DOI:10.12677/AAM.2022.1110792,PDF,HTML,XML,被引量下载: 285浏览: 520国家自然科学基金支持
作者:苏妍妍,彭定涛^*：贵州大学，数学与统计学院，贵州 贵阳
关键词:稀疏加组稀疏优化问题；非光滑非凸优化；折叠凹惩罚函数；光滑化方法；方向稳定点；Sparse plus Group Sparse Optimization Problem；Nonconvex and Nonsmooth Optimization；Folded Concave Penalty Function；Smoothing Method；Stationary Points

文章引用：苏妍妍, 彭定涛. 稀疏加组稀疏优化问题的方向稳定点及其光滑化方法[J]. 应用数学进展, 2022, 11(10): 7464-7477. https://doi.org/10.12677/AAM.2022.1110792

1. 引言

稀疏性是自然界中普遍存在的一种性质，利用该性质能够实现对信息的充分压缩，从而减少所需储存空间和信息的传输量，或是从海量的信息中过滤噪声影响提炼有效信息，随着大数据时代的到来，稀疏优化在压缩感知、变量选择、图像恢复等领域引起了广泛的关注，这些问题的主要目的是寻求欠定线性或非线性方程组的最稀疏解。

通常的稀疏性是对向量的每个分量而言的，而组稀疏是把向量的分量进行分组再考查各组是否整体为零，它体现为向量中非零分量或零分量会集中出现在某几个区域。组稀疏优化是一类具有分组结构的稀疏优化问题，利用变量之间结构关系的先验信息对变量进行分组分块，进而提高结果的可解释性和预测性能。对于一个n维变量 $x\in {ℝ}^n$，我们可以利用先验信息将x分成K组： $x={\left(x_{\left(1\right)}^{\top },\cdots ,x_{\left(K\right)}^{\top }\right)}^{\top }\in {ℝ}^n$，其中 $x_{\left(j\right)}={\left(x_{\left(j\right)1},\cdots ,x_{\left(j\right)n_j}\right)}^{\top }\in {ℝ}^{n_j}$表示x的第j个分组， $x_{\left(j\right)i}\in ℝ$表示 $x_{\left(j\right)}$的第i个元素，第j组中分量的个数为 $n_j$， ${\displaystyle {\sum }_{j=1}^k{n}_j}=n$。近年来，组稀疏结构在机器学习、信号图像处理、数理统计、模式识别、生物信息学等领域引起了广泛的关注 [1] [2] [3] [4] [5]。

在应用中，有时真实信号/数据不仅具有稀疏结构，同时也具有组稀疏结构，这就需要考虑如下稀疏加组稀疏优化问题：

$\underset{x\in {ℝ}^n}{\min }f\left(x\right):=\mathcal{L}\left(x\right)+{\lambda }_1{\Vert x\Vert }_0+{\lambda }_2{\Vert x\Vert }_{2,0}$(1)

其中 ${\lambda }_1,{\lambda }_2>0$， $\mathcal{L}:{ℝ}^n\to ℝ$连续可微， ${\Vert x\Vert }_0:=\#\left\{x_{\left(j\right)i}:x_{\left(j\right)i}\ne 0,j=1,\cdots ,K;i=1,\cdots ,n_j\right\}=${x中非零分量的个数}，而 ${\Vert x\Vert }_{2,0}:=\#\left\{x_{\left(j\right)}:\Vert x_{\left(j\right)}\Vert \ne 0,j=1,\cdots ,K\right\}=${x中非零组的个数}。

正则项中 ${\mathcal{l}}_0$-范数与 ${\mathcal{l}}_{2,0}$-范数的存在使得上述稀疏加组稀疏优化问题本质上是一类非凸、非光滑、非Lipschitz，甚至是不连续的NP难问题 [4] [6]。一种常见的处理方法是研究其松弛模型。 ${\mathcal{l}}_1$-范数是 ${\mathcal{l}}_0$-范数最常用的凸松弛，受到很多研究，人们提出了许多求解大规模 ${\mathcal{l}}_1$正则问题的高效算法 [7] [8] [9]。但 ${\mathcal{l}}_1$-松弛容易导致过度松弛而产生有偏估计，模型的解也不具备orcale性质 [10] [11] [12]。Fan和Li [11] [12] 提出了一种折叠凹惩罚，并证明了该非凸优化能够得到具有无偏性和orcale性质的解。随后研究者们给出了一系列连续的非凸惩罚函数，包括对数罚 [11]、SCAD罚(Smoothly Clipped Absolute Deviation Penalty) [11]、MCP (Minimax Concave Penalty) [13]、Capped- ${\mathcal{l}}_1$罚 [14] [15]、分数罚 [16]、硬阈值罚(HTP) [17] [18] 和桥罚 [19] 等。

对稀疏加组稀疏优化问题(1)，本文考虑以下松弛模型：

$\begin{array}{l}\underset{x\in {\text{R}}^n}{\min }f\left(x\right):=\mathcal{L}\left(x\right)+{\lambda }_1{\displaystyle \underset{i=1}{\overset{n}{\sum }}{\phi }_1}\left(\left|x_i\right|\right)+{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\phi }_2}\left(\Vert x_{\left(j\right)}\Vert \right)\\ =\mathcal{L}\left(x\right)+{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}{\lambda }_1}{\phi }_1\left(\left|x_{\left(j\right)i}\right|\right)+{\lambda }_2{\phi }_2\left(\Vert x_{\left(j\right)}\Vert \right)\right]},\end{array}$(2)

其中 ${\lambda }_1,{\lambda }_2>0$， ${\phi }_{\mathcal{l}}:R_{+}\to R_{+}\left(\mathcal{l}=1,2\right)$为满足以下条件的折叠凹罚函数：

${\phi }_{\mathcal{l}}$是局部Lipschitz连续的，在 $\left[0,\infty \right)$上单调不减，且 ${{\phi }^{\prime }}_{\mathcal{l}}\left(0\right):={{\phi }^{\prime }}_{\mathcal{l}}\left(0+\right)>0$；

${\phi }_{\mathcal{l}}\left(0\right)=0$，且当 $t>0$时， ${\phi }_{\mathcal{l}}\left(t\right)>0$。

本文结构如下，在第二节中，我们介绍方向稳定点和局部最优性质，并给出上述松弛模型的方向导数的表达式和方向稳定点的特征；第三节对松弛模型构造光滑化逼近函数，并给出了光滑化问题的一阶稳定点的刻画及性质；第四节研究光滑化问题一阶稳定点与松弛问题的方向稳定点的一致性，为使用光滑化方法求解松弛模型提供理论保证。

