University of Mazandaran, Iran http://www.cjms.umz.ac.ir
ISSN: 1735-0611 CJMS. 1(1)(2012), 7-12
CONE NORMED SPACES
M. ESHAGHI GORDJI1,∗, M. RAMEZANI2, H. KHODAEI3AND H. BAGHANI4
ABSTRACT. In this paper, we introduce the cone normed spaces and cone bounded linear mappings. Among other things, we prove the Baire category theorem and the Banach–Steinhaus theorem in cone normed spaces.
Keywords : One metric space; Baire category theorem; Banach Steinhaus theorem. Classification 2000 MSC: 64BXX.
1. INTRODUCTION
LetE be a Banach space andP be a subset ofE. P is called a cone whenever (1)P is a closed, non-empty set andP 6={0},
(2)ax+by ∈P for allx, y ∈P anda, b≥0, (3)P ∩(−P) = {0}.
For a given coneP ⊆E, we can define a partial ordering≤with respect toP byx≤y
if and only ify−x∈P. We shall writex < y ifx ≤yandx 6=y, whilex ywill stands fory−x∈intP, whereintP denoted the interior ofP. The coneP is normal if there is a numberM >0such that for allx, y ∈E
0≤x≤y =⇒ kxk ≤Mkyk.
The least positive number satisfying the above is called the normal constant ofP [1]. It is clear thatM ≥ 1. In the following we always suppose that E is a real Banach space andP is a cone inE withintP 6=∅and≤is a partial ordering with respect to
P.
Definition 1.1. ([1]) LetXbe non-empty set . Suppose that the mappingd :X×X →
Esatisfies:
(1)0≤d(x, y)for allx, y ∈Xandd(x, y) = 0if and only ifx=y, (2)d(x, y) = d(y, x)for allx, y ∈X,
(3)d(x, y)≤d(x, z) +d(z, y)for allx, y, z ∈X.
Received: 1 Oct 2011 Revised: 24 Oct 2011 Accepted: 20 Nov 2011
∗Corresponding author: [email protected].
Thendis called a cone metric onX, and(X, d)is called cone metric space.
Example 1.2. LetE =R2,P ={(x, y)∈E :x, y ≥0},X =Randd:X×X →E
defined byd(x, y) = (|x−y|, α|x−y|), whereα≥0is a constant. ThenP is a normal cone with normal constantM = 1and(X, d)is a cone metric space[2].
Definition 1.3. ([1]) Let(X, d)be a cone metric space,x ∈ X and{xn}a sequence inX. Then
(1) {xn} is said to convergent to x whenever for every c ∈ E, with 0 cthere is a positive integer N such that d(xn, x) c for all n ≥ N. We denote this by
limn→∞xn=xorxn→xasn → ∞.
(2) {xn} is said to be a Cauchy sequence whenever for every c ∈ E with 0 c there is a positive integerN such thatd(xn, xm)cfor alln, m≥N.
(3)(X, d)is called a complete cone metric space if every Cauchy sequence is con-vergent.
Let us recall [1] that ifP is a normal cone, then{xn} ⊆ X converges tox if and only ifd(xn, x)→0asn → ∞. Furthermore,{xn} ⊆X is a Cauchy sequence if and only ifd(xn, xm)→0asn, m→ ∞.
Definition 1.4. Let(X, d)be a cone metric space andB ⊆X.
(1) A pointb ∈ B is called an interior point of B whenever there exists a pointp,
0p, such thatBp(b)⊆BwhereBp(b) := {y∈X :d(b, y)p}.
(2) A subsetB ⊆Xis called open if each element ofB is an interior point ofB. The family β = {Be(x) : x ∈ X,0 e} is a sub-basis for a topology on X. We denote this cone topology byτc. The topologyτcis a Hausdorff and first countable [2].
In this paper we suppose thatP is a normal cone with normal constantM and fixed
c0 with0c0.
