# Arzela-Ascoli Theorem for Demi-Linear Mappings

Authors: Qiang Lei; Aihong Chen
Journal: Journal of Function Spaces
 (2014)

---

## Abstract

The Arzela-Ascoli theorem for demi-linear mappings is established. Some conclusions about preserving totally bounded sets are also obtained.


---

## Body

## 1. Introduction

In 2009, Li et al. introduced the definition of demi-linear mappings which is a meaningful extension of linear operators [1]. They also established a new equicontinuity theorem and a uniform boundedness principle for demi-linear mappings. After that, many related theories (such as demi-linear duality, demi-distributions, and matrix transformations) are obtained for demi-linear mappings which can be found in [2–4].Arzela-Ascoli theorem has a wide range of applications in many fields of mathematics. Many modern formulations of Arzela-Ascoli theorem have been obtained [5–10].In this paper, we will give some results on Arzela-Ascoli theorem for the space of demi-linear mappings.


## 2. Preliminary

LetX, Y be topological vector spaces over the scalar field 𝕂 and 𝒩(X) the family of neighborhoods of 0∈X. Let
(1)C(0)={γ∈𝕂𝕂:lim⁡t→0γ(t)=γ(0)=0,|γ(t)|≥|t|if|t|≤1}.Definition 1 ([1, Definition 2.1]).
A mappingf:X→Y is said to be demi-linear if f(0)=0 and there exist a γ∈C(0) and a U∈𝒩(X) such that x∈X, u∈U, and t∈{t∈𝕂:|t|≤1} yield r,s∈𝕂 for which |r-1|≤|γ(t)|, |s|≤|γ(t)|, and f(x+tu)=rf(x)+sf(u).We denote byℒγ,U(X,Y) the family of demi-linear mappings related to γ∈C(0) and U∈𝒩(X) and by 𝒦γ,U(X,Y) the subfamily of ℒγ,U(X,Y) satisfying the following property: if x∈X, u∈U, and |t|≤1, then f(x+tu)=f(x)+sf(u) for some s with |s|≤|γ(t)|.Note that, in general,ℒγ,U(X,Y) is a large extension of L(X,Y). For instance, if ∥·∥:X→[0,+∞) is a norm, then ∥·∥∈ℒγ,X(X,ℝ) for every γ∈C(0).Definition 2.
A familyΓ⊂YX is said to be equicontinuous at x∈X if for every W∈𝒩(Y), there is a V∈𝒩(X) such that f(x+V)⊂f(x)+W for all f∈Γ and Γ is equicontinuous on X or, simply, equicontinuous if Γ is equicontinuous at each x∈X.By the proof of [1, Theorem 3.1], the following conclusion can be easily obtained.Theorem 3.
LetΓ⊂ℒγ,U(X,Y). Then, Γ is equicontinuous at 0 if and only If Γ is equicontinuous on X.By the proof of [1, Theorem 3.2] and Definition 2, we also have the following result.Theorem 4.
Let{fα}⊂ℒγ,U(X,Y) and for every x∈X we have lim⁡fα(x)=f(x). Then, f∈ℒγ,U(X,Y). If, in addition, {fα} is equicontinuous, then f is continuous.Definition 5.
A familyΓ⊂YX is said to be uniformly continuous if for any neighborhood W of zero in Y, there exists a neighborhood V of zero in X such that x1-x2∈V implies f(x1)-f(x2)∈W.Definition 6.
LetE be a nonempty set and X a topological vector space, F⊂XE. A topology τ on E is said to be (E,F)-topology if there is an ℱ⊂2F such that ∪ℱ=F and τ is just the topology of uniform convergence on each M∈ℱ; that is, uα→u in (E,τ) if and only if for every M∈ℱ, f(uα)→f(u) uniformly with respect to f∈M.Letσ(E,F) be the weakest (E,F)-topology; that is, uα→u in (E,σ(E,F)) if and only if f(uα)→f(u), for all f∈F.LetE,X, and F be as in Definition 6. For x∈E, define x:XE→X by x(f)=f(x). Then, E⊂XF and by Definition 6 we have the (F,E)-topology on F. Also, we denote the weakest (F,E)-topology by σ(F,E), namely, the topology of pointwise convergence. That is,
(2)fα→σ(F,E)finFiff∀x∈X,lim⁡αfα(x)=f(x).