符号标记： $\mathcal{J}\left(x\right):=\left\{j\in \left\{1,\cdots ,K\right\}:x_{\left(j\right)}\ne 0\right\}$表示x中非零分组的下标， ${\mathcal{I}}_j\left(x\right):=\left\{i\in \left\{1,\cdots ,n_j\right\}:x_{\left(j\right)i}\ne 0\right\}$表示x的第j组中非零元素的下标， $\left|{\mathcal{I}}_j\left(x\right)\right|$表示第j组中非零元素的个数， $x_{{\mathcal{I}}_j\left(x\right)}$表示第j组中所有非零元素按原来的顺序组成的向量， $\mathcal{D}\left(x\right)=\left\{d\in {ℝ}^n:d_{\left(j\right)i}=0,\forall j=\left\{1,\cdots ,K\right\},i\notin {\mathcal{I}}_j\left(x\right)\right\}$， ${\nabla }_{\left(j\right)i}\mathcal{L}\left(x\right)=\frac{\partial \mathcal{L}\left(x\right)}{\partial x_{\left(j\right)i}}$， ${\nabla }_{\left(j\right)}\mathcal{L}\left(x\right)=\frac{\partial \mathcal{L}\left(x\right)}{\partial x_{\left(j\right)}}={\left(\frac{\partial \mathcal{L}\left(x\right)}{\partial x_{\left(j\right)1}},\cdots ,\frac{\partial \mathcal{L}\left(x\right)}{\partial x_{\left(j\right)n_j}}\right)}^{\top }$。

2. 问题(2)的方向稳定点

2.1. 方向稳定点的局部最优性

首先，我们介绍方向稳定点的定义 [20] [21] [22] [23]。

定义2.1 (方向稳定点)设 $f:{ℝ}^n\to ℝ$在给定点 $x^{*}\in {ℝ}^n$处方向可微，若 $\forall d\in {ℝ}^n$，f的方向导数

$f\prime \left(x^{*};d\right)=\underset{t\downarrow 0}{\lim }\frac{f\left(x^{*}+td\right)-f\left(x^{*}\right)}{t}\ge 0,$

则称 $x^{*}$为f的一个方向稳定点。

注意到，若f在 $x^{*}$处可微，则 $\nabla f\left(x^{*}\right)=0$，进而 $f\prime \left(x^{*};d\right)=\langle \nabla f\left(x^{*}\right),d\rangle =0,\forall d\in {ℝ}^n$。

为了方便表示，记 $m\left(x\right)=\Vert x\Vert$，则任意 $x,d\in {ℝ}^n$， $m\left(x\right)$的方向导数为：

$m\prime \left(x;d\right)=\{\begin{array}{ll}\Vert d\Vert ,\hfill & \Vert x\Vert =0\hfill \\ \frac{\langle x,d\rangle }{\Vert x\Vert },\hfill & \Vert x\Vert \ne 0\hfill \end{array}.$(3)

特别的，当 $n=1$时， $m\left(x\right)=\Vert x\Vert =\left|x\right|$，因此

$m\prime \left(x;d\right)=\{\begin{array}{ll}\left|d\right|,\hfill & x=0\hfill \\ \frac{xd}{\left|x\right|},\hfill & x\ne 0\hfill \end{array}.$(4)

文献 [24] 中指出，方向稳定点有以下局部最优性质。

引理2.1 (方向稳定点的局部最优性)设 $f:{ℝ}^n\to ℝ$在 $x\in {ℝ}^n$上是局部Lipschitz连续且方向可微的，则有下列结论：

(i) 如果 $\hat{x}$是f的局部极小点，则 $\hat{x}$也是f的方向稳定点。

(ii) f在 $\hat{x}$满足一阶增长性条件，即存在 $\hat{x}$的一个邻域 $\mathcal{V}$和 $\delta >0$，使得

$f\left(x\right)\ge f\left(\hat{x}\right)+\delta \Vert x-\hat{x}\Vert ,\forall x\in \mathcal{V},$

等价于在 $\hat{x}$满足

$f\prime \left(\hat{x};d\right)>0,\forall d\in {ℝ}^n\backslash \left\{0\right\}.$

2.2. 方向稳定点的特征

在本小节中，我们给出了问题(2)方向导数的计算公式和方向稳定点的特征。

前述折叠凹惩罚函数大多可以写成凸差(Difference-of-Convex)形式 [20]，即两个凸函数的差。我们的分析基于惩罚函数的凸差表示。因此，引进以下假设。

假设2.1罚函数 ${\phi }_{\mathcal{l}}:{ℝ}_{+}\to {ℝ}_{+},\left(\mathcal{l}=1,2\right)$具有凸差形式： ${\phi }_{\mathcal{l}}\left(t\right)\triangleq g_{\mathcal{l}}\left(t\right)-h_{\mathcal{l}}\left(t\right)$，其中 $h_{\mathcal{l}}\left(t\right)\triangleq {\max }_{1\le v\le \overline{v_{\mathcal{l}}}}\left\{h_{\mathcal{l}v}\left(t\right)\right\}$， $g_{\mathcal{l}},h_{\mathcal{l}v}\left(1\le v\le \overline{v_{\mathcal{l}}}\right)$在 $t\in \left(0,\infty \right)$上都是可微凸函数，且满足：

${g^{\prime }}_{\mathcal{l}}\left(0\right):={g^{\prime }}_{\mathcal{l}}\left(0+\right),{h^{\prime }}_{\mathcal{l}v}\left(0\right)={h^{\prime }}_{\mathcal{l}v}\left(0+\right)\left(1\le v\le \overline{v_{\mathcal{l}}}\right).$

在假设2.1下，问题(2)可以改写成以下形式：

$\underset{x\in {ℝ}^n}{\min }f\left(x\right):=\mathcal{L}\left(x\right)+\left\{{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[g_1\left(\left|x_{\left(j\right)i}\right|\right)-h_1\left(\left|x_{\left(j\right)i}\right|\right)\right]}}+{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[g_2\left(\Vert x_{\left(j\right)}\Vert \right)-h_2\left(\Vert x_{\left(j\right)}\Vert \right)\right]}\right\}.$(5)