2. CONE NORMED SPACES
Definition 2.1. LetX be real vector space. Suppose that the mappingk.kp : X →E satisfies:
(1)kxkp ≥0for allx∈X andkxkp = 0if and only ifx= 0, (2)kαxkp =|α| kxkp, for allx∈Xandα ∈R,
(3)kx+ykp ≤ kxkp+kykp for allx, y ∈X.
Thenk.kp is called a cone norm onXand(X,k.kp)is called a cone normed space.
It is easy to see that every normed space is a cone normed space by puttingE :=R,
P := [0.∞).
Example 2.2. LetE = l1, P = {{xn} ∈ E : xn ≥ 0, f or all n}and(X,k.k)be a normed space andk.kp : X → E defined bykxkp ={
kxk
2n}. ThenP is a normal cone with constant normalM = 1and(X,k.kp)is a cone normed space.
Let(X,k.kp)be a cone normed space. Setd(x, y) = kx−ykp. It is easy to see that
(X, d)is a cone metric space.dis called getting cone metric of cone normk.kp.
Lemma 2.4. Let(X,k.kp)and(Y,k.kp)be cone normed spaces, and letT be a linear
map fromXintoY. If T has any one of the five following properties, it has all five of them:
(a) (continuity at a point) For some fixed x0 ∈ X we have: Given 0 c there is a
0tsuch thatkT x−T x0kp cwheneverkx−x0kp t.
(b) (continuity at zero) For 0 cthere is a 0 t such that kT xkp cwhenever
x∈Xandkxkp t.
(c) (continuity at every point of x) For any x ∈ X we have: Given0 cthere is a 0tsuch thatkT x−T ykp cwhenevery∈X andky−xkp t.
(d) (uniform continuity) Given0 cthere is a 0 t such that kT x−T ykp c
wheneverx, y ∈Xandkx−ykp t.
(e) (sequential continuity) Given any sequence {xn} ⊆ X which is convergent to a
pointx0 ∈X, the sequence{T xn} ⊆Y is convergent to the pointT x0 ∈Y.
Proof: First assume thatT has property (a). So for somex0 ∈ X and any0 cwe
can choose0tsuch thatkT x−T x0kp cwheneverkx−x0kp t. Then for any
w∈Xwithkwkp twe havekT(w+x0)−T x0k cbecausek(w+x0)−x0kp t. ButT is linear this says thatkT wkp cwheneverkwkp tand we have shown that (a) implies (b).
Now suppose thatT has property (b), letx∈Xand0cbe given. There is a0t
such that kT wkp cwhenever kwkp t. We have kT(y−x)kp cwhenever ky−xkp t; just usey−xin place ofw. Again recalling thatT is linear we see that (b) implies (c). Clearly (c) implies (a). Thus (a), (b) and (c) are equivalent.
Let us show that (b) implies (d). Given0cwe may choose0tso thatkwkp t implieskT wkp c. Now if x, y ∈ X and kx−ykp t then kT(x−y)kp c. SinceT is linear (b) implies (d) and clearly (d) implies (b). Thus (a) through (d) are equivalent.
We will complete the proof by showing that (b) and (e) are equivalent. First suppose that T has property (b) and let 0 c be given. Then we have choose 0 t that kT wkp c whenever kwkp t. Now suppose {xn} ⊆ X converges x0. Then
we can find positive integerN such that kxn −x0kp t whenever n ≥ N. Thus kT(xn−x0)kp cwhenevern≥N. Clearly this says that{T xn}converges toT x0.
Now assume (e) and negation of (b). So we are supposing that there is a0 csuch that for any0 twe can find wt ∈ X withkwtkp t andkT wtkp 6 c. Thus for thiscwe can find{wn} ⊆ X such thatkwnkp 2cn andkT wnkp 6 cfor alln. But {wn}is converges to 0 becausek kwnkpk ≤ M
kck
2n → 0 asn → ∞. By (e) {T wn}
must converge toT0 = 0and this impossible.
Proposition 2.5. Let(X,k.kp)be a cone normed space, andx∈X,0c. Then
y∈Bc(x)⇐⇒(∃{zn} ⊆Bc(x) ; zn→y).
