UA MATH567 高维统计II 随机向量7 Grothendieck不等式
UA MATH567 高维统计II 随机向量7 Grothendieck不等式
上一讲我们介绍了用半正定规划近似一个整数规划的方法,要证明这种近似与原整数规划解的大小关系,我们需要Grothendieck不等式,所以这一讲我们证明这个不等式:
Grothendieck不等式
AAA是m×nm \times nm×n的实矩阵,xi,yj∈{−1,1}x_i,y_j \in \{-1,1\}xi,yj∈{−1,1},假设∣∑i,jAijxiyj∣≤1|\sum_{i,j}A_{ij}x_iy_j| \le 1∣∑i,jAijxiyj∣≤1,则∀H\forall H∀H(Hilbert space),∀ui,vj∈H\forall u_i,v_j \in H∀ui,vj∈H,∥ui∥=∥vj∥=1\left\| u_i \right\|=\left\| v_j \right\|=1∥ui∥=∥vj∥=1,
∣∑i,jAi,j⟨ui,vj⟩∣≤K,K≤1.783|\sum_{i,j}A_{i,j}\langle u_i,v_j \rangle| \le K,K \le 1.783∣i,j∑Ai,j⟨ui,vj⟩∣≤K,K≤1.783
证明
首先我们分析一下这个不等式的条件,xi,yj∈{−1,1}x_i,y_j \in \{-1,1\}xi,yj∈{−1,1},∣∑i,jAijxiyj∣≤1|\sum_{i,j}A_{ij}x_iy_j| \le 1∣∑i,jAijxiyj∣≤1,它等价于
∣∑i,jAijxiyj∣≤max∣xi∣max∣yj∣|\sum_{i,j}A_{ij}x_iy_j| \le \max |x_i| \max |y_j|∣i,j∑Aijxiyj∣≤max∣xi∣max∣yj∣
如果xi,yj∈{−1,1}x_i,y_j \in \{-1,1\}xi,yj∈{−1,1},∣∑i,jAijxiyj∣≤1|\sum_{i,j}A_{ij}x_iy_j| \le 1∣∑i,jAijxiyj∣≤1成立,显然
∣∑i,jAijxiyj∣≤max∣xi∣max∣yj∣|\sum_{i,j}A_{ij}x_iy_j| \le \max |x_i| \max |y_j|∣i,j∑Aijxiyj∣≤max∣xi∣max∣yj∣
如果∣∑i,jAijxiyj∣≤max∣xi∣max∣yj∣|\sum_{i,j}A_{ij}x_iy_j| \le \max |x_i| \max |y_j|∣∑i,jAijxiyj∣≤max∣xi∣max∣yj∣成立,则
∣∑i,jAijximax∣xi∣yjmax∣yj∣∣≤1|\sum_{i,j}A_{ij}\frac{x_i}{\max |x_i|} \frac{y_j}{\max |y_j|}| \le 1∣i,j∑Aijmax∣xi∣ximax∣yj∣yj∣≤1
显然ximax∣xi∣,yjmax∣yj∣\frac{x_i}{\max |x_i|}, \frac{y_j}{\max |y_j|}max∣xi∣xi,max∣yj∣yj都在[−1,1][-1,1][−1,1]上取值,注意到∑i,jAijxiyj\sum_{i,j}A_{ij}x_iy_j∑i,jAijxiyj是一个线性函数,它的最优解一定是角点解,所以xi,yj∈{−1,1}x_i,y_j \in \{-1,1\}xi,yj∈{−1,1},∣∑i,jAijxiyj∣≤1|\sum_{i,j}A_{ij}x_iy_j| \le 1∣∑i,jAijxiyj∣≤1;
与之类似,我们可以分析一下结论,∀H\forall H∀H(Hilbert space),∀ui,vj∈H\forall u_i,v_j \in H∀ui,vj∈H,∥ui∥=∥vj∥=1\left\| u_i \right\|=\left\| v_j \right\|=1∥ui∥=∥vj∥=1,
∣∑i,jAi,j⟨ui,vj⟩∣≤K,K≤1.783|\sum_{i,j}A_{i,j}\langle u_i,v_j \rangle| \le K,K \le 1.783∣i,j∑Ai,j⟨ui,vj⟩∣≤K,K≤1.783
等价于∀H\forall H∀H(Hilbert space),∀ui,vj∈H\forall u_i,v_j \in H∀ui,vj∈H
∣∑i,jAi,j⟨ui,vj⟩∣≤Kmaxi∥ui∥maxj∥vj∥|\sum_{i,j}A_{i,j}\langle u_i,v_j \rangle| \le K\max_i\left\| u_i \right\| \max_j \left\| v_j \right\|∣i,j∑Ai,j⟨ui,vj⟩∣≤Kimax∥ui∥jmax∥vj∥
第一步,Reduction (先证明一个更弱的结果,即KKK与AAA相关)
引入K(A)K(A)K(A),
K(A)=inf{K:∣∑i,jAi,j⟨ui,vj⟩∣≤Kmaxi∥ui∥maxj∥vj∥}K(A) = \inf\{K:|\sum_{i,j}A_{i,j}\langle u_i,v_j \rangle| \le K\max_i\left\| u_i \right\| \max_j \left\| v_j \right\|\}K(A)=inf{K:∣i,j∑Ai,j⟨ui,vj⟩∣≤Kimax∥ui∥jmax∥vj∥}