定理2.1若假设2.1成立，问题(2)的方向导数有以下计算公式：

$\begin{array}{l}f\prime \left(x;d\right):=\langle \nabla \mathcal{L}\left(x\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{g^{\prime }}_1\left(\left|x_{\left(j\right)i}\right|\right)-\underset{v_{\left(j\right)i}\in \mathcal{A}\left(x_{\left(j\right)i}\right)}{\max }h{\prime }_{1v_{\left(j\right)i}}\left(\left|x_{\left(j\right)i}\right|\right)\right]m\prime }}\left(x_{\left(j\right)i};d_{\left(j\right)i}\right)\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{g^{\prime }}_2\left(\Vert x_{\left(j\right)}\Vert \right)-\underset{v_{\left(j\right)}\in \mathcal{A}\left(x_{\left(j\right)}\right)}{\max }{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert x_{\left(j\right)}\Vert \right)\right]m\prime }\left(x_{\left(j\right)};d_{\left(j\right)}\right),\end{array}$(6)

其中 $\mathcal{A}\left(x_{\left(j\right)i}\right)=\left\{v_{\mathcal{l}}\in \left\{1,2,\cdots ,\overline{v_{\mathcal{l}}}\right\}:h_{v_{\mathcal{l}}}\left(\left|x_{\left(j\right)i}\right|\right)=h\left(\left|x_{\left(j\right)i}\right|\right)\right\}$， $\mathcal{A}\left(x_{\left(j\right)}\right)=\left\{v_{\mathcal{l}}\in \left\{1,2,\cdots ,\overline{v_{\mathcal{l}}}\right\}:h_{v_{\mathcal{l}}}\left(\Vert x_{\left(j\right)}\Vert \right)=h\left(\Vert x_{\left(j\right)}\Vert \right)\right\}$ $\left(\mathcal{l}=1,2\right)$。

证明. [24] (定理2.3)中给出了组稀疏惩罚问题折叠凹松弛模型的方向导数的表达式，在该表达式基础上，加上稀疏惩罚项折叠凹松弛的方向导数表达式即可，而后者可以通过方向导数的定义直接计算得到。 $\square$

下面引理说明，在问题(2)的任意方向稳定点 $\hat{x}$处，对应于 $\hat{x}$的非零元素的支撑集上损失函数 $\mathcal{L}$的梯度值可以用 $g_{\mathcal{l}},h_{\mathcal{l}v}\left(\mathcal{l}=1,2\right)$的导数刻画，而对应于 $\hat{x}$的非零组中元素的位置上， $\mathcal{L}$的梯度值受到 $g_{\mathcal{l}}\left(0\right),h_{\mathcal{l}v}\left(0\right)$的控制。

引理2.2假设2.1成立， $\hat{x}\in {ℝ}^n$是问题(2)的一个方向稳定点，则可以得到以下结论： $\forall v_{\left(j\right)}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)}\right)$， $v_{\left(j\right)i}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)i}\right)$，

(i) $\forall j\in \mathcal{J}\left(\hat{x}\right),i\in {\mathcal{I}}_j(\; x\; ^\; )$

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|=\left|{\hat{x}}_{\left(j\right)i}\right|\left|{\lambda }_1\frac{\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right|.$

特别的，当 $\bar{v}=1$时， $\forall j\in \mathcal{J}\left(\hat{x}\right),i\in {\mathcal{I}}_j(\; x\; ^\; )$

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|=\left|{\hat{x}}_{\left(j\right)i}\right|\left|{\lambda }_1\frac{{{\phi }^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{{{\phi }^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right|.$

(ii) 任取 $j\in \mathcal{J}\left(\hat{x}\right)$，满足 $\left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)\ne \varnothing$(相当于 $\left|{\mathcal{I}}_j\left(\hat{x}\right)\right|<n_j$或 $x_{{\mathcal{I}}_j\left(\hat{x}\right)}\ne x_{\left(j\right)}$)，则

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1\left[{g^{\prime }}_1\left(0\right)-h{\prime }_{1v_{\left(j\right)i}}\left(0\right)\right],\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right),$

进而

$\left|{\sum }_{i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)}{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1{\sum }_{i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)}\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(0\right)\right].$

特别的，当 $\overline{v_{\mathcal{l}}}=1\left(\mathcal{l}=1,2\right)$时， $\forall j\in \mathcal{J}\left(\hat{x}\right)$，若 $\left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)\ne \varnothing$，则

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1{{\phi }^{\prime }}_1\left(0\right),\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right),$(7)

进而

$\left|{\sum }_{i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)}{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1\left(n_j-\left|I_j\left(\hat{x}\right)\right|\right){{\phi }^{\prime }}_1\left(0\right).$

证明. (i) 根据定理2.1， $\forall d\in {ℝ}^n$， $v_{\left(j\right)}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)}\right)$， $v_{\left(j\right)i}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)i}\right)$，可以得到：

$\begin{array}{c}0\le f^{\prime }\left(\hat{x};d\right)\\ =\langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-\underset{v_{\left(j\right)i}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)i}\right)}{\max }{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]m^{\prime }}}\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-\underset{v_{\left(j\right)}\in \mathcal{A}\left({\hat{x}}_{\left(j\right)}\right)}{\max }{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]m^{\prime }}\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)\\ \le \langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]m^{\prime }}}\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]m^{\prime }}\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right).\end{array}$(8)

特别地，任取 $d\in \mathcal{D}\left(\hat{x}\right)$，则根据(3)和(4)知

$m^{\prime }\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)=\{\begin{array}{ll}\left|d_{\left(j\right)i}\right|=0,\hfill & {\hat{x}}_{\left(j\right)i}=0\hfill \\ \frac{{\hat{x}}_{\left(j\right)i}{d}_{\left(j\right)i}}{\left|{\hat{x}}_{\left(j\right)i}\right|},\hfill & {\hat{x}}_{\left(j\right)i}\ne 0\hfill \end{array},m^{\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)=\{\begin{array}{ll}\Vert d_{\left(j\right)}\Vert =0,\hfill & \Vert {\hat{x}}_{\left(j\right)}\Vert =0\hfill \\ \frac{\langle {\hat{x}}_{\left(j\right)},d_{\left(j\right)}\rangle }{\Vert {\hat{x}}_{\left(j\right)}\Vert },\hfill & \Vert {\hat{x}}_{\left(j\right)}\Vert \ne 0\hfill \end{array}.$(9)

