趙 方 浩
(南京財(cái)經(jīng)大學(xué) 應(yīng)用數(shù)學(xué)學(xué)院,江蘇 南京 210003)
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Hilbert空間中一類(lèi)上下界均衡問(wèn)題解的存在性
趙 方 浩
(南京財(cái)經(jīng)大學(xué) 應(yīng)用數(shù)學(xué)學(xué)院,江蘇 南京 210003)
在Hilbert空間中討論了一類(lèi)上下界均衡問(wèn)題解的存在性,在研究過(guò)程中,把上下界均衡問(wèn)題轉(zhuǎn)化為帶對(duì)稱(chēng)界均衡問(wèn)題,在定義域?yàn)榫o集和非緊集兩種情況下,分別給出了相關(guān)存在性定理,推廣了文獻(xiàn)[1,6]中的相應(yīng)結(jié)果。
上下界;均衡問(wèn)題;緊性;非緊性
總假設(shè)X為Hilbert空間,X上的內(nèi)積記為〈·,·〉,由該內(nèi)積誘導(dǎo)的范數(shù)記為‖·‖,由該范數(shù)誘導(dǎo)的度量記為d。 X中非空子集構(gòu)成的集族記為2X,D為X的子集,0為X中的零元,R為實(shí)數(shù)集。
注1帶上下界均衡問(wèn)題總可以通平移變換轉(zhuǎn)化成帶對(duì)稱(chēng)界均衡問(wèn)題,因?yàn)?/p>
c1≤f(x,y)≤c2?
由注1可知帶上下界均衡問(wèn)題總可以通過(guò)平移變換轉(zhuǎn)化成帶對(duì)稱(chēng)界均衡問(wèn)題,故下面只需研究帶對(duì)稱(chēng)界均衡問(wèn)題(SBEP)。
下面給出了Hilbert空間中均衡問(wèn)題解的存在性,即引理1。
引理1設(shè)D為X的閉子集,如果映射f∶D×D→R滿(mǎn)足下列條件:
(1)對(duì)?x∈D,f(x,·)下半連續(xù)且下有界,
(2)對(duì)?x∈D,f(x,x)=0,
(3)對(duì)?x,y,z∈D,都有f(z,x)≤f(z,y)+f(y,x),
證明不失一般性,只證明ε=1的情況。記
F(x)={y∈D∶f(x,y)+‖y-x‖≤0}
由(1)知,對(duì)?x∈D,F(xiàn)(x)是閉集;由(2)知,
x∈F(x),故F(x)是非空集合。設(shè)y∈F(x),即f(x,y)+‖y-x‖≤0,令z∈F(x)[即f(y,z)+‖y-z‖≤0]不等式兩邊相加,由(3)可得,
0≥f(x,y)+‖y-x‖+f(y,z)+‖y-z‖≥
f(x,z)+‖z-x‖,即z∈F(x),因此y∈F(x)意味著F(y)?F(x)。
即
‖x-z‖≤-v(x),?z∈F(x)。
特別地,如果x1,x2∈F(x),‖x1-x2‖≤‖x-x1‖+‖x-x2‖≤-v(x)-v(x)=
-2v(x),則diam[F(x)]≤-2v(x),?x∈D。
固定x0∈D,存在x1∈F(x0)使得f(x0,x1)≤v(x0)+2-1。記x2為F(x1)的任意一點(diǎn)有
f(x1,x2)≤v(x1)+2-2。重復(fù)上述步驟,可以定義一個(gè)序列{xn}∈D,且xn+1∈F(xn)使得
f(xn,xn+1)≤v(xn)+2-(n+1)。
注意到
因此v(xn+1)≥v(xn)-f(xn,xn+1),
且-v(xn)≤-f(xn,xn+1)+2-(n+1)≤
[v(xn+1)-v(xn)]+2-(n+1),
定理1設(shè)D為X的緊子集,如果映射f∶D×D→R滿(mǎn)足下列條件:
(1) 對(duì)?x∈D,f(x,·)下半連續(xù),
(2) 對(duì)?x∈D,f(x,x)=0,
(3) 對(duì)?x,y,z∈D,都有f(z,x)≤f(z,y)+f(y,x),
證明對(duì)?n∈N+,令xn∈D是一個(gè)ε均衡點(diǎn) (由引理1可知該點(diǎn)存在),即
注2定理1推廣了文獻(xiàn)[1]中推論3.2。
推論1設(shè)D為X的緊子集,如果映射f∶
D×D→R滿(mǎn)足下列條件:
(1)對(duì)?x∈D,f(x,·)下半連續(xù),
(2)對(duì)?y∈D,f(·,y)上半連續(xù),
(3)對(duì)?x∈D,f(x,x)=c,
(4)對(duì)?x,y,z∈D,都有|f(z,x)|+c≥
證明令F(x,y)=c-|f(x,y)|,由(1)可得對(duì)?x∈D,F(xiàn)(x,·)下半連續(xù)。由(2)得,對(duì)?y∈D,F(xiàn)(·,y)上半連續(xù),由(3)可得,對(duì)?x∈D,F(xiàn)(x,x)=c-c=0;下證對(duì)?x,y,z∈D,都有F(z,x)≤F(z,y)+F(y,x),由(4)可知
|f(z,x)|+c≥|f(z,y)|+|f(y,x)| ,
-|f(z,x)|-c≤-|f(z,y)|-|f(y,x)| ,
所以
