原始问题
假设\(f(x),c_i(x),h_j(x)\)是定义在\(R^n\)上的连续可微函数,考虑约束最优化问题
\[\begin{aligned} \mathop{min}\limits_{x \in R^n}\ &f(x) \\ s.t.\ &c_i(x) \leq 0,i = 1,2,\cdots,k \\ &h_j(x) = 0,j = 1,2,\cdots,l \end{aligned} \]称此约束最优化问题为原始最优化问题或原始问题,首先引入广义拉格朗日函数:
\[L(x,\alpha,\beta) = f(x) + \sum_{i = 1}^k\alpha_ic_i(x) + \sum_{j = 1}^l\beta_jh_j(x) \]这里,\(x = (x^{(1)},x^{(2)},\cdots,x^{(n)})^T \in R^n,\alpha_i,\beta_j\)是拉格朗日乘子,\(\alpha_i \geq 0\),考虑\(x\)的函数:
\[\theta_P(x) = \mathop{max}\limits_{\alpha,\beta,\alpha_i \geq 0}L(x,\alpha,\beta) \]这里的下标\(P\)表示原始问题,因为\(c_i(x) \leq 0,\alpha_i \geq 0\),所以\(\sum_{i = 1}^k\alpha_ic_i(x) \leq 0\),类似的\(\sum_{j = 1}^l\beta_jh_j(x) = 0\),故可以得到以下结论:
\[\theta_P(x) = \begin{cases} f(x) & x满足原始约束条件 \\ +\infty & otherwise \end{cases} \]考虑极小化问题:
\[\mathop{min}\limits_x\theta_P(x) = \mathop{min}\limits_{x} \mathop{max}\limits_{\alpha,\beta,\alpha_i \geq 0} L(x,\alpha,\beta) \]上式和原始的最优化问题是等价的,也就是和原问题有相同的解,\(\mathop{min}\limits_{x} \mathop{max}\limits_{\alpha,\beta,\alpha_i \geq 0} L(x,\alpha,\beta)\)称为广义拉格朗日函数的极小极大问题,为了方便,定义原始问题的最优值\(p^* = \mathop{min}\limits_{x}\theta_P(x)\)称为原始问题的值
对偶问题
定义
\[\theta_D(\alpha,\beta) = \mathop{min}\limits_xL(x,\alpha,\beta) \]再考虑极大化\(\theta_D(\alpha,\beta) = \mathop{min}\limits_xL(x,\alpha,\beta)\),即:
\[\mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0}\theta_D(\alpha,\beta) = \mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0} \mathop{min}\limits_xL(x,\alpha,\beta) \]问题\(\mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0} \mathop{min}\limits_xL(x,\alpha,\beta)\)称为广义拉格朗日函数的极大极小问题,可以将极大极小问题表示为约束最优化问题:
\[\begin{aligned} \mathop{max}\limits_{\alpha,\beta}\ &\theta_D(\alpha,\beta) = \mathop{max}\limits_{\alpha,\beta} \mathop{min}\limits_xL(x,\alpha,\beta) \\ s.t.\ &\alpha_i \geq 0,i = 1,2,\cdots,k \end{aligned} \]对偶问题的最优值\(d^* = \mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0}\ \theta_D(\alpha,\beta)\)
原始问题和对偶问题的关系
定理1
若原始问题和对偶问题都有最优值,则:
\[d^* = \mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0} \mathop{min}\limits_xL(x,\alpha,\beta) \leq \mathop{min}\limits_{x} \mathop{max}\limits_{\alpha,\beta,\alpha_i \geq 0} L(x,\alpha,\beta) = p^* \]证明如下:
对任意的\(\alpha,\beta,x\)有
所以:
\[\theta_D(\alpha,\beta) \leq \theta_P(x) \]由于原始问题和对偶问题均有最优值,所以:
\[ \mathop{max}\limits_{\alpha,\beta;\alpha_i \geq 0} \theta_D(\alpha,\beta) \leq \mathop{min}\limits_x \theta_P(x) \]得证。
定理2
考虑原始问题和对偶问题,假设函数\(f(x)\)和\(c_i(x)\)是凸函数,\(h_j(x)\)是仿射函数,假设不等式约束\(c_i(x)\)是严格可行的,及存在\(x\)对所有的\(i\)有\(c_i(x) \lt 0\),则存在\(x^*,\alpha^*,\beta^*\),使得\(x^*\)是原问题的解,\(\alpha^*,\beta^*\)是对偶问题的解,并且有:
\[p^* = d^* = L(x^*,\alpha^*,\beta^*) \]定理3
假设函数\(f(x)\)和\(c_i(x)\)是凸函数,\(h_j(x)\)是仿射函数,不等式约束\(c_i(x)\)是严格可行的,那么有\(x^*\)是原问题的解,\(\alpha^*,\beta^*\)是对偶问题的解的充分必要条件是\(x^*,\alpha^*,\beta^*\)满足下面的KKT条件(Karush-Kuhn-Tucker):
\[\nabla_xL(x^*,\alpha^*,\beta^*) = 0 \\ \alpha^*_ic_i(x^*) = 0,i = 1,2,\cdots,k \\ c_i(x^*) \leq 0,i = 1,2,\cdots,k \\ \alpha_i^* \geq 0,i = 1,2,\cdots,k \\ h_j(x^*) = 0,j = 1,2,\cdots,l \]\(\alpha^*_ic_i(x^*) = 0,i = 1,2,\cdots,k\)称为KKT的对偶互补条件,由此条件可知:若\(\alpha_i^* \gt 0\),则\(c_i(x^*) = 0\).
标签:拉格朗,geq,limits,学习,beta,对偶性,mathop,theta,alpha From: https://www.cnblogs.com/eryoyo/p/16704750.html