显然
K(A)≤∑∣Aij∣K(A) \le \sum |A_{ij}|K(A)≤∑∣Aij∣
假设{ui},{vj}\{u_i\},\{v_j\}{ui},{vj}是使得K(A)K(A)K(A)成为最小的ui,vj∈Hu_i,v_j \in Hui,vj∈H,从而
∑i,jAi,j⟨ui,vj⟩=K(A)\sum_{i,j}A_{i,j}\langle u_i,v_j \rangle = K(A)i,j∑Ai,j⟨ui,vj⟩=K(A)
第二步,Randomness
定义g∼N(0,IN)g \sim N(0,I_N)g∼N(0,IN),N=n+mN = n+mN=n+m,定义Ui=⟨g,ui⟩,Vj=⟨g,vj⟩U_i = \langle g,u_i\rangle,V_j = \langle g,v_j\rangleUi=⟨g,ui⟩,Vj=⟨g,vj⟩,则Ui,VjU_i,V_jUi,Vj都是标准正态随机变量,
E[UiVj]=⟨ui,vj⟩E[U_iV_j] = \langle u_i,v_j\rangleE[UiVj]=⟨ui,vj⟩
所以
K(A)=∑i,jAi,j⟨ui,vj⟩=∑i,jAi,jE[UiVj]=E∑i,jAijUiVjK(A) = \sum_{i,j}A_{i,j}\langle u_i,v_j \rangle = \sum_{i,j}A_{i,j}E[U_iV_j] = E \sum_{i,j}A_{ij}U_iV_jK(A)=i,j∑Ai,j⟨ui,vj⟩=i,j∑Ai,jE[UiVj]=Ei,j∑AijUiVj
第三步,Truncation
定义R>1R>1R>1,
Ui=Ui−+Ui+=Ui1∣Ui∣≤R+Ui1∣Ui∣>RVj=Vj−+Vj+=Vj1∣Vj∣≤R+Vj1∣Vj∣>RU_i = U_i^-+U_i^+ = U_i1_{|U_i| \le R}+U_i1_{|U_i|>R} \\ V_j= V_j^-+V_j^+ = V_j1_{|V_j| \le R}+V_j1_{|V_j|>R}Ui=Ui−+Ui+=Ui1∣Ui∣≤R+Ui1∣Ui∣>RVj=Vj−+Vj+=Vj1∣Vj∣≤R+Vj1∣Vj∣>R
显然Ui−,Vj−U_i^-,V_j^-Ui−,Vj−都是有界的,考虑
∥Ui+∥22≤2(R+1R)12πe−R2/2≤4R2\left\| U_i^+ \right\|_2^2 \le 2(R+\frac{1}{R})\frac{1}{\sqrt{2\pi}}e^{-R^2/2} \le \frac{4}{R^2}∥∥Ui+∥∥22≤2(R+R1)2π1e−R2/2≤R24
同样的
∥Vj+∥22≤4R2\left\| V_j^+ \right\|_2^2 \le \frac{4}{R^2}∥∥Vj+∥∥22≤R24
第四步,计算
K(A)=E∑i,jAijUiVj=E∑i,jAij(Ui−+Ui+)(Vj−+Vj+)=∑i,jAijEUi−Vj−+∑i,jAijEUi−Vj++∑i,jAijEUi+Vj−+∑i,jAijEUi+Vj+K(A) = E \sum_{i,j}A_{ij}U_iV_j = E \sum_{i,j}A_{ij}(U_i^-+U_i^+)(V_j^-+V_j^+) \\ = \sum_{i,j}A_{ij}EU_i^-V_j^-+\sum_{i,j}A_{ij}EU_i^-V_j^++\sum_{i,j}A_{ij}EU_i^+V_j^-+\sum_{i,j}A_{ij}EU_i^+V_j^+ K(A)=Ei,j∑AijUiVj=Ei,j∑Aij(Ui−+Ui+)(Vj−+Vj+)=i,j∑AijEUi−Vj−+i,j∑AijEUi−Vj++i,j∑AijEUi+Vj−+i,j∑AijEUi+Vj+
先分析一下第三项,根据Cauchy-Schwarz不等式
∑i,jAijEUi+Vj−≤∑i,jAij∥Ui+∥2∥Vj−∥2≤K(A)max∥Ui+∥2max∥Vj−∥2≤2K(A)R\sum_{i,j}A_{ij}EU_i^+V_j^- \le \sum_{i,j}A_{ij}\left\| U_i^+ \right\|_2 \left\| V_j^- \right\|_2 \\ \le K(A) \max \left\| U_i^+ \right\|_2 \max \left\| V_j^- \right\|_2 \le \frac{2K(A)}{R}i,j∑AijEUi+Vj−≤i,j∑Aij∥∥Ui+∥∥2∥∥Vj−∥∥2≤K(A)max∥∥Ui+∥∥2max∥∥Vj−∥∥2≤R2K(A)
第二项与之类似,所以
K(A)≤R2+6K(A)RK(A) \le R^2+\frac{6K(A)}{R}K(A)≤R2+R6K(A)
注意到这时的K(A)K(A)K(A)就和AAA没有关系了,于是我们记为KKK,考虑RRR的不同取值,我们可以得到KKK的不同的上界。
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