Banach空間中可數(shù)簇全擬-φ-漸近非擴張非自映射的強收斂定理

李小蓉

(宜賓學院 數(shù)學學院, 四川 宜賓 644000)

1 預備知識

本文假設E是實Banach空間,E*是E的對偶,C是E的非空閉凸子集,J:E→2E*是按照如下方式定義的賦范對偶映射

J(x)={f*∈E*:〈x,f*〉=

‖x‖2=‖f*‖2,x∈E}.

都有

設U={x∈E:‖x‖=1}是單位球面,稱Banach空間E是光滑的,如果對?x,y∈U,極限

存在.如果對?x,y∈U,極限一致存在,則稱E是一致光滑的.

設C是Banach空間E的一非空閉凸子集,稱映射T:C→E是非擴張的,如果對?x,y∈C都有

‖Tx-Ty‖≤‖x-y‖.

‖Tnx-Tny‖≤kn‖x-y‖, ?x,y∈C.

本文用F(T)表示T的不動點集,即F(T)={x∈C:x=Tx}.設C是Banach空間E的子集,稱映射C是E的收縮核,如果存在連續(xù)的映射P:E→C,使得Px=x,?x∈C.顯然一致凸Banach空間的每個非空閉凸子集都是E的收縮核.稱映射P:E→C是非擴張的收縮映射,如果P是非擴張的,且是C到E的收縮的映射.

現(xiàn)在假設E是光滑的、嚴格凸、自反的Banach空間,C是E的非空閉凸子集.本文用φ:E×E→R+={a∈R|a>0}表示Lyapunov函數(shù)

φ(x,y)=‖x‖2-2〈x,Jy〉+‖y‖2, ?x,y∈E.

由φ的定義可得

(‖x‖-‖y‖)2≤φ(x,y)≤

(‖x‖+‖y‖)2, ?x,y∈E,

且

φ(x,J-1(λJy+(1-λ)Jz)≤

λφ(x,y)+(1-λ)φ(x,z), ?x,y∈E.

?x∈E.

引理1.1[2]設E是嚴格凸、光滑的Banach空間,則φ(x,y)=0當且僅當x=y.

引理1.2[2]E是自反、嚴格凸、光滑的Banach空間,D是E的非空閉凸子集,則有

?x∈D,y∈E.

引理1.3[2]E是自反、嚴格凸、光滑的Banach空間,D是E的非空閉凸子集,則有

?〈z-y,J(x)-J(z)〉≥0, ?y∈D.

定義1.5設E是實Banach空間,C是E的非空閉凸子集,

1) 稱C是E的收縮核,如果存在連續(xù)函數(shù)P:E→C,使得Px=x,?x∈C;

2) 稱P:E→C為保核收縮映射,如果P2=P;

3) 稱P:E→C為非擴張的保核收縮映射,如果P是非擴張的,且為保核收縮映射.

定義1.6設P:C→E是非擴張的收縮映射,

1) 稱自映射T:C→C為擬-φ-非擴張映射,如果F(T)≠?且

φ(u,Tnx)≤φ(u,x),

?x∈C,u∈F(T),n≥1;

2) 稱T:C→E為擬-φ-非擴張非自映射,如果F(T)≠?且

φ(u,T(PT)n-1x)≤φ(u,x),

?x∈C,u∈F(T),n≥1;

3) 稱T:C→E為擬-φ-漸近非擴張非自映射,如果F(T)≠?,且存在實序列{kn}?[1,∞),kn→1使得

φ(u,T(PT)n-1x)≤knφ(u,x),

?x∈C,u∈F(T),n≥1.

注1.7由定義1.5可知,如果T:C→E是擬-φ-非擴張非自映射,則T為擬-φ-漸近非擴張非自映射(取kn=1).

引理1.8[3]設E是一致凸、光滑、自反的Banach空間,序列{xn}和{yn}?E.如果φ(xn,yn)→0,且{xn}或{yn}有界,則‖xn-yn‖→0.

