一類變分包含組解的強收斂定理

史 杰,馮世強,何中全

(西華師范大學(xué)數(shù)學(xué)與信息學(xué)院，中國 南充 637009)

設(shè)E是Banach空間,E*是其對偶空間,正規(guī)對偶映射J:E→2E*如下定義:

J(x)=f∈E*:〈x,f〉=‖x‖2=‖f‖2,x∈E，

其中,當(dāng)E是嚴格凸的光滑的Banach空間時,J和J-1是單值的[1].

設(shè)E是光滑的Banach空間,函數(shù)φ:E×E→R如下定義:

φ(x,y)=‖x‖2-2〈x,Jy〉+‖y‖2,?x,y∈E.

由于〈x,Jy〉≤‖x‖‖Jy‖=‖x‖‖y‖,于是(‖x‖-‖y‖)2≤φ(x,y),即φ(x,y)≥0.

設(shè)E是嚴格凸且光滑的自反Banach空間,C是E中非空閉凸集.那么對任意的x∈E,存在唯一的x0∈C使得[1]

φ(x0,x)=minφ(y,x),y∈C.

本文稱x0是x在C上的投影.定義投影算子∏C:E→2C如下

∏C(x)=y∈C:φ(y,x)=minφ(z,x),z∈C,x∈E.

設(shè)E是Banach空間,E*是其對偶空間,集值映射M:E→2E*,若滿足

〈x-y,f-g〉≥0,?x,y∈E,f∈Mx,g∈My，

則稱M是單調(diào)的.若M滿足

(1)M是單調(diào)的;

(2)對(x,f)∈E×E*,〈x-y,f-g〉≥0,?y∈E,g∈My,必有f∈Mx.

則稱M是極大單調(diào)的.

設(shè)E是Banach空間,E*是其對偶空間.若映射T:E→E滿足

φ(Tx,Ty)≤〈Tx-Ty,Jx-Jy〉,?x,y∈E，

則稱T為確定非擴張的.易得,T為確定非擴張的等價于

〈Tx-Ty,JTx-JTy〉≤〈Tx-Ty,Jx-Jy〉,?x,y∈E.

若映射B:E→E*滿足

(1)η:E→E為任意一個確定非擴張映射;

(2)對任意λ≥0,有

〈η(J-1(Jx-λBx))-η(J-1(Jy-λBy)),Bx-By〉≥0,?x,y∈E.

則稱B為廣義單調(diào)的.顯然,廣義單調(diào)映射必是單調(diào)的.

設(shè)Bi:E→E*是單值映射,Mi:E→2E*是多值映射,i=1,2,…,N,θ是零元素.本文研究如下的變分包含組問題(VISP):

VISP 求x∈E,使得

當(dāng)Bi≡B,Mi≡M,VISP問題變?yōu)槿缦伦兎职瑔栴}(Ⅵ):

VI 設(shè)B:E→E*是單值映射,M:E→2E*是多值映射,θ是零元素,求x∈E,使得θ∈B(x)+M(x)成立.

下面給出本文所需的一些引理.

引理1[2]設(shè)M:E→2E*是極大單調(diào)映射,B:E→E*是Lipshitz連續(xù)映射,則S=M+B:E→2E*是極大單調(diào)映射.

引理2[1]設(shè)E是嚴格凸的光滑實自反Banach空間,C是E中非空閉凸集,令x∈E,那么對任意y∈C,有φ(y,∏Cx)+φ(∏Cx,x)≤φ(y,x).

引理3[3]設(shè)E是嚴格凸的光滑實Banach空間，xn、yn都是E中子列，xn或yn是有界的且

φ(xn,yn)→0,n→∞,那么有xn-yn→0,n→∞.

引理4[1]令C是一光滑實Banach空間E凸集,令x∈E,那么x0∈∏Cx當(dāng)且僅當(dāng)

〈z-x0,Jx0-Jx〉≥0,?z∈C.

1 主要結(jié)果

本節(jié)將通過構(gòu)造非擴張映射,得到新的迭代算法，使之產(chǎn)生的序列收斂到變分包含組問題的解.本文工作推廣和改進了文獻[2]、[4～10]中的一些結(jié)果.

其中Mi:E→2E*是極大單調(diào)映射,Bi:E→E*是Lipshitz連續(xù)映射,λi>0.

即

再由Mi的極大單調(diào)性得

(1)

又由于φ(z2,z1)+φ(z1,z2)≥0得

0≤‖z2‖2+‖z1‖2-2〈z2,Jz1〉+‖z1‖2+‖z2‖2-2〈z1,Jz2〉=2(‖z2‖2+‖z1‖2-

〈z2,Jz1〉-〈z1,Jz2〉)=2(〈z2,Jz2〉+〈z1,Jz1〉-〈z2,Jz1〉-〈z1,Jz2〉)=2〈z2-z1,Jz2-Jz1〉.

即

〈z2-z1,Jz2-Jz1〉≥0.

(2)

引理6設(shè)E是Banach空間,E*是其對偶空間,對一切i=1,…,N,?x∈E,有下面結(jié)論成立:

對任意x,y∈E,由Mi的極大單調(diào)性得

由Mi的極大單調(diào)性得

即

因此

證畢.

接下來構(gòu)造如下算法W：

其中,0≤ηn≤e對任意0≤e<1,i=0,1，…，n.

證分4步來證明.

(3)

由于xn=∏Cn(x0),根據(jù)引理2,得

φ(xn,x0)≤φ(p,x0)-φ(p,xn)≤φ(p,x0),φ(xn,x0)≤φ(xn+1,x0),?n≥0.

φ(xn+m,xn)=φ(xn+m,∏Cn(x0))≤φ(xn+m,x0)-φ(xn,x0).

故有

2‖p‖‖Jzn-Jxn‖).

任取(vj,gj)∈Bj+Mj,即gj-Bj(vj)∈Mj(vj).由Mj的極大單調(diào)性得

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