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UA SIE545 优化理论基础1 例题1 常见的凸集

發布時間:2025/4/14 编程问答 35 豆豆
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UA SIE545 優化理論基礎1 例題1 常見的凸集

    • 一些例題

在優化理論中,我們主要討論下面幾種凸集:

  • 超平面:{x:pTx=β}\{x:p^Tx=\beta\}{x:pTx=β}
  • 半空間:{x:pTx≤β}\{x:p^Tx \le \beta\}{x:pTxβ}
  • 范數錐:{(x,t):∥x∥≤t}\{(x,t):\left\| x\right\| \le t\}{(x,t):xt}
  • 多面體:{x:Ax≤b}\{x:Ax \le b\}{x:Axb}
  • 與超平面和半空間相關的結論是分離定理,主要討論點與凸集、凸集與凸集的分離;與多面體相關的結論是多面體的表示定理以及由此導出的線性規劃的單純形法。這兩類結論我們在例題2和例題3中討論,這次例題我們討論一般性的凸集的判定與性質。

    一些例題

    例1 S1={x:A1x≤b1},S2={x:A2x≤b2}S_1 = \{x:A_1x \le b_1\},S_2 = \{x:A_2x \le b_2\}S1?={x:A1?xb1?},S2?={x:A2?xb2?}是兩個非空集合,記S=S1∪S2S = S_1\cup S_2S=S1?S2?, S^={y+z:A1y≤b1λ1,A2z≤b2λ2,λ1+λ2=1,λ1,λ2≥0}\hat S = \{y+z:A_1y \le b_1\lambda_1,A_2z \le b_2\lambda_2,\lambda_1+\lambda_2 = 1,\lambda_1,\lambda_2 \ge 0\}S^={y+z:A1?yb1?λ1?,A2?zb2?λ2?,λ1?+λ2?=1,λ1?,λ2?0}。假設S1,S2S_1,S_2S1?,S2?有界,證明conv(S)=S^conv(S)=\hat Sconv(S)=S^

    證明
    1)conv(S)?S^conv(S) \subset \hat Sconv(S)?S^:
    ?x∈conv(S)\forall x \in conv(S)?xconv(S), ?x1,?,xm∈S\exists x_1,\cdots,x_m \in S?x1?,?,xm?S, ?λ1,λm≥0,λ1+?+λm=1\exists \lambda_1,\lambda_m \ge 0,\lambda_1+\cdots+\lambda_m=1?λ1?,λm?0,λ1?+?+λm?=1,
    x=∑i=1mλixix = \sum_{i=1}^m \lambda_i x_ix=i=1m?λi?xi?

    不失一般性,我們假設x1,?,xk∈S1,xk+1,?,xm∈S2,k∈{0,?,m}x_1,\cdots,x_k \in S_1,x_{k+1},\cdots,x_m \in S_2,k \in \{0,\cdots,m\}x1?,?,xk?S1?,xk+1?,?,xm?S2?,k{0,?,m}
    x=∑i=1kλixi+∑i=k+1mλixi=(∑i=1kλi)∑i=1kλixi(∑i=1kλi)+(∑i=k+1mλi)∑i=k+1mλixi(∑i=k+1mλi)∈S^x = \sum_{i=1}^k \lambda_i x_i+\sum_{i=k+1}^m \lambda_i x_i \\= \left(\sum_{i=1}^k \lambda_i \right) \sum_{i=1}^k \frac{\lambda_ix_i}{\left(\sum_{i=1}^k \lambda_i \right)}+\left(\sum_{i=k+1}^m \lambda_i \right) \sum_{i=k+1}^m \frac{\lambda_ix_i}{\left(\sum_{i=k+1}^m \lambda_i \right)} \in \hat Sx=i=1k?λi?xi?+i=k+1m?λi?xi?=(i=1k?λi?)i=1k?(i=1k?λi?)λi?xi??+(i=k+1m?λi?)i=k+1m?(i=k+1m?λi?)λi?xi??S^
    2)conv(S)?S^conv(S) \supset \hat Sconv(S)?S^:
    ?x∈S^\forall x \in \hat S?xS^, ?y∈S1,z∈S2\exists y \in S_1,z \in S_2?yS1?,zS2?, x=λ1y+λ2z,λ1+λ2=1,λ1,λ2≥0x = \lambda_1 y + \lambda_2 z, \lambda_1+\lambda_2 = 1,\lambda_1,\lambda_2 \ge 0x=λ1?y+λ2?z,λ1?+λ2?=1,λ1?,λ2?0,根據下面例9的結論,命題得證。