## 3. Main Results

Throughout this section, letX and Y be topological vector spaces on ℂ. Denote by ℒγ,U0(X,Y) the family of demi-linear mappings related to γ∈C(0) and U0∈𝒩(X). For U0∈𝒩(X), there is a balanced closed U∈𝒩(X) such that U⊂U0. Then, ℒγ,U0(X,Y)⊂ℒγ,U(X,Y). Without generality, we will assume that U0 is balanced closed in ℒγ,U0(X,Y).Recall that a setS in a topological vector space is called totally bounded if for each neighborhood U of zero there is a finite set F with S⊂F+U.Theorem 7.
Letf∈ℒγ,U0(X,Y). If f is continuous at zero, then f preserves totally bounded sets.Proof.
Suppose thatB is a totally bounded subset of X. Since Y is a topological vector space, for every balanced closed neighborhood V∈𝒩(Y), there exists a balanced closed neighborhood V1∈𝒩(Y) such that V1+V1⊂V. For V1, there is a U∈𝒩(X) with U⊂U0 such that
(3)f(U)⊂11+|γ(1)|V1
by the continuity of f at zero. For U, there exists a finite subset {x1,x2,…,xm} of B such that B⊂⋃i=1m‍(xi+U); that is, for every x∈B there exists an i0 with 1≤i0≤m and u∈U such that x=xi0+u. Then,
(4)f(x)=f(xi0+u)=rxf(xi0)+sxf(u),
where |rx-1|≤|γ(1)| and |sx|≤|γ(1)|. Let A={f(xi):i=1,…,m} and A~ be the balanced hull of A. Obviously, the finite set A is totally bounded and hence A~ is totally bounded. From
(5)M={rxf(xi):i=1,…,m,x∈B}=(1+|γ(1)|){rx1+|γ(1)|f(xi):i=1,…,m,x∈B}⊂(1+|γ(1)|)A~,
it follows that M is totally bounded and hence there exists a finite subset {y1,y2,…,yn} of Y such that
(6)M⊂{y1,y2,…,yn}+V1.
Then, from (3), (4), and (6), we have
(7)f(x)∈M+sxf(U)⊂{y1,y2,…,yn}+V1+sx1+|γ(1)|V1⊂{y1,y2,…,yn}+V1+V1⊂{y1,y2,…,yn}+V
for all x∈B. Thus, f(B) is totally bounded.Recall that a mapping with uniform continuity is also equicontinuous. The following theorem shows that the converse holds true for mappings in𝒦γ,U0(X,Y).Theorem 8.
LetΓ⊂ℒγ,U0(X,Y). Then, Γ is uniformly continuous if and only if Γ is equicontinuous at zero.Proof.
We only need to prove the “if” part. For everyW∈𝒩(Y), there exists a balanced closed W1∈𝒩(Y) with W1⊂W. For (1/|γ(1)|)W1∈𝒩(Y), since Γ is equicontinuous at zero, there is a V∈𝒩(X) such that
(8)f(V)⊂(1|γ(1)|)W1
for all f∈Γ. For arbitrary x1,x2∈X with x1-x2∈V, we have
(9)f(x1)=f(x2+x1-x2)=f(x2)+sff(x1-x2),
where |sf|≤|γ(1)| and f∈Γ. It follows that
(10)f(x1)-f(x2)=sff(x1-x2)∈sff(V)⊂sf(1|γ(1)|)W1⊂|γ(1)|(1|γ(1)|)W1=W1⊂W
for all f∈Γ, which shows that Γ is uniformly continuous.Theorem 9 (Ascoli Theorem).
LetX,Y be topological vector spaces, Y be Hausdorff, and ℰ be the family of all compact subsets of X. Let τℰ be the topology on ℒγ,U0(X,Y) of uniform convergence on each K∈ℰ. If Γ⊂ℒγ,U0(X,Y) is equicontinuous and Γ¯σ(ℒγ,U0(X,Y),X) is compact, then Γ¯τℰ is equicontinuous and compact.Proof.
