo-linear functional defined on a subspace Z of X, where Z 6= Ω. Then there exists a bounded pseudo-linear functionalfˆ on X such that fˆ(z) =f(z) for all z ∈Z and kfˆkX=kf kZ. If Z = Ω then f(z) = 0 for all z ∈Z and its linear extension is fˆ(x) = 0 for all x ∈X. In this case, we have kf kZ= 0 =kfˆkX.
Proof. Under the assumptions, the conditions in Theorem 6.2 will be satisfied automatically.
Next, we also provide an interesting application of using Hahn-Banach theorems. As a matter of fact, the following proposition is also very useful in the future study.
Proposition6.1. Let(X,k · k)be a nonstandard normed space such that the following conditions are satisfied:
(i) for anyα, β ∈R andx, y∈X, (α+iβ)x=αx⊕iβx, i(x⊕y) =ix⊕iy and i(αx) = (iα)x = α(ix); if F is taken as the field of real numbers, then this condition is not needed;
(ii) The null set Ω is closed under the vector addition;
(iii) the norm k · k satisfies the null condition;
(iv) the norm k · k satisfies the null inequality;
(v) the distributive law for vector addition and associative law for scalar mul- tiplication hold true;
(vi) (α+β)x=αx⊕βx for anyx∈X and α, β >0.
(vii) Let xˆ∈X\Ω be given such that k α1xˆ⊕ · · · ⊕αnxˆk=|Pn
i=1αi|· k xˆk for any finite sequence {a1,· · · , αn} in F.
Then there exists a bounded pseudo-linear functional fˆon X such that kfˆk= 1 andfˆ(ˆx) =kxˆk.
Proof. We consider the subset Z of X defined by Z = span({ˆx} ∪Ω).
Then Z is a subspace of X. By conditions (ii) and (iii) and Remark 3.1, we also see that Ω is a subspace ofX. Therefore, for anyz∈Z,zhas the form of ω⊕α1xˆ⊕ · · · ⊕αnxˆ orα1xˆ⊕ · · · ⊕αnxˆ for some finite sequence{α1,· · · , αn} inF and some ω∈Ω. We define a real-valued functionf onZ by
f(z) =
(Pn
i=1αi)kxˆk ifz∈Z\Ω with z=ω⊕α1xˆ⊕ · · · ⊕αnxˆ orz=α1xˆ⊕ · · · ⊕αnxˆ
0 ifz∈Ω.
The function f is well-defined, which can be realized after proving the pseudo-linearity.
We first show that f is a pseudo-linear functional on Z. For z1 =ω1⊕ α1xˆ⊕ · · · ⊕αnxˆ and z2 =ω2⊕αˆ1xˆ⊕ · · · ⊕αˆmx, we haveˆ
f(z1⊕z2) =
n
X
i=1
αi+
m
X
j=1
ˆ αj
kxˆk=f(z1) +f(z2)
For z1 = α1xˆ⊕ · · · ⊕αnxˆ and z2 = ˆα1xˆ⊕ · · · ⊕αˆmx, we similarly haveˆ f(z1 ⊕z2) = f(z1) +f(z2). For z1 ∈ Ω and z2 = ω ⊕αˆ1xˆ⊕ · · · ⊕αˆmxˆ or z2= ˆα1xˆ⊕ · · · ⊕αˆmx, we similarly haveˆ f(z1⊕z2) =f(z1) +f(z2). Forα6= 0, we have
f(αz1) =f(αω1⊕(αα1)ˆx⊕ · · · ⊕(ααn)ˆx)
=α
n
X
i=1
αi
!
kxˆk=αf(z1)
by condition (v). We similarly havef(αz1) =αf(z1) forz1 =α1xˆ⊕ · · · ⊕αnx.ˆ We also need to claim that ifz1 =z2∈Z, thenf(z1) =f(z2). Sincez1 z2∈Ω, we have
0 =f(z1 z2) =f(z1⊕(−z2)) =f(z1) +f(−z2) =f(z1)−f(z2), which implies f(z1) =f(z2). Therefore, we conclude thatf is a pseudo-linear functional onZ. From Proposition 3.1, we also see that the normk · ksatisfies the null equality. Now, forz=ω⊕α1xˆ⊕ · · · ⊕αnx, by condition (vii), we haveˆ
kzk=kω⊕α1xˆ⊕ · · · ⊕αnxˆk=kα1xˆ⊕ · · · ⊕αnxˆk=
n
X
i=1
αi
· kxˆk. Therefore, we obtain
|f(z)|=
kzk ifz∈Z\Ω 0 ifz∈Ω
This shows that f is a bounded pseudo-linear functional with norm k f kZ= 1. Corollary 6.2 says that there exists a bounded pseudo-linear func- tional ˆf on X such that k fˆkX=k f kZ= 1 and ˆf(z) = f(z) for z ∈ Z. In particular, ˆf(ˆx) =f(ˆx) =kxˆk. We complete the proof.
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Received 4 October 2011 National Kaohsiung Normal University, Department of Mathematics,
Kaohsiung 802, Taiwan hcwu@nknucc.nknu.edu.tw