Proof: Lety ∈ Bc(x). Then for any positive integern, zn ∈ B2cn(y)∩Bc(x) 6= ∅. We obtainzn →yasn → ∞. Now we suppose that{zn} ∈ Bc(x)is a sequence that
Hence,zn∈Bp(y)andW ∩Bc(x)6=∅. Soy ∈Bc(x).
We need the following lemma to prove the Baire category theorem.
Lemma 2.6. Let(X,k.kp)be a cone normed space,x∈Xand0c. ThenBc2(x)⊆
Bc(x).
Proof: Lety ∈ Bc
2(x). Then there is a sequence{zn} ⊆ B
c
2(x)such thatzn → yas
n → ∞. We can choose the positive integern such thatkzn −ykp 2c. We obtain thatkx−ykp ≤ kx−znkp+kzn−ykp 2c+c2 =c. Lettinga= (kx−znkp+kzn−
ykp)− kx−ykp, by attention that action+is continuous inE, we have
c− kx−ykp = (c−(kx−znkp+kzn−ykp) + a∈a+intP =int(a+P)⊆intP.
Sokx−ykp cand thusy∈Bc(x).
Theorem 2.7. (Baire Category Theorem) Let(X,k.kp)be a cone Banach space. Then
every countable intersection of dense and open sets is dense.
Proof: Let{An} ⊆Xbe a sequence of dense and open sets. Supposex∈X, andWis a open set such thatW consists ofx. Then there is0rsuch thatBr(x)⊆W. Since
A1is dense inX, we obtainz1 ∈A1∩Br(x)6=∅. ButA1∩Br(x)is open and hence there exists0 r0 such thatBr0(z1) ⊆ A1∩Br(x). We can choose the positive real
numberk1 such thatk1 < min{12,2kr101k}. By settingr1 =k1r01, we haver1 r 0
2 and
Br1(z1)⊆Br10(z1). SinceA2 is dense inX, we havez2 ∈A2∩Br1(z1)6=∅, but this
set is open. Thus there exists0r02 such thatBr02(z2)⊆A2∩Br1(z1). By choosing 0 < k2 < min{12,22k1r02k}, we haver2 = k2r02 r
0 2
2 and henceBr2(z2) ⊆ Br02(z2).
Since A3 is dense in X then let z3 ∈ A3 ∩ Br2(z2). Since A3 is open, there is a 0 r03 such thatBr30(z3) ⊆ A3 ∩Br2(z2). We can choose the real number0 < k3
for which k3 < min{12,23k1r0
3k}. Setting r3 = k3r 0
3, we conclude that r3
r0 3
2 and
Br3(z3)⊆Br03(z3). Repeating the above argument we obtain
...Br3(z3)⊆Br2(z2)⊆Br1(z1).
We claimrn → 0, becausern =knrn0
1 2nkr0
nk r
0
n. Sokrnk ≤ 2nMkr0
nkkr
0
nk= M
2n → 0
asn → ∞. Moreover , we can show that{zn}is Cauchy sequence inX. To see this, let > 0is given, there is a positive integerN such that Mkrnk < for alln ≥ N. Sokkzm −znkpk ≤ Mkrnk< for allm > n≥N. This means that{zn}is Cauchy sequence. SinceX is cone Banach space, there isz ∈ X such that lim
n→∞zn = z. For
any positive integerN, ifn > N thenzn∈BrN(zN), soz ∈BrN(zN)and hence
z ∈
∞
\
N=1
BrN(zN)⊆
∞
\
N=1
AN \
Br(x)⊆
∞ \ N=1 AN \ W.
This says thatT∞
N=1AN is dense inX.
Definition 2.9. A subsetA ⊆Eis upper bounded, if there exists0≤tsuch thata≤t
for alla∈ A,tis an upper bound forA. We say thatP has supremum property, if for every upper bounded setAinP least upper bound exists in P, we show this element tosupA.
Example 2.10. Let E = R2 , P = {(x, y) ∈ E : x, y ≥ 0}. P is a normal cone
with normal constantM = 1, and P has supremum property. Because ifA ⊆ P is upper bounded set, then supA = (supz∈Aπ1(z), supz∈Aπ2(z)). Whereπ1 andπ2 are
projections on first and second components, respectively.