因此，得到

$\begin{array}{c}0\le {\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left\{{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]m^{\prime }\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)\right\}}}\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]m^{\prime }}\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)\\ ={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left\{{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]\frac{{\hat{x}}_{\left(j\right)i}{d}_{\left(j\right)i}}{\left|{\hat{x}}_{\left(j\right)i}\right|}\right\}}}\\ +{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}\frac{\langle {\hat{x}}_{\left(j\right)},d_{\left(j\right)}\rangle }{\Vert {\hat{x}}_{\left(j\right)}\Vert }\end{array}$

$\begin{array}{c}={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{d}_{\left(j\right)i}}}{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\\ +{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{\hat{x}}_{(j)i}\{\frac{{\lambda }_1\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]}{\left|{\hat{x}}_{\left(j\right)i}\right|}+\frac{{\lambda }_2\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\}}}\end{array}$

根据 $d\in \mathcal{D}\left(\hat{x}\right)$的任意性，可得 $\forall j\in \mathcal{J}\left(\hat{x}\right),i\in {\mathcal{I}}_j(\; x\; ^\; )$

$0={\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)+{\hat{x}}_{\left(j\right)i}\left\{\frac{{\lambda }_1\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]}{\left|{\hat{x}}_{\left(j\right)i}\right|}+\frac{{\lambda }_2\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right\}.$

因此 $\forall j\in \mathcal{J}\left(\hat{x}\right),i\in {\mathcal{I}}_j(\; x\; ^\; )$

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|=\left|{\hat{x}}_{\left(j\right)i}\right|\left|{\lambda }_1\frac{\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_{1v_{\left(j\right)i}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j\right)}}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right|.$

当 $\bar{v}=1$时，可得 $\forall j\in \mathcal{J}\left(\hat{x}\right),i\in {\mathcal{I}}_j(\; x\; ^\; )$

$0={\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)+{\hat{x}}_{\left(j\right)i}\left[{\lambda }_1\frac{{{\phi }^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{{{\phi }^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right],$(10)

进而

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|=\left|{\hat{x}}_{\left(j\right)i}\right|\left|{\lambda }_1\frac{{{\phi }^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{{{\phi }^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right|.$

(ii) 任取满足 中条件的 $j_0$， ${\hat{x}}_{\left(j_0\right)}={\left({\hat{x}}_{\left(j_0\right)1},\cdots ,{\hat{x}}_{\left(j_0\right)n_{j_0}}\right)}^{\top }$和任意取定 $i_0\in \left\{1,\cdots ,n_{j_0}\right\}\backslash {\mathcal{I}}_{j_0}\left(\hat{x}\right)$，定义 $d^{+},d^{-}$如下：

$d_{\left(j\right)i}^{+}=\{\begin{array}{ll}1,\hfill & j=j_0,i=i_0\hfill \\ 0,\hfill & 其它\hfill \end{array},d_{\left(j\right)i}^{-}=\{\begin{array}{ll}-1,\hfill & j=j_0,i=i_0\hfill \\ 0,\hfill & 其它\hfill \end{array},$

则有 $\langle x_{\left(j_0\right)},d_{\left(j_0\right)}^{\pm }\rangle =0$且 $\Vert d_{\left(j\right)}^{\pm }\Vert =0,\forall j\ne j_0$。从而由(3)和(4)知

$m\prime \left({\hat{x}}_{\left(j_0\right)};d_{\left(j_0\right)}^{\pm }\right)=\frac{\langle x_{\left(j_0\right)},d_{\left(j_0\right)}^{\pm }\rangle }{\Vert x_{\left(j_0\right)}\Vert }=0,m\prime \left({\hat{x}}_{\left(j_0\right)i_0};d_{\left(j_0\right)i_0}^{\pm }\right)=\left|d_{\left(j_0\right)i}^{\pm }\right|=1.$

根据(8)，可得

$\begin{array}{c}0\le f^{\prime }\left(\hat{x};d^{\pm }\right)\\ \le {\nabla }_{\left(j_0\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j_0\right)i_0}^{\pm }+{\lambda }_1\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j_0\right)i_0}\right|\right)-{h^{\prime }}_{1v_{\left(j_0\right)i_0}}\left(\left|{\hat{x}}_{\left(j_0\right)i_0}\right|\right)\right]m^{\prime }\left({\hat{x}}_{\left(j_0\right)i_0};d_{\left(j_0\right)i_0}^{\pm }\right)\\ +{\lambda }_2\left[g_2\left(\Vert {\hat{x}}_{\left(j_0\right)}\Vert \right)-{h^{\prime }}_{2v_{\left(j_0\right)}}\left(\Vert {\hat{x}}_{\left(j_0\right)}\Vert \right)\right]m^{\prime }\left({\hat{x}}_{\left(j_0\right)};d_{\left(j_0\right)}^{\pm }\right)\\ \le \pm {\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right)+{\lambda }_1\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_{1v_{\left(j_0\right)i_0}}\left(0\right)\right],\end{array}$

因此

$\left|{\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_{1v_{\left(j_0\right)i_0}}\left(0\right)\right].$

进而

$\left|{\sum }_{i\in \left\{1,\cdots ,n_{j_0}\right\}\backslash {\mathcal{I}}_{j_0}\left(\hat{x}\right)}{\nabla }_{\left(j_0\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\sum }_{i\in \left\{1,\cdots ,n_{j_0}\right\}\backslash {\mathcal{I}}_{j_0}\left(\hat{x}\right)}\left|{\nabla }_{\left(j_0\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1{\sum }_{i\in \left\{1,\cdots ,n_{j_0}\right\}\backslash {\mathcal{I}}_{j_0}\left(\hat{x}\right)}\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_{1v_{\left(j_0\right)i}}\left(0\right)\right].$

当 $\overline{v_{\mathcal{l}}}=1\left(\mathcal{l}=1,2\right)$时， $\forall j\in \mathcal{J}\left(\hat{x}\right)$，若 $\left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)\ne \varnothing$，则

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1{{\phi }^{\prime }}_1\left(0\right),$

进而

$\left|{\sum }_{i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)}{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_1\left(n_j-\left|I_j\left(\hat{x}\right)\right|\right)\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_1\left(0\right)\right]={\lambda }_1\left(n_j-\left|I_j\left(\hat{x}\right)\right|\right){{\phi }^{\prime }}_1\left(0\right).$