    例2 SSS是歐式空間的非空子集,xˉ∈S\bar{x} \in SxˉS, C={y:y=λ(x?xˉ),λ≥0,x∈S}C = \{y:y = \lambda(x-\bar{x}),\lambda \ge 0,x \in S\}C={y:y=λ(x?xˉ),λ0,xS}

  • 證明CCC是凸錐
  • 如果SSS是凸集,則CCC是凸集
  • 證明
    1)?μ≥0,?y∈C?{0}\forall \mu \ge 0,\forall y \in C \setminus \{0\}?μ0,?yC?{0}, ?x∈S?{xˉ},λ>0\exists x \in S\setminus \{\bar x\},\lambda >0?xS?{xˉ},λ>0, y=λ(x?xˉ)y=\lambda (x-\bar{x})y=λ(x?xˉ)μy=(μλ)(x?xˉ)∈C\mu y = (\mu \lambda)(x-\bar{x}) \in Cμy=(μλ)(x?xˉ)C
    2)?y1,y2∈C\forall y_1,y_2 \in C?y1?,y2?C, ?x1,x2∈S,λ1,λ2≥0\exists x_1,x_2 \in S,\lambda_1,\lambda_2 \ge 0?x1?,x2?S,λ1?,λ2?0, y1=λ1(x1?xˉ),y2=λ2(x2?xˉ)y_1 = \lambda_1(x_1-\bar x),y_2 = \lambda_2 (x_2-\bar x)y1?=λ1?(x1??xˉ),y2?=λ2?(x2??xˉ). ?μ∈[0,1]\forall \mu \in [0,1]?μ[0,1],
    μy1+(1?μ)y2=μλ1(x1?xˉ)+(1?μ)λ2(x2?xˉ)=μλ1x1+(1?μ)λ2x2?[μλ1+(1?μ)λ2]xˉ=[μλ1μλ1+(1?μ)λ2x1+(1?μ)λ2μλ1+(1?μ)λ2x2?xˉ][μλ1+(1?μ)λ2]∈C\mu y_1 + (1-\mu)y_2 = \mu \lambda_1 (x_1-\bar x)+(1-\mu)\lambda_2 (x_2 -\bar x) \\ = \mu \lambda_1 x_1 + (1-\mu)\lambda_2 x_2 - [\mu \lambda_1+ (1-\mu)\lambda_2 ]\bar x \\ = \left[ \frac{\mu \lambda_1}{\mu \lambda_1+ (1-\mu)\lambda_2}x_1 + \frac{ (1-\mu)\lambda_2}{\mu \lambda_1+ (1-\mu)\lambda_2}x_2-\bar x \right][\mu \lambda_1+ (1-\mu)\lambda_2 ] \in Cμy1?+(1?μ)y2?=μλ1?(x1??xˉ)+(1?μ)λ2?(x2??xˉ)=μλ1?x1?+(1?μ)λ2?x2??[μλ1?+(1?μ)λ2?]xˉ=[μλ1?+(1?μ)λ2?μλ1??x1?+μλ1?+(1?μ)λ2?(1?μ)λ2??x2??xˉ][μλ1?+(1?μ)λ2?]C

    例3 C1,C2C_1,C_2C1?,C2?是歐式空間中的凸錐,證明C1⊕C2C_1 \oplus C_2C1?C2?也是凸錐,并且C1⊕C2=conv(C1∪C2)C_1 \oplus C_2=conv(C_1 \cup C_2)C1?C2?=conv(C1?C2?)
    證明
    1)?y+z∈C1⊕C2\forall y+z \in C_1 \oplus C_2?y+zC1?C2?, y∈C1,z∈C2y \in C_1,z \in C_2yC1?,zC2?, ?λ≥0\forall \lambda \ge 0?λ0,
    λ(y+z)=λy+λz∈C1⊕C2\lambda (y+z) = \lambda y + \lambda z \in C_1 \oplus C_2λ(y+z)=λy+λzC1?C2?