φ(u,Ti(PTi)n-1x)≤φ(u,x)+νnζ(φ(u,x))+μn,

?n≥1,i≥1, ?x∈C,u∈F.

多值的和單值的全擬-φ-漸近非擴張映像的例子見文獻[4],該文中已指出,通常的廣義漸近非擴張映像是全擬-φ-漸近非擴張映像的特例.

定義1.10稱非自映射T:C→E為一致L-Lipschitz連續(xù),如果存在常數(shù)L>0使得

‖T(PT)n-1x-T(PT)n-1y‖≤L‖x-y‖,

?x,y∈C, ?n≥1.

引理1.11設E是一致光滑、嚴格凸,且具有Kadec-Klee性質的Banach空間,C是E的非空閉凸子集.T:C→E是全擬-φ-漸近非擴張非自映射,ζ:R+={a∈R|a>0}→R+={a∈R|a>0}是嚴格增的連續(xù)函數(shù),其中,ζ(0)=0,且當n→∞時,非負實序列νn→0,μn→0,如果μ1=0,則T的不動點集F(T)是閉集合.

證明令序列{un}?F(T),其中當n→∞時,un→u.由于T是全擬-φ-漸近非擴張非自映射,且μ1=0,故可得

故φ(u,Tu)=0,即u∈F(T).因此F(T)是閉集合.

關于漸近非擴張自映射或非自映射的強弱收斂、相對非擴張、擬-φ-非擴張、擬-φ-漸近非擴張自映射和非自映射的強弱收斂性,參見文獻[5-29].

2 主要結論

定理2.1設E是一致光滑、嚴格凸、自反,且具有Kadec-Klee性質的Banach空間,C是E的非空閉凸子集.令{Ti:C→E,i=1,2,3,…}是一簇一致全擬-φ-漸近非擴張非自映射,對?i≥1,Ti都是一致Li-Lipschitz連續(xù)映射.設實序列{αn}?[0,1],{βn}?(0,1)滿足以下條件:

設xn是按以下方式生成的序列

?x1∈E,C1=C;

yn,i=J-1[αnJx1+(1-αn)(βnJxn+

(1-βn)JTi(PTi)n-1xn],i≥1;

φ(z,x1)+(1-αn)φ(z,xn)+ξn};

其中

證明分5步證明此定理.

1) 首先證F和Cn是C的閉凸子集.

由引理1.11知F(Ti)是閉集合,又已知F是C的有界凸子集,故F是C的閉凸子集.

設序列{un}?F(T),且un→u.由于Ti:C→E是一簇全擬-φ-漸近非擴張非自映射,故

由已知C1=C是閉凸的.設當n≥2時Cn是閉凸集,下面證Cn+1是閉凸集.

φ(z,x1)+(1-αn)φ(z,xn)+ξn}=

(1-αn)φ(z,xn)+ξn}∩Cn=

2(1-αn)〈z,Jxn〉-2〈z,Jyn,i〉≤

αn‖x1‖2+(1-αn)‖xn‖2-‖yn,i‖2}∩Cn,

故Cn+1是閉凸集.

2) 證明對?n≥1有F?C.

顯然有F?C1=C.設對某個n≥2有F?Cn,令

wn,i=J-1(βnJxn+(1-βn)JTi(PTi)n-1xn),

對任何u∈F?Cn有

φ(u,yn,i)=φ(u,J-1(αnJx1+(1-αn)Jwn,i))≤

αnφ(u,x1)+(1-αn)φ(u,wn,i),

和

φ(u,wn,i)=φ(u,J-1(βnJxn+

(1-βn)JTi(PTi)n-1xn))≤

βnφ(u,xn)+(1-βn)φ(u,Ti(PTi)n-1xn)≤

βnφ(u,xn)+(1-βn)(φ(u,xn)+

νnζ(φ(u,xn))+μn)=

φ(u,xn)+(1-βn)(νnζ(φ(u,xn))+μn).