结论得证。 $\square$

当 ${\phi }_{\mathcal{l}}\left(\mathcal{l}=1,2\right)$是 $\left(0,\infty \right)$上两个可微凸函数的差(例如SCAD罚、MCP、Capped- ${\mathcal{l}}_1$罚)时，下述推论给出问题(2)的方向导数在其方向稳定点 $\hat{x}$处更加简洁的表达式，以及当 $f\prime \left(\hat{x};d\right)=0$时，方向d的特征。

推论2.1若假设2.1成立，且 ${\phi }_{\mathcal{l}}\left(t\right)=g_{\mathcal{l}}\left(t\right)-h_{\mathcal{l}}\left(t\right)$，其中 $g_{\mathcal{l}},h_{\mathcal{l}}\left(\mathcal{l}=1,2\right)$是 $\left(0,\infty \right)$上的可微凸函数。若 $\hat{x}$是问题(2)的方向稳定点，则下述结论成立：

(i) $\forall d\in {ℝ}^n$，

$f\prime \left(\hat{x};d\right)={\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}}+{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_2}\left(0\right)\Vert d_{\left(j\right)}\Vert .$(11)

(ii) 若 $j\notin \mathcal{J}\left(\hat{x}\right)$，则下述不等式成立：

$\begin{array}{l}\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_2{{\phi }^{\prime }}_2\left(0\right)+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right),\forall i\in \left\{1,\cdots ,n_j\right\},\\ 0\le {\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert +{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]},\forall d\in {ℝ}^n.\end{array}$

(iii) 若 ${\lambda }_1{{\phi }^{\prime }}_1\left(0\right)>\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|,\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)$，则当方向 $d\in {ℝ}^n$满足 $f\prime \left(\hat{x};d\right)=0$时，必有

$d_{\left(j\right)i}=0,\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right).$

(iv) 若 ${{\phi }^{\prime }}_2\left(0\right)>0$， ${\lambda }_1{{\phi }^{\prime }}_1\left(0\right)>\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|,\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}$，则当方向 $d\in {ℝ}^n$满足 $f\prime \left(\hat{x};d\right)=0$时，必有

$d_{\left(j\right)}=0,\forall j\notin \mathcal{J}\left(\hat{x}\right).$

(v) 若 ${{\phi }^{\prime }}_2\left(0\right)>0$， ${\lambda }_1{{\phi }^{\prime }}_1\left(0\right)>\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|,\forall j\in \left\{1,\cdots ,K\right\},i\in {\mathcal{I}}_j\left(\hat{x}\right)$，则当方向 $d\in {ℝ}^n$满足 $f\prime \left(\hat{x};d\right)=0$时，必有 ${\hat{x}}_{\left(j\right)i}=0\Rightarrow d_{\left(j\right)i}=0$.

证明. (i) 因为 $\bar{v}=1$，根据(6)知： $\forall d\in {ℝ}^n$，

$\begin{array}{l}f\prime \left(\hat{x};d\right)=\langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]m\prime }}\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]m\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right).\end{array}$

由(3)和(4)可得

$\begin{array}{c}{f}^{\prime }\left(\hat{x};d\right)=\langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left[{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)\right]}}\frac{{\hat{x}}_{\left(j\right)i}{d}_{\left(j\right)i}}{\left|{\hat{x}}_{\left(j\right)i}\right|}\\ +{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{\lambda }_1}}\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_1\left(0\right)\right]\left|d_{\left(j\right)i}\right|+{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }\left[{g^{\prime }}_2\left(0\right)-{h^{\prime }}_2\left(0\right)\right]}\Vert d_{\left(j\right)}\Vert \\ +{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }\left[{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)\right]}\frac{\langle {\hat{x}}_{\left(j\right)},d_{\left(j\right)}\rangle }{\Vert {\hat{x}}_{\left(j\right)}\Vert }\end{array}$

$\begin{array}{c}={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{d}_{\left(j\right)i}}}\left\{{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)+{\lambda }_1{\hat{x}}_{\left(j\right)i}\frac{{g^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)-{h^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)}{\left|{\hat{x}}_{\left(j\right)i}\right|}\right\}\\ +{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left\{{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1\left[{g^{\prime }}_1\left(0\right)-{h^{\prime }}_1\left(0\right)\right]\left|d_{\left(j\right)i}\right|\right\}}}+{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }\left[{g^{\prime }}_2\left(0\right)-{h^{\prime }}_2\left(0\right)\right]}\Vert d_{\left(j\right)}\Vert \\ +{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{\hat{x}}_{\left(j\right)i}}}{d}_{\left(j\right)i}\frac{{g^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)-{h^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\\ ={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{d}_{\left(j\right)i}}}\left\{{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)+{\hat{x}}_{\left(j\right)i}\left[{\lambda }_1\frac{{{\phi }^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)}{\left|{\hat{x}}_{\left(j\right)i}\right|}+{\lambda }_2\frac{{{\phi }^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)}{\Vert {\hat{x}}_{\left(j\right)}\Vert }\right]\right\}\\ +{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}}+{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_2}\left(0\right)\Vert d_{\left(j\right)}\Vert \end{array}$

根据(10)，可得

$f\prime \left(\hat{x};d\right)={\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}}+{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_2}\left(0\right)\Vert d_{\left(j\right)}\Vert .$

(ii) 根据等式(11)， $\forall d\in {ℝ}^n$，

$\begin{array}{l}f\prime \left(\hat{x};d\right)={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}}\\ +{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }\left\{{\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert +{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}\right\}}.\end{array}$(12)

任取 $j_0\notin \mathcal{J}\left(\hat{x}\right)$，任取 $i_0\in \left\{1,\cdots ,n_{j_0}\right\}$，定义 $d^0$如下：

$d_{\left(j\right)i}^0=\{\begin{array}{ll}-{\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right),\hfill & j=j_0,i=i_0\hfill \\ 0,\hfill & 其它\hfill \end{array}.$

注意到 $\hat{x}$是问题(2)的方向稳定点，可得

$\begin{array}{c}0\le f^{\prime }\left(\hat{x};d^0\right)={\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j_0\right)}^0\Vert +{\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right)d_{\left(j_0\right)i_0}^0+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j_0\right)i_0}^0\right|\\ =\left|{\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right)\right|\left[{\lambda }_2{{\phi }^{\prime }}_2\left(0\right)-\left|{\nabla }_{\left(j_0\right)i_0}\mathcal{L}\left(\hat{x}\right)\right|+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\right].\end{array}$

再根据 $j_0,i_0$的任意性，可得

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\le {\lambda }_2{{\phi }^{\prime }}_2\left(0\right)+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right),\forall j\notin \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}.$