From now on, we suppose thatP has supremum property.
Definition 2.11. Let (X,k.kp) be a cone normed space. A subset A in X is cone bounded, if{kxkp;x∈A}is upper bounded.
Definition 2.12. Let(X,k.kp)and(Y,k.kp)be cone normed spaces, andΛ :X →Y be a linear mapping. Λis cone bounded if the setΛ(Bc0(0)) is a cone bounded set.
LetB(X, Y)denotes the set of all cone bounded linear mappings fromX intoY. It is easy to see thatkΛkp =sup{kΛ(x)kp :kxkp c0}is a cone norm onB(X, Y).
In the following, we obtain a linear mapping that is not cone bounded.
Example 2.13. LetE = R2, P ={(x, y) ∈ E : x, y ≥ 0}, c0 = (1,1)and letX be
the set of all real- valued polynomials on interval[0,1]andk.ku is supremum norm on
X, that iskfku = sup{|f(x)|: x ∈[0,1]}for allf ∈ X. Letk.kp :X → E defined bykfkp = kfkuc0. It is easy to see that (X,k.kp)is a cone normed space. Suppose
D : X → X defined by D(f) = f0. Then D is a linear mapping that is not cone bounded.
Theorem 2.14. (The Uniform Boundness Principle) Let(X,k.kp)be a cone Banach
space and (Y,k.kp) be a cone normed space. Suppose A ⊆ B(X, Y) is pointwise
bounded, that is for eachx ∈X, the set{T x:T ∈A}is cone bounded. ThenAis a cone bounded set inB(X, Y).
Proof: Let
En ={x∈X :kT xkp ≤nc0 f orall T ∈A}
for alln ∈ N. It is easy too see that the set{y ∈ Y : kykp ≤ nc0}is closed. Hence,
for each positive integern, Enis a closed set. We claimX =
S∞
n=0En. Too see this,
since0c0, choose0< δ such that
c0+{x∈E :kxk ≤δ} ⊆P.
Ifx ∈ X is given. Settingαx = sup{kT xkp : T ∈ A}. Choose a positive integern such thatkαx
n k< δ. Soc0− αx
n ∈c0+{x∈E :kxk ≤δ} ⊆P andαx nc0. This shows thatx ∈ En. SoX = S∞
n=0En. ButX is a cone Banach space and haenceX
is second category. Then there is a positive integerksuch that intEk = intEk 6= ∅. Suppose y ∈ intEk. We can choose 0 r such that Br(y) ⊆ Ek. Repeating the above method, there exists a positive integermsuch that c0
m r. Now, we letT ∈A
andx ∈X,kxkp c0. Then k(y+
x
m)−ykp =k x mkp
c0
y+mx ∈Br(y). We have
kT xkp =mkT
x
mkp =mkT(y+ x
m)−T ykp
≤m(kT(y+ x
m)kp+kT ykp)
≤m(k+k) = 2mk.
ThuskTkp ≤2mk. In the other wordsAis a cone bounded set inB(X, Y).
The question arises here is whether Hahn Banach theorem and open mapping theo-rem can be extended similar to cone normed spaces or not?
REFERENCES
[1] Huang Long-Guang and Zhang Xian, Cone metric spaces and fixed point theorems on con-tractive mappings, Journal of Mathematical Analysis and Applications 332(2007), 1468-1476.
[2] S. Rezapour, M. Derafshpour and R. Hamlbarani, A review on topological properties on cone metric spaces, Bulletin of the Korean Mathematical Society, to appear.
1Department of Mathematics, Semnan University, P. O. Box 35195-363, Semnan, Iran,
2Department of Mathematics, Semnan University, P. O. Box 35195-363, Semnan, Iran,
3 Department of Mathematics, Semnan University, P. O. Box 35195-363, Semnan, Iran,
4 Department of Mathematics, Semnan University, P. O. Box 35195-363, Semnan, Iran,
