东北大学学报:自然科学版   2015, Vol. 36 Issue (2): 301-304   PDF (284 KB)    
乘积空间中凹泛函型锥拉伸与压缩不动点定理
张国伟, 张秀萍    
东北大学 理学院,辽宁 沈阳 110819
摘要:考虑赋范线性空间的乘积空间, 由因子空间中的锥生成乘积空间中的锥. 全连续算子定义在乘积空间中锥与两个闭球相交得到的有界闭集上, 并且值域在锥中. 在由锥上一类非负正齐次凹泛函表示的混合型锥拉伸与压缩条件下, 利用构造性方法将其转化为Schauder型问题, 证明了几个全连续算子的不动点定理. 通过例子说明这里所需要的凹泛函在常用的空间及其锥上是容易构造的.
关键词不动点     全连续算子     锥拉伸与压缩     凹泛函     乘积空间    
Fixed Point Theorems of Cone Expansion and Compression of Concave Functional Type in Product Space
ZHANG Guo-wei , ZHANG Xiu-ping    
School of Sciences, Northeastern University, Shenyang 110819, China.
Corresponding author: ZHANG Guo-wei, professor, E-mail: gwzhangneum@sina.com
Abstract: A product space of normed linear spaces is considered, and the cone in the product space is produced by the cones in its factor spaces. A completely continuous operator is in the product space defined on the bounded closed set which is the intersection of the cone with two closed balls, and the range is in the cone. Under the mixed cone expansion and compression conditions that are expressed through a class of nonnegative, positively homogeneous, concave functionals on the cone, some fixed point theorems about the completely continuous operator are proved by constructing methods and converting them into the problems of Schauder type. It is illustrated by example that the concave functionals needed here are easily constructed in a common space and on a cone in it.
Key words: fixed point     completely continuous operator     cone expansion and compression     concave functional     product space    

E是实赋范线性空间,θ表示E中的零元素,KE中的锥,见文献[1]. 在乘积空间E×E={(u1,u2)|u1,u2E}中定义范数为‖(u1,u2)‖=‖u1‖+‖u2‖. 设K1K2E中的两个锥,易证K1×K2为乘积空间E×E中的锥. 如果0<ri<ri(i=1,2),定义Kr,R=(K1)r1,R1×(K2)r2,R2,其中

如果算子Ai:K1×K2E(i=1,2)是全连续的,则称A=(A1,A2)是全连续的. 文献[2]在乘积空间中证明了由锥导出的半序型锥拉伸与压缩不动点定理. 泛函型锥拉伸与压缩全连续算子的不动点定理及其应用见文献[3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]. 本文在乘积空间证明了几个凹泛函型混合锥拉伸与压缩条件下的不动点定理. 设α:K→[0,+∞)为凹泛函,如果α(λx)=λα(x),∀λ≥0,∀x∈K,则称α正齐次.

1 主要结论

以下均假设A=(A1,A2):Kr,RK1× K2全连续,αi:Ki→[0,+∞)是非负正齐次凹泛函,αi(x)>0,∀xKi\{θ} (i=1,2).

定理1 如果下列条件满足:

u1‖=r1α1(A1(u))≥α1(u1),

u1‖=R1α1(A1(u))≤α1(u1),

u2‖=r2α2(A2(u))≥α2(u2),

u2‖=R2α2(A2(u))≤α2(u2),
其中u=(u1,u2)∈Kr,R,那么AKr,R中存在不动点.

证明 令h=(h1,h2)∈K1×K2,hiθ (i=1,2). 定义A :K1×K2K1×K2如下(其中记ri]= ri/‖ui,[Ri]=Ri/‖ui (i=1,2)):

如果u1=0或u2=0,则A (u)=h;

如果0<‖u1‖<r1且0<‖u2‖<r2,则A(u)=min{[r1]-1,[r2]-1}A([r1]u1,[r2]u2)+ (1-min{[r1]-1,[r2]-1})h;

如果0<‖u1‖<r1r2≤‖u2‖≤R2,则

A(u)=[r1]-1A([r1]u1,u2)+(1-[r1]-1)h;

如果0<‖u1‖<r1且‖u2‖>R2,则

A(u)=[r1]-1A([r1]u1,[R2]u2)+(1-[r1]-1)h;