任取 $j_g\notin \mathcal{J}\left(\hat{x}\right)$，定义 $d^g$如下：

$d_{\left(j\right)}^g=\{\begin{array}{ll}{d}_{j_g},\hfill & j=j_g\hfill \\ 0,\hfill & 其它\hfill \end{array}.$

根据(12)，可得

$\begin{array}{c}0\le f\prime \left(\hat{x};d^g\right)={\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j_g\right)}^g\Vert +{\displaystyle \underset{i=1}{\overset{n_{j_g}}{\sum }}\left[{\nabla }_{\left(j_g\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j_g\right)i}^g+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j_g\right)i}^g\right|\right]}\\ ={\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j_g\right)}\Vert +{\displaystyle \underset{i=1}{\overset{n_{j_g}}{\sum }}\left[{\nabla }_{\left(j_g\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j_g\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j_g\right)i}\right|\right]}.\end{array}$

再由 $j_g$的任意性，，知

$0\le {\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert +{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]},\forall j\notin \mathcal{J}\left(\hat{x}\right),d\in {ℝ}^n.$

(iii) 根据(12)知

$\begin{array}{c}f\prime \left(\hat{x};d\right)\ge {\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left|d_{\left(j\right)i}\right|}}\left[{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)-\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\right]\\ +{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }\left\{{\lambda }_2{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert +{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left[{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)d_{\left(j\right)i}+{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)\left|d_{\left(j\right)i}\right|\right]}\right\}}.\end{array}$(13)

根据不等式(7)和推论3.1 (ii)，知

$f\prime \left(\hat{x};d\right)\ge {\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left|d_{\left(j\right)i}\right|}}\left[{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)-\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\right]\ge 0.$

再由条件 $f\prime \left(\hat{x};d\right)=0$，得

$0={\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left|d_{\left(j\right)i}\right|}}\left[{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)-\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\right].$

因 ${\lambda }_1{{\phi }^{\prime }}_1\left(0\right)>\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|,\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right)$，故

$d_{\left(j\right)i}=0,\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}\backslash {\mathcal{I}}_j\left(\hat{x}\right).$

(iv) 由(12)得

$\begin{array}{l}f\prime \left(\hat{x};d\right)\ge {\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\left|d_{\left(j\right)i}\right|}}\left[{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)-\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\right]+{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}\left|d_{\left(j\right)i}\right|}}\left[{\lambda }_1{{\phi }^{\prime }}_1\left(0\right)-\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|\right]\\ +{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{\lambda }_2}{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert .\end{array}$

根据不等式(7)以及条件 ${\lambda }_1{{\phi }^{\prime }}_1\left(0\right)>\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left(\hat{x}\right)\right|,\forall j\notin \mathcal{J}\left(\hat{x}\right),\forall i\in \left\{1,\cdots ,n_j\right\}$，知

$f\prime \left(\hat{x};d\right)\ge {\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{\lambda }_2}{{\phi }^{\prime }}_2\left(0\right)\Vert d_{\left(j\right)}\Vert \ge 0.$

再由条件 $f\prime \left(\hat{x};d\right)=0$和 ${{\phi }^{\prime }}_2\left(0\right)>0$，得 $d_{\left(j\right)}=0,\forall j\notin \mathcal{J}\left(\hat{x}\right)$。

(v)综合(iii)和(iv)即得。 $\square$

3. 问题(2)的光滑化模型

我们已知问题(2)的方向稳定点具有良好的局部最有性质，但考虑到正则项的非凸非光滑性质，直接计算问题(2)的方向稳定点是困难的。在实际应用中，光滑逼近的方法广泛应用于科学计算和实验，例如 [24] [25] [26] [27]。因此本文探讨使用光滑化方法来计算非光滑问题(2)的方向稳定点。

首先构造其光滑化模型。考虑到光滑函数的复合函数仍然是光滑函数，可以通过分别将 ${\phi }_{\mathcal{l}}\left(t\right)\left(\mathcal{l}=1,2\right)$和 $m\left(x\right)$进行光滑化，得到连续可微的函数 ${\tilde{\phi }}_{\mathcal{l}\mu }\left(t\right)\left(\mathcal{l}=1,2\right)$和 ${\tilde{m}}_{\mu }\left(x\right)$，再将它们重新复合，就得到了 ${\phi }_{\mathcal{l}}\left(\Vert x\Vert \right)$的光滑逼近函数 ${\tilde{\phi }}_{\mathcal{l}\mu }°{\tilde{m}}_{\mu }\left(x\right)$。

$m\left(x\right)$的光滑化：取 ${\tilde{m}}_{\mu }\left(x\right)=\sqrt{{\Vert x\Vert }^2+\mu }\ge 0$作为 $m\left(x\right)=\Vert x\Vert$的参数为 $\mu$的光滑化函数，其中 $\mu \in \left(0,\infty \right)$。显然 ${\tilde{m}}_{\mu }\left(x\right)$关于x二阶连续可微，关于 $\mu$单调递增。注意到， $\forall x,d\in {ℝ}^n$， ${{\tilde{m}}^{\prime }}_{\mu }\left(x;d\right)$的方向导数和梯度为：

${{\tilde{m}}^{\prime }}_{\mu }\left(x;d\right)=\langle \nabla {\tilde{m}}_{\mu }\left(x\right),d\rangle ,\nabla {\tilde{m}}_{\mu }\left(x\right)=\frac{x}{\sqrt{{\Vert x\Vert }^2+\mu }}.$(14)

特别地，对 $x_{\left(j\right)i}\in ℝ$，有

$m\left(x_{\left(j\right)i}\right)=\left|x_{\left(j\right)i}\right|,{\tilde{m}}_{\mu }\left(x_{\left(j\right)i}\right)=\sqrt{{\left(x_{\left(j\right)i}\right)}^2+\mu },$(15)

${{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(j\right)i}\right)=\frac{x_{\left(j\right)i}}{\sqrt{{\left(x_{\left(j\right)i}\right)}^2+\mu }},{{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(j\right)i};d_{\left(j\right)i}\right)=\frac{x_{\left(j\right)i}{d}_{\left(j\right)i}}{\sqrt{{\left(x_{\left(j\right)i}\right)}^2+\mu }}.$(16)

容易验证， ${\tilde{m}}_{\mu }\left(x\right)$满足如下收敛性质： $\forall x,d\in {ℝ}^n$，

(i) (光滑化函数的一致收敛性)： $\underset{x\prime \to x,\mu \downarrow 0}{\lim }{\tilde{m}}_{\mu }\left(x^{\prime }\right)=m\left(x\right)$；

