凸优化4：Operations that preserve convexity

1、交集 Intersection

若S1,S2S_1,S_2S1,S2为凸，则S1∩S2S_1 \cap S_2S1∩S2为凸

扩展：若SαS_{\alpha}Sα为凸集，则∀a∈A,∩⁡α∈ASα\forall \space a \in A, {\underset {\alpha \in A}{\operatorname {\cap} }} S_{\alpha}∀ a∈A,α∈A∩Sα为凸集

2、仿射函数 Affine function

定义：afunctionf:Rn→Rmisaffineifitisasumofalinearfunctionandaconstranta \space function \space f:R^n \rightarrow R^m \space is \space affine \space if \space it \space is \space a \space sum \space of \space a \space linear \space function \space and \space a \space constranta function f:Rn→Rm is affine if it is a sum of a linear function and a constrant

数学描述为：
f:Rn→Rm是仿射函数，当f(x)=Ax+b,A∈Rm×n,b∈Rmf:R^n \rightarrow R^m是仿射函数，当f(x)=Ax+b,A \in R^{m \times n},b \in R^m f:Rn→Rm是仿射函数，当f(x)=Ax+b,A∈Rm×n,b∈Rm
定理：若S∈RnS \in R^nS∈Rn为凸集，f:Rn→Rmf:R^n \rightarrow R^mf:Rn→Rm为仿射函数，则：f(S)={f(x)∣x∈S}f(S)=\{f(x) \mid x \in S\}f(S)={f(x)∣x∈S}为凸

逆仿射(inverse affine)：g:Rn→Rmg:R^n \rightarrow R^mg:Rn→Rm是仿射函数，则g−1(S)={x∣g(x)∈S}g^{-1}(S)=\{x \mid g(x) \in S\}g−1(S)={x∣g(x)∈S}

定理：缩放与移位是保持凸性的
S为凸集，αS={αx∣x∈S},S+a={x+a∣x∈S}都为凸S1,S2为凸集，S1+S2={x+y∣x∈S1,y∈S2},S1×S2={(x,y)∣x∈S1,y∈S2}都为凸S为凸集，\alpha S=\{\alpha x \mid x \in S\},S+a=\{x+a \mid x \in S\}都为凸 \\ S_1,S_2为凸集，S_1+S_2=\{x+y\mid x \in S_1,y \in S_2\},\\ S_1 \times S_2=\{(x,y)\mid x\in S_1,y\in S_2\} 都为凸 S为凸集，αS={αx∣x∈S},S+a={x+a∣x∈S}都为凸S1,S2为凸集，S1+S2={x+y∣x∈S1,y∈S2},S1×S2={(x,y)∣x∈S1,y∈S2}都为凸
Ex1：线性矩阵的不变形 LMI( linear matrix inequality)
A(X)=x1A1+⋯+xnAn⪯B⇔A(X)−B⪯0半负定B,Ai,xi∈Sm(对称矩阵)A(X)=x_1A_1+\cdots +x_nA_n \preceq B \\ \Leftrightarrow A(X)-B \preceq 0\space\space 半负定\\ B,A_i,x_i \in S^m(对称矩阵) A(X)=x1A1+⋯+xnAn⪯B⇔A(X)−B⪯0 半负定B,Ai,xi∈Sm(对称矩阵)
定义仿射变换f(x)=B−A(x)f(x)=B-A(x)f(x)=B−A(x),S+nS^n_+S+n为凸集，f−1(S+n)={x∣B−A(x)⪰0}f^{-1}(S^n_+)=\{x \mid B-A(x) \succeq 0\}f−1(S+n)={x∣B−A(x)⪰0}为凸