SinceΓ is equicontinuous, for every balanced closed V∈𝒩(Y), there exists a balanced closed U∈𝒩(X) such that f(U)⊂V for all f∈Γ. For every f∈Γ¯τℰ, there is a net {fα}⊂Γ such that fα→τℰf. Since the singleton is compact, that is, {x}∈ℰ for all x∈X, we have lim⁡αfα(x)=f(x) for all x∈X. Then, by the balanced closed property of V, we have
(11)f(x)=lim⁡αfα(x)∈V,∀x∈U;
that is, f(U)⊂V. As f runs over the set Γ¯τℰ, we have that Γ¯τℰ is equicontinuous at zero and hence Γ¯τℰ is equicontinuous on X by Theorem 3.
Let{fα:D} be a net in Γ¯τℰ. Since the topology σ(ℒγ,U0(X,Y),X) is weaker than τℰ, we have Γ¯τℰ⊂Γ¯σ(ℒγ,U0(X,Y),X) and hence {fα:D} is also a net in Γ¯σ(ℒγ,U0(X,Y),X). Since Γ¯σ(ℒγ,U0(X,Y),X) is compact, the net {fα:D} has a convergent subnet of {fα:D}, which we also denote by {fα:D}. Let f0 be the limit of {fα:D} in Γ¯σ(ℒγ,U0(X,Y),X); that is,
(12)lim⁡αfα(x)=f0(x),∀x∈X.
Then, we have f0∈ℒγ,U0(X,Y) and f0 is continuous by Theorem 4.
To proveΓ¯τℰ is compact, we only need to prove fα→τℰf0. Assume this is not true; then, there exists a K∈ℰ such that fα(x)→f0(x) is not uniform for x∈K. Then, there are V∈𝒩(Y), increasing sequence {αk}⊂D, and {xk}⊂K such that
(13)fαk(xk)-f0(xk)∉V.
For V, there exists a balanced V1∈𝒩(Y) such that V1+V1+V1⊂V. Since K is compact, the sequence {xn} has a convergent subnet {xnδ} with xnδ→x0∈K. For V1, there exists a U∈𝒩(X) such that
(14)f(x0+U)⊂f(x0)+V1,∀f∈Γ¯τℰ
by the equicontinuity of Γ¯τℰ. Then,
(15)fα(x0+U)⊂fα(x0)+V1,∀α∈D.
For U, there exists a δ0 such that xnδ∈x0+U, for all δ≥δ0. Then,
(16)fα(xnδ)∈fα(x0)+V1,∀α∈D,∀δ≥δ0.
Since f0 is continuous, there is a δ1≥δ0 such that
(17)f0(xnδ)∈f0(x0)+V1,∀δ≥δ0.
Since fα(x0)→f0(x0), there exists an α0 such that
(18)fα(x0)∈f0(x0)+V1,∀α≥α0.
Then, (16) and (18) imply that
(19)fα(xnδ)∈fα(x0)+V1⊂fα(x0)+V1+V1,∀δ≥δ0,∀α≥α0.
It follows from (17) that
(20)fα(xnδ)-f0(xnδ)∈f0(x0)+V1+V1-f0(x0)-V1=V1+V1-V1⊂V,∀δ≥δ1,∀α≥α0.
This contracts (13) and hence fα→τℰf0, which yields the desired result.In fact, the converse of Theorem9 is also true. The following theorem gives the Arzela Theorem for demi-linear mappings.Theorem 10 (Arzela Theorem).
LetX,Y be topological vector spaces, Y be Hausdorff, and ℰ be the family of all compact subsets of X. Let τℰ be the topology on ℒγ,U0(X,Y) of uniform convergence on each K∈ℰ. If Γ is a family of continuous mappings in ℒγ,U0(X,Y) and Γ¯τℰ is compact, then(I)
Γ is equicontinuous;(II)
Γ¯σ(ℒγ,U0(X,Y),X) is compact.Proof.
For conclusion (I), we just need to prove thatΓ is equicontinuous at zero. Suppose that Γ is not equicontinuous at zero; that is, there exists a V∈𝒩(Y) such that for every U∈𝒩(X) there are xU∈U and fU∈Γ such that
(21)fU(xU)∉V.
Since 𝒩(X) is directed by the relation “⊂,” we have that {xU:𝒩(X)} and {fU:𝒩(X)} are nets, respectively, in X and ℒγ,U0(X,Y). Since {fU:U∈𝒩(X)}⊂Γ and Γ¯τℰ is compact, the net {fU:𝒩(X)} has a convergent subnet, which we also denote by {fU:𝒩(X)}. Let f0 be the limit of {fU:𝒩(X)} in Γ¯τℰ; that is, for each K∈ℰ we have fU(x)→f0(x) uniformly for x∈K. Obviously,