因此可得

{φ(u,xn)+(1-βn)(νnζ(φ(u,xn))+μn)}≤

αnφ(u,x1)+(1-αn){φ(u,xn)+

αnφ(u,x1)+(1-αn)φ(u,xn)+

αnφ(u,x1)+(1-αn)φ(u,xn)+ξn,

其中

即u∈Cn+1,因此F?Cn+1.

3) 證明序列{xn}?C強收斂于C中一點u*.

〈xn-y,Jx1-Jxn〉≥0, ?y∈Cn.

又因為對?n≥1,F?Cn,故可得

〈xn-u,Jx1-Jxn〉, ?u∈F.

由引理1.2知,對?n≥1,?u∈F有

φ(u,x1)-φ(u,xn)≤φ(u,x1).

φ(xni,x1)≤φ(u*,x1), ?ni≥1.

由于范數(shù)‖·‖是弱下半連續(xù)的,故可得

‖u*‖2-2〈u*,Jx1〉+‖x1‖2=φ(u*,x1),

故

則有

且‖xni‖→‖u*‖.因為xni?u*和E具有Kadec-Klee性質可得

由φ(xn,x1)收斂和

可得

φ(xn,x1)=φ(u*,x1).

現(xiàn)設存在序列{xnj}?{xn}也滿足xnj→q,則由引理1.2可得

φ(u*,x1)-φ(u*,x1)=0,

故u*=q且

因此

4)證明u*∈F.

因為xn+1∈Cn+1和αn→0,故

(1-αn)φ(xn+1,xn)+ξn→0,n→∞.

由于xn→u*,且由引理1.7可得,對?i≥1有

φ(u,Ti(PTi)n-1xn)≤φ(u,xn)+

νnζ(φ(u,xn))+μn,

故{Ti(PTi)n-1xn}是一致有界的.

‖wn,i‖=‖J-1(βnJxn+

(1-βn)JTi(PTi)n-1xn)‖≤

βn‖xn‖+(1-βn)‖Ti(PTi)n-1xn‖≤

‖xn‖+‖Ti(PTi)n-1xn‖,

即{wn,i}是一致有界序列.

由假設αn→0,對?i≥1可得

因為J在E*的每個有界閉子集下是一致連續(xù)的,對?i≥1可得

J在E的每個子集下是一致連續(xù)的可得

(1-βn)(JTi(PTi)n-1xn-Ju*)‖=

由條件(ii)可得

由于J是一致連續(xù)的,故

?i≥1.

對?i≥1,Ti是一致Li-Lipschitz連續(xù)可得

‖Ti(PTi)nxn-Ti(PTi)n-1xn‖≤

‖Ti(PTi)nxn-Ti(PTi)n-1xn+1‖+

‖Ti(PTi)nxn+1-xn+1‖+

‖xn+1-xn‖+‖xn-Ti(PTi)n-1xn‖≤

(Li+1)‖xn+1-xn‖+‖Ti(PTi)nxn+1-xn+1‖+

‖xn-Ti(PTi)n-1xn‖.

因為

且xn→u*,因此可得

且

即

由TiP的連續(xù)性,可得TiPu*=u*.因為u*∈C,Pu*=u*,故Tiu*=u*.由于i的任意性知u*∈F.

注2.2定理2.1與參考文獻中的結果不同之處在于:本文在具有Kadec-Klee性質的一致光滑和嚴格凸Banach空間中研究了一類完全擬-φ-漸近非擴張非自映像簇的公共不動點的迭代逼近問題.而在參考文獻中討論的是:在一致凸和一致光滑的Banach空間中漸近非擴張非自映像(或廣義漸近非擴張非自映像簇)的公共不動點的迭代逼近問題.本文的結果改進和推廣了這些文獻中的相應的結果.

致謝宜賓學院青年基金項目(2010Q29)對本文給予了資助,謹致謝意.

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