如果r1≤‖u1‖≤R1且0<‖u2‖<r2,则

A(u)=[r2]-1A(u1,[r2]u2)+(1-[r2]-1)h;

如果r1≤‖u1‖≤R1r2≤‖u2‖≤R2,则 A(u)=A(u);

如果r1≤‖u1‖≤R1且‖u2‖>R2,则

A(u)=A(u1,[R2]u2);

如果‖u1‖>R1且0<‖u2‖<r2,则

A(u)=[r2]-1A([R1]u1,[r2]u2)+(1-[r2]-1)h;

如果‖u1‖>R1r2≤‖u2‖≤R2,则

A(u)=A([R1]u1,u2);

如果‖u1‖>R1且‖u2‖>R2,则

A(u)=A([R1]u1,[R2]u2).

易见A是连续的,由于它的值域包含于紧集co {A(Kr,R)∪{h}},所以又是紧的. 根据Schauder不动点定理[1],存在uK1×K2使得A(u)=u. 下面证明u∈Kr,R,否则与条件矛盾,于是A(u)=u. 由于hi≠0<(i=1,2),显然‖u1‖>0<且‖u2‖>0.

如果0<‖u1‖<r1且0<‖u2‖<r2,不妨设min{[r1]-1,[r2]-1}=[r1]-1,那么

u=A(u)=[r1]-1A([r1]u1,[r2]u2)+(1-[r1]-1)h,

u1=[r1]-1A1([r1]u1,[r2]u2)+(1-[r1]-1)h1.

α1的凹性知

α1(u1)≥[r1]-1α1(A1([r1]u1,[r2]u2))+(1-[r1]-1)α1(h1)>[r1]-1α1(A1([r1]u1,[r2]u2)),

α1([r1]u1)=[r1]α1(u1)>α1(A1([r1]u1,[r2]u2)).

如果0<‖u1‖<r1r2≤‖u2‖≤R2,则

u=A(u)=[r1]-1A([r1]u1,u2)+(1-[r1]-1)h,

u1=[r1]-1A1([r1]u1,u2)+(1-[r1]-1)h1.

α1的凹性知

α1(u1)≥[r1]-1α1(A1([r1]u1,u2))+(1-[r1]-1)α1(h1)>[r1]-1α1(A1([r1]u1,u2)),

α1([r1]u1)=[r1]α1(u1)>α1(A1([r1]u1,u2)).

如果0<‖u1‖<r1且‖u2‖>R2,那么

u=A(u)=[r1]-1A([r1]u1,[R2]u2)+(1-[r1]-1)h,

u1=[r1]-1A1([r1]u1,[R2]u2)+(1-[r1]-1)h1.

α1的凹性知

α1(u1)≥[r1]-1α1(A1([r1]u1,[R2]u2))+(1-[r1]-1)α1(h1)>[r1]-1α1(A1([r1]u1,[R2]u2)),

α1([r1]u1)=[r1]α1(u1)>α1(A1([r1]u1,[R2]u2)).

如果r1≤‖u1‖≤R1且0<‖u2‖<r2,则

u=A(u)=[r2]-1A(u1,[r2]u2)+(1-[r2]-1)h,

u2=[r2]-1A2(u1,[r2]u2)+(1-[r2]-1)h2.

α2的凹性知

α2(u2)≥[r2]-1α2(A2(u1,[r2]u2))+(1-[r2]-1)α2(h2)>[r2]-1α2(A2(u1,[r2]u2)),

α2([r2]u2)=[r2]α2(u2)>α2(A2(u1,[r2]u2)).

如果r1≤‖u1‖≤R1且‖u2‖>R2,那么

u=A(u)=A(u1,[R2]u2),从而

u2=A2(u1,[R2]u2). 于是

α2(A2(u1,[R2]u2))=α2(u2)>[R2]α2(u2)=α2([R2]u2).

如果‖u1‖>R1且0<‖u2‖<r2,那么

u=A(u)=[r2]-1A([R1]u1,[r2]u2)+(1-[r2]-1)h,

u2=[r2]-1A2([R1]u1,[r2]u2)+(1-[r2]-1)h2,

α2的凹性知

α2(u2)≥[r2]-1α2(A2([R1]u1,[r2]u2))+(1-[r2]-1)α2(h2)> [r2]-1α2(A2([R1]u1,[r2]u2)),

α2([r2]u2)=[r2]α2(u2)>α2(A1([R1]u1,[r2]u2)).