    2)Check C1⊕C2=conv(C1∪C2)C_1 \oplus C_2=conv(C_1 \cup C_2)C1?C2?=conv(C1?C2?)
    i) C1⊕C2?conv(C1∪C2)C_1 \oplus C_2\subset conv(C_1 \cup C_2)C1?C2??conv(C1?C2?): ?x∈C1⊕C2,?y∈C1,z∈C2,λ≥0\forall x \in C_1 \oplus C_2, \exists y \in C_1,z \in C_2,\lambda \ge 0?xC1?C2?,?yC1?,zC2?,λ0, x=λy+λz=122λy+122λz∈conv(C1∪C2)x = \lambda y + \lambda z = \frac{1}{2} 2\lambda y+ \frac{1}{2}2\lambda z \in conv(C_1 \cup C_2)x=λy+λz=21?2λy+21?2λzconv(C1?C2?)
    ii) C1⊕C2?conv(C1∪C2)C_1 \oplus C_2\supset conv(C_1 \cup C_2)C1?C2??conv(C1?C2?): ?x∈conv(C1∪C2)\forall x \in conv(C_1 \cup C_2)?xconv(C1?C2?), ?y∈C1,z∈C2,λ∈[0,1]\exists y \in C_1,z \in C_2,\lambda \in [0,1]?yC1?,zC2?,λ[0,1], x=λy+(1?λ)z∈C1⊕C2x = \lambda y + (1-\lambda) z \in C_1 \oplus C_2x=λy+(1?λ)zC1?C2?

    例4 假設A,BA,BA,B是兩個凸集,證明conv(A∪B)={λA+(1?λ)B:λ∈[0,1]}conv(A \cup B)=\{\lambda A+(1-\lambda)B:\lambda \in [0,1]\}conv(AB)={λA+(1?λ)B:λ[0,1]}

    證明
    1)conv(A∪B)?{λA+(1?λ)B:λ∈[0,1]}?Sconv(A \cup B) \subset \{\lambda A+(1-\lambda)B:\lambda \in [0,1]\}\triangleq Sconv(AB)?{λA+(1?λ)B:λ[0,1]}?S:
    ?x∈conv(A∪B)\forall x \in conv(A \cup B)?xconv(AB), ?x1,?,xk∈A,xk+1,?,xm∈B,k∈{0,?,m}\exists x_1,\cdots,x_k \in A,x_{k+1},\cdots,x_m \in B,k \in \{0,\cdots,m\}?x1?,?,xk?A,xk+1?,?,xm?B,k{0,?,m}, ?λ1,λm≥0,λ1+?+λm=1\exists \lambda_1,\lambda_m \ge 0,\lambda_1+\cdots+\lambda_m=1?λ1?,λm?0,λ1?+?+λm?=1,
    x=∑i=1kλixi+∑i=k+1mλixi=(∑i=1kλi)∑i=1kλixi(∑i=1kλi)+(∑i=k+1mλi)∑i=k+1mλixi(∑i=k+1mλi)∈Sx = \sum_{i=1}^k \lambda_i x_i+\sum_{i=k+1}^m \lambda_i x_i \\ =\left(\sum_{i=1}^k \lambda_i \right) \sum_{i=1}^k \frac{\lambda_ix_i}{\left(\sum_{i=1}^k \lambda_i \right)}+\left(\sum_{i=k+1}^m \lambda_i \right) \sum_{i=k+1}^m \frac{\lambda_ix_i}{\left(\sum_{i=k+1}^m \lambda_i \right)} \in Sx=i=1k?λi?xi?+i=k+1m?λi?xi?=(i=1k?λi?)i=1k?(i=1k?λi?)λi?xi??+(i=k+1m?λi?)i=k+1m?(i=k+1m?λi?)λi?xi??S

    2)conv(A∪B)?{λA+(1?λ)B:λ∈[0,1]}?Sconv(A \cup B) \supset \{\lambda A+(1-\lambda)B:\lambda \in [0,1]\}\triangleq Sconv(AB)?{λA+(1?λ)B:λ[0,1]}?S:
    ?x∈S\forall x \in S?xS, ?y∈A,z∈B,λ∈[0,1]\exists y \in A,z \in B,\lambda \in [0,1]?yA,zB,λ[0,1], x=λy+(1?λ)z∈conv(A∪B)x = \lambda y+(1-\lambda)z \in conv(A \cup B)x=λy+(1?λ)zconv(AB) (y,z∈A∪By,z \in A\cup By,zAB)

    總結

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