(ii) (方向导数的一致收敛与弱一致收敛性)：

在 $x\ne 0$点， $\underset{x\prime \to x,\mu \downarrow 0}{\lim }\langle \nabla {\tilde{m}}_{\mu }\left(x\prime \right),d\rangle =\langle \nabla m\left(x\right),d\rangle =m\prime \left(x;d\right)$；

对 $x=0$点， $\lim {\text{sup}}_{x\prime \to 0,\mu \downarrow 0}\langle \nabla {\tilde{m}}_{\mu }\left(x^{\prime }\right),d\rangle =\Vert d\Vert =m\prime \left(0;d\right)$。

受Peng和Chen [24] 的启发，对于 ${\phi }_{\mathcal{l}}\left(t\right)\left(\mathcal{l}=1,2\right)$的光滑化函数，本文作出下述假设。

假设3.1设 ${\phi }_{\mathcal{l}}\left(t\right)\left(\mathcal{l}=1,2\right)$的光滑化函数 ${\phi }_{\mathcal{l}\mu }\left(t\right)\left(\mathcal{l}=1,2\right)$满足以下条件：

(i) $\underset{\mu \downarrow 0,t\to s}{\lim }{\tilde{\phi }}_{\mathcal{l}\mu }\left(t\right)={\phi }_{\mathcal{l}}\left(s\right)$。

(ii) 当 $s=0$时， $\underset{\mu \downarrow 0,t\downarrow 0}{\lim \text{sup}}{{\tilde{\phi }}^{\prime }}_{\mathcal{l}\mu }\left(t\right)={{\phi }^{\prime }}_{\mathcal{l}}\left(0+\right)={{\phi }^{\prime }}_{\mathcal{l}}\left(0\right)\ge 0$；当 $s\ne 0$时， $\underset{\mu \downarrow 0,t\to s}{\lim }{{\tilde{\phi }}^{\prime }}_{\mathcal{l}\mu }\left(t\right)={{\phi }^{\prime }}_{\mathcal{l}}\left(s\right)$。

需要强调的是，在许多情况下，满足假设3.1的光滑化函数是存在的，例如Peng和Chen在 [24] 中给出了MCP、SCAD、Capped- ${\mathcal{l}}_1$三类典型罚函数的光滑化函数，并验证了它们满足假设3.1。

若光滑化函数 ${\phi }_{\mathcal{l}}\left(t\right)\left(\mathcal{l}=1,2\right)$满足假设3.1，则可得问题(2)的光滑化模型：

$\underset{x\in {ℝ}^n}{\min }{\tilde{f}}_{\mu }\left(x\right):=\mathcal{L}\left(x\right)+{\displaystyle \underset{j=1}{\overset{K}{\sum }}\left[{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}{\lambda }_1}{\tilde{\phi }}_{1\mu }°{\tilde{m}}_{\mu }\left(x_{\left(j\right)i}\right)+{\lambda }_2{\tilde{\phi }}_{2\mu }°{\tilde{m}}_{\mu }\left(x_{\left(j\right)}\right)\right]}.$(17)

基于其连续可微性，当 $\mu >0$时，对于任意方向 $d\in {ℝ}^n$， ${\tilde{f}}_{\mu }\left(x\right)$的方向导数为：

${{\tilde{f}}^{\prime }}_{\mu }\left(x;d\right)=\langle \nabla {\tilde{f}}_{\mu }\left(x\right),d\rangle ,$(18)

其中

$\nabla {\tilde{f}}_{\mu }\left(x\right)=\nabla \mathcal{L}\left(x\right)+{\lambda }_2\left(\begin{array}{c}{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left(x_{\left(1\right)}\right)\nabla {\tilde{m}}_{\mu }\left(x_{(1)}\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left(x_{\left(k\right)}\right)\nabla {\tilde{m}}_{\mu }\left(x_{\left(k\right)}\right)\end{array}\right)+{\lambda }_1\left(\begin{array}{c}{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left(x_{\left(1\right)1}\right){{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(1\right)1}\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left(x_{\left(1\right)n_1}\right){{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(1\right)n_1}\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left(x_{\left(K\right)1}\right){{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(K\right)1}\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left(x_{\left(K\right)n_K}\right){{\tilde{m}}^{\prime }}_{\mu }\left(x_{\left(K\right)n_K}\right)\end{array}\right).$(19)

问题(17)是光滑优化问题，其稳定点 ${\hat{x}}^{\mu }\in {\text{R}}^n$定义为使 $\nabla {\tilde{f}}_{\mu }\left({\hat{x}}^{\mu }\right)=0$的点。

以下定理给出了光滑化问题(17)稳定点的特征，即在其稳定点处损失函数的梯度值 $\nabla \mathcal{L}$可以用对应的惩罚函数 ${\tilde{\phi }}_{l\mu }°{\tilde{m}}_{\mu }$的值来刻画。

定理3.1 (光滑化问题稳定点的特征)记 ${\hat{x}}^{\mu }$为光滑化参数为 $\mu >0$时光滑化问题(17)的稳定点，则

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left({\hat{x}}^{\mu }\right)\right|=\left|{\hat{x}}_{\left(j\right)i}^{\mu }\right|\left|{\lambda }_1\frac{{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(j\right)i}^{\mu }\right)}{\sqrt{{\left({\hat{x}}_{\left(j\right)i}^{\mu }\right)}^2+\mu }}+{\lambda }_2\frac{{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(j\right)}^{\mu }\right)}{\sqrt{{\Vert {\hat{x}}_{\left(j\right)}^{\mu }\Vert }^2+\mu }}\right|,\forall j\in \left\{1,\cdots ,K\right\},\forall i\in \left\{1,\cdots ,n_j\right\}.$

证明. 因为 ${\hat{x}}_{\mu }$满足 $0=\nabla {\tilde{f}}_{\mu }\left({\hat{x}}^{\mu }\right)$，根据(14)和(19)，可得