结论：故A(X)−B⪯0A(X)-B \preceq 0A(X)−B⪯0为凸集

Ex2：椭球是球的仿射映射
ε={x∣(x−xc)Tp−1(x−xc)≤1},P∈S++n单位球{u∣∣∣u∣∣2≤1}令f(u)=p12u+xc,其中p12p12=p\varepsilon = \{x \mid (x-x_c)^Tp^{-1}(x-x_c) \leq 1\},P \in S^n_{++}\\ 单位球 \space\space \{u \mid ||u||_2 \leq 1\}\\ 令f(u)=p^{\frac{1}{2}}u+x_c,其中p^{\frac{1}{2}}p^{\frac{1}{2}}=p\\ ε={x∣(x−xc)Tp−1(x−xc)≤1},P∈S++n单位球 {u∣∣∣u∣∣2≤1}令f(u)=p21u+xc,其中p21p21=p
构造新集合
{f(u)∣∣∣u∣∣2≤1}={p12u+xc∣∣∣u∣∣2≤1}\{f(u)\mid ||u||_2 \leq 1\}\\ =\{p^{\frac{1}{2}}u+x_c \mid ||u||_2 \leq 1\}\\ {f(u)∣∣∣u∣∣2≤1}={p21u+xc∣∣∣u∣∣2≤1}
令x=p12u+xcx=p^{\frac{1}{2}}u+x_cx=p21u+xc，即u=p−12(x−xc),P∈S++nu=p^{-\frac{1}{2}}(x-x_c),P\in S^n_{++}u=p−21(x−xc),P∈S++n，则上式：
={x∣∣∣p−12(x−xc)∣∣2≤1}={x∣(x−xc)TP−1(x−xc)≤1}=\{x\mid ||p^{-\frac{1}{2}}(x-x_c)||_2 \leq 1\} \\ =\{x\mid (x-x_c)^TP^{-1}(x-x_c) \leq 1\} ={x∣∣∣p−21(x−xc)∣∣2≤1}={x∣(x−xc)TP−1(x−xc)≤1}

这里无法证明下式：
∣∣p−12(x−xc)∣∣2≤1(x−xc)T(p−12)T(p−12)(x−xc)≤1无法证明(p−12)T(p−12)=P−1||p^{-\frac{1}{2}}(x-x_c)||_2 \leq 1 \\ (x-x_c)^T(p^{-\frac{1}{2}})^T(p^{-\frac{1}{2}})(x-x_c) \leq 1\\ 无法证明(p^{-\frac{1}{2}})^T(p^{-\frac{1}{2}})=P^{-1} ∣∣p−21(x−xc)∣∣2≤1(x−xc)T(p−21)T(p−21)(x−xc)≤1无法证明(p−21)T(p−21)=P−1

3、透视函数 Perspection function

透视函数P:Rn+1→RnP:R^{n+1} \rightarrow R^nP:Rn+1→Rn, domP=Rn×R++dom P=R^n \times R_{++}domP=Rn×R++， R++为整数，dom为domain，定义域R_{++}为整数，dom为domain，定义域R++为整数，dom为domain，定义域

数学描述为：
P(z,t)=zt,z∈Rn,t∈R++P(z,t)=\frac{z}{t},z\in R^n,t\in R_{++} \\ P(z,t)=tz,z∈Rn,t∈R++