如果‖u1‖>R1r2≤‖u2‖≤R2,那么

u=A(u)=A([R1]u1,u2),

从而u1=A1([R1]u1,u2). 于是

α1(A1([R1]u1,u2))=α1(u1)>[R1]α1(u1)=α1([R1]u1).

如果‖u1‖>R1且‖u1‖>R1,那么

u=A(u)=A([R1]u1,[R2]u2),

从而ui=Ai([R1]u1,[R2]u2). 于是

αi(Ai([R1]u1,[R2]u2))=αi(ui)>[Ri]αi(ui)=αi([Ri]ui)(i=1,2).

定理2 如果下列条件满足:

u1‖=r1α1(A1(u))≥α1(u1),

u1‖=R1α1(A1(u))≤α1(u1),

u2‖=r2α2(A2(u))≤α2(u2),

u2‖=R2α2(A2(u))≥α2(u2),
其中u=(u1,u2)∈Kr,R,那么AKr,R中存在不动点.

证明 因为∀u2∈(K2)r2,R2,

([R2]+[r2]-1)u2∈(K2)r2,R2,

所以可设A1*:Kr,RK1×K2

A1*(u)=A1(u1,([R2]+[r2]-1)u2);

可设A2*:Kr,RK1×K2

显然当‖u2‖=r2时,‖([R2]+[r2]-1)× u2‖=R2;当‖u2‖=R2时,‖([R2]+[r2]-1)u2‖=r2. 因此算子A*=(A1*,A2*)满足定理1的条件,从而A*存在不动点v=(v1,v2)∈Kr,R.

u=(u1,u2)为u1=v1,

u2=(R2/‖v2‖+r2/‖v2‖-1)v2,

易见u=(u1,u2)∈Kr,RA的不动点.

定理3 如果下列条件满足:

u1‖=r1α1(A1(u))≤α1(u1),

u1‖=R1α1(A1(u))≥α1(u1),

u2‖=r2α2(A2(u))≥α2(u2),

u2‖=R2α2(A2(u))≤α2(u2),

其中u=(u1,u2)∈Kr,R,那么AKr,R中存在不动点.

证明 因为∀u1∈(K1)r1,R1,

([R1]+[r1]-1)u1∈(K1)r1,R1,

所以可设A1 **:Kr,RK1×K2

可设A2 **:Kr,RK1×K2

A2 **(u)=A2(([R1]+[r1]-1)u1,u2).

显然当 ‖u1‖=r1时,‖([R1]+[r1]-1)× u1‖=R1; 当‖u1‖=R1时,‖([R1]+[r1]-1)u1‖=r1. 因此算子A* *=(A1 **,A2 **)满足定理1的条件,从而A* *存在不动点v=(v1,v2)∈Kr,R.

u=(u1,u2)为u2=v2,

u1=(R1/‖v1‖+r1/‖v1‖-1)v1,

易见u=(u1,u2)∈Kr,RA的不动点.

定理4 如果下列条件满足:

u1‖=r1α1(A1(u))≤α1(u1),

u1‖=R1α1(A1(u))≥α1(u1),

u2‖=r2α2(A2(u))≤α2(u2),

u2‖=R2α2(A2(u))≥α2(u2),

其中u=(u1,u2)∈Kr,R,那么AKr,R中存在不动点.

证明 因为∀u1∈(K1)r1,R1,

([R1]+[r1]-1)u1∈(K1)r1,R1,

u2∈(K2)r2,R2,

([R2]+[r2]-1)u2∈(K2)r2,R2,

所以可设A1* * *:Kr,RK1×K2

可设A2* * *:Kr,RK1×K2

因此算子A* * *=(A1* * *,A2* * *)满足定理1的条件,从而A* * *存在不动点v=(v1,v2)∈Kr,R.

u=(u1,u2)为

u1=(R1/‖v1‖+r1/‖v1‖-1)v1,

u2=(R2/‖v2‖+r2/‖v2‖-1)v2,

易见u=(u1,u2)∈Kr,RA的不动点.

2 结 语

E=C(G)N维欧氏空间RN中非空有界闭集G上的连续函数空间,令

其中ε>0,G0<是G的非空闭子集. 显然KE中的锥,定义
于是α是锥K上的非负正齐次凹泛函,并且α(x)>0,∀x∈K\{θ}.

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