(22)lim⁡UxU=0
and hence {xU:U∈𝒩(X)} is compact; that is, {xU:U∈𝒩(X)}∈ℰ. Therefore, we have fU(x)→f0(x) uniformly for {xU:U∈𝒩(X)}. For V, there exists a V1∈𝒩(Y) such that
(23)V1+V1⊂V.
Then for V1, there exists a U1∈𝒩(X) such that
(24)fU(x)-f0(x)∈V1,∀U⊂U1,∀x∈{xV:V∈𝒩(X)}.
It follows that
(25)fU(xU)-f0(xU)∈V1,∀U⊂U1.
By (22) and the continuity of f0, for V1, there is a U2∈𝒩(X) such that
(26)f0(xU)∈V1,∀U⊂U2.
From (23), (25), and (26), we have
(27)fU(xU)=(fU(xU)-f0(xU))+f0(xU)∈V1+V1⊂V
which is in contradiction with (21) and hence the desired result holds.
For conclusion (II), we will prove thatΓ¯σ(ℒγ,U0(X,Y),X)=Γ¯τℰ. And we just need to show that Γ¯σ(ℒγ,U0(X,Y),X)⊂Γ¯τℰ. For each f∈Γ¯σ(ℒγ,U0(X,Y),X), there exists a net {fα} in Γ such that lim⁡αfα(x)=f(x), for all x∈X. We only need to prove that fα→τℰf. If this is not true, then there is a K∈ℰ such that fα(x)→f(x) is not uniform for x∈K. Then, there are V∈𝒩(Y), increasing sequence {αk}, and {xk}⊂K such that
(28)fαk(xk)-f(xk)∉V.
Since K is compact, there exist an x0∈K and a convergent subnet {xkδ} of {xk} such that lim⁡δxkδ=x0. By the equicontinuity of Γ, the continuity of f, and (28), we have
(29)fkδ(x0)-f(x0)∉V.
However,
(30)lim⁡δfkδ(x0)=lim⁡αfα(x0)=f(x0)
which leads a contradiction to (29) and hence the desired result holds.Theorem 11.
LetX,Y be topological vector spaces and X be of second category. If Γ is a family of continuous mappings in ℒγ,U0(X,Y) and for every x∈X, the set {f(x):f∈Γ} is totally bounded, then for every totally bounded subset B of X, we have that {f(x):x∈B,f∈Γ} is totally bounded.Proof.
Since totally bounded set is bounded, it follows thatΓ is equicontinuous on X by the equicontinuity theorem [1, Theorem 3.1]. For every balanced closed V∈𝒩(Y), there exists a balanced closed V1∈𝒩(Y) such that V1+V1⊂V. For V1, there is a U∈𝒩(X) with U⊂U0 such that
(31)f(U)⊂11+γ(1)V1,∀f∈Γ.
For U, there exist x1,x2,…,xn∈B such that B⊂⋃i=1n‍(xi+U); that is, for every x∈B, there exists an i0 with 1≤i0≤n and u∈U such that x=xi0+u. Then,
(32)f(x)=f(xi0+u)=rff(xi0)+sff(u),
where |rf-1|≤|γ(1)|, |sf|≤|γ(1)|, and f∈Γ. Let A={f(xi):f∈Γ,i=1,2,…,n}. Since for every x∈X, the set {f(x):f∈Γ} is totally bounded, we see that A is totally bounded. Let A~ be the balanced hull of A; that is, A~=⋃|t|≤1‍tA. Then, A~ is totally bounded. It follows from
(33){rfA:f∈Γ}=(1+|γ(1)|){rf1+|γ(1)|A:f∈Γ}⊂(1+|γ(1)|)A~
that {rfA:f∈Γ} is totally bounded. Hence, there exist y1,y2,…,ym∈Y such that
(34){rfA:f∈Γ}⊂{y1,y2,…,ym}+V1,∀f∈Γ.
Then, (31), (32), and (34) imply that
(35)f(x)=rff(xi0)+sff(u)∈rfA+sf1+|γ(1)|V1⊂{y1,y2,…,ym}+V1+V1⊂{y1,y2,…,ym}+V
for all x∈B and f∈Γ. Thus, {f(x):x∈B,f∈Γ} is totally bounded.



---

*Source: 679825-2014-03-02.xml*