定理：任意一个凸集经过透视函数之后仍为凸集

Ex1：Rn+1R^{n+1}Rn+1内的线段
X={x~,xn+1}∈{Rn,R++},Y={y~,yn+1}∈{Rn,R++}θ≥0,线段可表示为:θx+(1−θ)y证明：Rn+1线段⟶PRn线段x⟶PP(x),y⟶PP(y)θx+(1−θ)y⟶PP(θx+(1−θ)y)P(θx+(1−θ)y)=θx~+(1−θ)y~θxn+1+(1−θ)yn+1=θxn+1θxn+1+(1−θ)yn+1x~xn+1+(1−θ)yn+1θxn+1+(1−θ)yn+1y~yn+1⟶PμP(x)+(1−μ)P(y),0≤μ≤1上式属于n维空间中的线段，线段是凸集X=\{\widetilde{x},x_{n+1}\} \in \{R^n,R_{++}\},Y=\{\widetilde{y},y_{n+1}\}\in \{R^n,R_{++}\}\\ \theta \geq 0,线段可表示为:\theta x+(1-\theta)y\\ 证明：R^{n+1}线段 \stackrel{P}{\longrightarrow} R^n线段\\ x \stackrel{P}{\longrightarrow} P(x),y \stackrel{P}{\longrightarrow} P(y)\\ \theta x+(1-\theta)y \stackrel{P}{\longrightarrow} P(\theta x+(1-\theta)y)\\ P(\theta x+(1-\theta)y)=\frac{\theta \widetilde{x}+(1-\theta)\widetilde{y}}{\theta x_{n+1}+(1-\theta)y_{n+1}}\\ =\frac{\theta x_{n+1}}{\theta x_{n+1}+(1-\theta)y_{n+1}} \frac{\widetilde{x}}{x_{n+1}}+\frac{(1-\theta) y_{n+1}}{\theta x_{n+1}+(1-\theta)y_{n+1}} \frac{\widetilde{y}}{y_{n+1}}\\ \stackrel{P}{\longrightarrow} \mu P(x)+(1-\mu)P(y),0 \leq \mu \leq 1 \\ 上式属于n维空间中的线段，线段是凸集 X={x,xn+1}∈{Rn,R++},Y={y,yn+1}∈{Rn,R++}θ≥0,线段可表示为:θx+(1−θ)y证明：Rn+1线段⟶PRn线段x⟶PP(x),y⟶PP(y)θx+(1−θ)y⟶PP(θx+(1−θ)y)P(θx+(1−θ)y)=θxn+1+(1−θ)yn+1θx+(1−θ)y=θxn+1+(1−θ)yn+1θxn+1xn+1x+θxn+1+(1−θ)yn+1(1−θ)yn+1yn+1y⟶PμP(x)+(1−μ)P(y),0≤μ≤1上式属于n维空间中的线段，线段是凸集
Ex2：任意凸集的反透视映射仍是凸集
P−1(C)={(x,t)∈Rn+1∣xt∈C,t>0}(x,t)∈P−1(C),(y,s)∈P−1(C),0≤θ≤1P−1((x,t),(y,s))=θx+(1−θ)yθt+(1−θ)s=(θtθt+(1−θ)s)xt+(1−θtθt+(1−θ)s)ys⟶P−1μP(x,t)+(1−μ)P(y,s)∈CP^{-1}(C)=\{(x,t) \in R^{n+1} \mid \frac{x}{t} \in C,t>0\} \\ (x,t) \in P^{-1}(C), (y,s) \in P^{-1}(C), 0 \leq \theta \leq 1 \\ P^{-1}((x,t),(y,s))=\frac{\theta x+(1-\theta)y}{\theta t+(1-\theta)s}\\ =(\frac{\theta t}{\theta t+(1-\theta)s})\frac{x}{t}+(1-\frac{\theta t}{\theta t+(1-\theta)s})\frac{y}{s}\\ \stackrel{P^{-1}}{\longrightarrow} \mu P(x,t)+(1-\mu)P(y,s) \in C P−1(C)={(x,t)∈Rn+1∣tx∈C,t>0}(x,t)∈P−1(C),(y,s)∈P−1(C),0≤θ≤1P−1((x,t),(y,s))=θt+(1−θ)sθx+(1−θ)y=(θt+(1−θ)sθt)tx+(1−θt+(1−θ)sθt)sy⟶P−1μP(x,t)+(1−μ)P(y,s)∈C

Ex3：将仿射函数与透视函数结合⟶\longrightarrow⟶线性分段函数
g:Rn⟶Rm+1为仿射映射g(x)=[ACT]x+[bd],A∈Rm×n,b∈Rm,C∈Rn,d∈RP:Rm+1⟶Rm线性分段函数：P:Rn⟶Rm=P∙gf(x)=Ax+bCTx+d,domf={x∣CTx+d>0}两个线性变换的除法也可以保持凸性g:R^n \longrightarrow R^{m+1} 为仿射映射\\ g(x)=\left[ \begin{matrix} A \\ C^T \end{matrix} \right]x+\left[ \begin{matrix} b \\ d \end{matrix} \right],A\in R^{m \times n},b\in R^m,C\in R^n,d\in R \\ P:R^{m+1} \longrightarrow R^m\\ 线性分段函数：P:R^n \longrightarrow R^m=P \bullet g\\ f(x)=\frac{Ax+b}{C^Tx+d},dom \space f=\{x \mid C^Tx+d > 0\}\\ 两个线性变换的除法也可以保持凸性 g:Rn⟶Rm+1为仿射映射g(x)=[ACT]x+[bd],A∈Rm×n,b∈Rm,C∈Rn,d∈RP:Rm+1⟶Rm线性分段函数：P:Rn⟶Rm=P∙gf(x)=CTx+dAx+b,dom f={x∣CTx+d>0}两个线性变换的除法也可以保持凸性