c-|f(z,x)|≤c-|f(z,y)|+c-|f(y,x)| ,
F(z,x)≤F(z,y)+F(y,x)。
注3推論1同時(shí)是對(duì)文獻(xiàn)[6]中定理5.1的改進(jìn)和推廣 。
定理2設(shè)D為X的子集,‖·‖為X上的內(nèi)積誘導(dǎo)的范數(shù),若映射f∶D×D→R滿(mǎn)足:
(1)對(duì)?x∈D,f(x,·)下半連續(xù)且下有界,
(2)對(duì)?x∈D,f(x,x)=0,
(3)對(duì)?x,y,z∈D,都有f(z,x)≤f(z,y)+f(y,x),
(4)對(duì)?y∈D,f(·,y)上半連續(xù),
證明不失一般性,不妨設(shè)Kr非空,對(duì)?x∈D,考慮下面的非空集合
S(x)={y∈D∶‖y‖≤‖x‖∶f(x,y)≤0},
對(duì)?x,y∈D,y∈S(x),有S(y)?S(x)。事實(shí)上,對(duì)z∈S(y),有‖z‖≤‖y‖≤‖x‖,由(3)得f(x,z)≤f(x,y)+f(y,z)≤0;另一方面,由于Kr是緊的,由(1)可得,對(duì)?x∈D,S(x)?Kr是緊集,進(jìn)一步,由定理1,?xr∈Kr使得
f(xr,y)≥0,?y∈Kr,
(1)
下面分兩種情況討論。
情況1a≤r,令y0∈S(x),使得‖y0‖=a≤r,則有f(x,y0)≤0。由于f(xr,x)≤0且由條件(3)可得f(xr,y0)≤f(xr,x)+f(x,y0)≤0,此不等式與(1)式矛盾。
注4定理2是對(duì)文獻(xiàn)[1]中定理4.1的推廣。
推論2設(shè)D為X的緊子集,‖·‖為X上的內(nèi)積誘導(dǎo)的范數(shù),若映射f∶D×D→R滿(mǎn)足:
(1) 對(duì)?x∈D,f(x,·)下半連續(xù)且下有界,
(2) 對(duì)?x∈D,f(x,x)=c,
(3) 對(duì)?x,y,z∈D,都有|f(z,x)|+c≥
|f(z,y)|+|f(y,x)| ,
(4) 對(duì)?y∈D,f(·,y)上半連續(xù),
證明令F(x,y)=c-|f(x,y)|,易證F(x,y)滿(mǎn)足定理2中的條件(1)~(4)其證法與推論1的相同,此處省略。下面證明F(x,y)滿(mǎn)足定理2的條件(5),因?yàn)?r>0,對(duì)?x∈DKr都?y∈D,‖y‖≤‖x‖,使得
f(x,y)≥c,f(x,y)≥c,可知|f(x,y)|≥c,所以
注5定理2是對(duì)文獻(xiàn)[6]中定理5.2的改進(jìn)和推廣。
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Existence of Equilibrium Problems with Lower and Upper Bounds in Hilbert Space
ZHAO Fang-hao
(Department of Applied Mathematics, Nanjing University of Finance and Economics,Nanjing, Jiangsu 210003, China)
In this paper, we discuss the existence of equilibrium problems with the upper and lower bounds. In the process of study, the equilibrium problem of the upper and lower bounds is transformed into symmetric bounded equilibrium one. In both cases of the domain of compact set and non-compact set, some related theorems are given respectively. In addition, the corresponding results in literature [1,6] are generalized.
upper and lower bounds; equilibrium problems; compact; non-compact
2015-12-29
趙方浩,男,河南開(kāi)封人,南京財(cái)經(jīng)大學(xué)應(yīng)用數(shù)學(xué)學(xué)院碩士研究生,研究方向?yàn)榉蔷€(xiàn)性分析與經(jīng)濟(jì)應(yīng)用。E-mail: zfhzs2015@163.com
時(shí)間:2016-8-17 11:31
http://www.cnki.net/kcms/detail/34.1150.N.20160817.1131.002.html
O29
A
1007-4260(2016)03-0005-03
10.13757/j.cnki.cn34-1150/n.2016.03.002