$0=\nabla \mathcal{L}\left({\hat{x}}^{\mu }\right)+{\lambda }_2\left(\begin{array}{c}{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)}^{\mu }\right)\nabla {\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)}^{\mu }\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)}^{\mu }\right)\nabla {\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)}^{\mu }\right)\end{array}\right)+{\lambda }_1\left(\begin{array}{c}{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)1}^{\mu }\right){{\tilde{m}}^{\prime }}_{\mu }\left({\hat{x}}_{\left(1\right)1}^{\mu }\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)n_1}^{\mu }\right){{\tilde{m}}^{\prime }}_{\mu }\left({\hat{x}}_{\left(1\right)n_1}^{\mu }\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)1}^{\mu }\right){{\tilde{m}}^{\prime }}_{\mu }\left({\hat{x}}_{\left(K\right)1}^{\mu }\right)\\ ⋮\\ {{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)n_K}^{\mu }\right){{\tilde{m}}^{\prime }}_{\mu }\left({\hat{x}}_{\left(K\right)n_K}^{\mu }\right)\end{array}\right)$

$=\nabla \mathcal{L}\left({\hat{x}}^{\mu }\right)+\left(\begin{array}{c}{\lambda }_2{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)}^{\mu }\right)\frac{{\hat{x}}_{\left(1\right)1}^{\mu }}{\sqrt{{\Vert {\hat{x}}_{\left(1\right)}^{\mu }\Vert }^2+\mu }}+{\lambda }_1{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)1}^{\mu }\right)\frac{{\hat{x}}_{\left(1\right)1}^{\mu }}{\sqrt{{\left({\hat{x}}_{\left(1\right)1}^{\mu }\right)}^2+\mu }}\\ ⋮\\ {\lambda }_2{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)}^{\mu }\right)\frac{{\hat{x}}_{\left(1\right)n_1}^{\mu }}{\sqrt{{\Vert {\hat{x}}_{\left(1\right)}^{\mu }\Vert }^2+\mu }}+{\lambda }_1{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(1\right)n_1}^{\mu }\right)\frac{{\hat{x}}_{\left(1\right)n_1}^{\mu }}{\sqrt{{\left({\hat{x}}_{\left(1\right)n_1}^{\mu }\right)}^2+\mu }}\\ ⋮\\ {\lambda }_2{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)}^{\mu }\right)\frac{{\hat{x}}_{\left(K\right)1}^{\mu }}{\sqrt{{\Vert {\hat{x}}_{\left(K\right)}^{\mu }\Vert }^2+\mu }}+{\lambda }_1{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)1}^{\mu }\right)\frac{{\hat{x}}_{\left(K\right)1}^{\mu }}{\sqrt{{\left({\hat{x}}_{\left(K\right)1}^{\mu }\right)}^2+\mu }}\\ ⋮\\ {\lambda }_2{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)}^{\mu }\right)\frac{{\hat{x}}_{\left(K\right)n_K}^{\mu }}{\sqrt{{\Vert {\hat{x}}_{\left(K\right)}^{\mu }\Vert }^2+\mu }}+{\lambda }_1{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(K\right)n_K}^{\mu }\right)\frac{{\hat{x}}_{\left(K\right)n_K}^{\mu }}{\sqrt{{\left({\hat{x}}_{\left(K\right)n_K}^{\mu }\right)}^2+\mu }}\end{array}\right).$

因此，

$\left|{\nabla }_{\left(j\right)i}\mathcal{L}\left({\hat{x}}^{\mu }\right)\right|=\left|{\hat{x}}_{\left(j\right)i}^{\mu }\right|\left|{\lambda }_1\frac{{{\tilde{\phi }}^{\prime }}_{1\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(j\right)i}^{\mu }\right)}{\sqrt{{\left({\hat{x}}_{\left(j\right)i}^{\mu }\right)}^2+\mu }}+{\lambda }_2\frac{{{\tilde{\phi }}^{\prime }}_{2\mu }°{\tilde{m}}_{\mu }\left({\hat{x}}_{\left(j\right)}^{\mu }\right)}{\sqrt{{\Vert {\hat{x}}_{\left(j\right)}^{\mu }\Vert }^2+\mu }}\right|,\forall j\in \left\{1,\cdots ,K\right\},\forall i\in \left\{1,\cdots ,n_j\right\}.$

$\square$

4. 光滑化前后稳定点的一致性

使用光滑化方法求解非光滑优化问题的一个重要的问题是：随着光滑参数趋于零，光滑化问题的解能否趋向于原问题的解，即光滑化前后解的一致性问题。只有当解的一致性成立时，才能保证光滑化问题的解趋向于原问题的解。

下面的定理说明本文的光滑化方法能够保证光滑化问题(17)的稳定点与问题(2)的方向稳定点具有一致性。

定理4.1 (稳定点的一致性)设光滑化函数 ${\tilde{\phi }}_{1\mu }$和 ${\tilde{\phi }}_{2\mu }$均满足假设3.1，并设 ${\hat{x}}^{{\mu }_k}$是光滑化问题(17)在光滑化参数 $\mu ={\mu }_k$时的稳定点，则当 ${\mu }_k\to 0^{+}$时， ${\left\{{\hat{x}}^{{\mu }_k}\right\}}_{k=1}^{\infty }$的任意聚点都是问题(2)的方向稳定点。

证明. 设 $\hat{x}$为 ${\left\{{\hat{x}}^{{\mu }_k}\right\}}_{k=1}^{\infty }$的任一聚点，不妨设

$\underset{k\to \infty }{\lim }{\hat{x}}^{{\mu }_k}=\hat{x}.$

因 ${\hat{x}}^{{\mu }_k}$是 ${\tilde{f}}_{{\mu }_k}\left(x\right)$的稳定点，根据 ${\tilde{f}}_{{\mu }_k}\left(x\right)$连续可微性和(18)和(19)可知， $\forall d\in {ℝ}^n$，有

$\begin{array}{c}0=\langle \nabla {\tilde{f}}_{{\mu }_k}\left({\hat{x}}^{{\mu }_k}\right),d\rangle \\ =\langle \nabla \mathcal{L}\left({\hat{x}}^{{\mu }_k}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \\ =\langle \nabla \mathcal{L}\left({\hat{x}}^{{\mu }_k}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin I_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}+{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \\ +{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle .\end{array}$(20)

令 $k\to \infty$，则 ${\mu }_k\to 0$， ${\hat{x}}^{{\mu }_k}\to \hat{x}$。根据第三节中 ${\tilde{m}}_{\mu }\left(x\right)$的一致和弱一致收敛性质以及假设3.1，可得下述极限：

$\begin{array}{l}\forall j\in \mathcal{J}\left(\hat{x}\right),\forall i\in {\mathcal{I}}_j\left(\hat{x}\right):\underset{k\to \infty }{\lim }{{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}=m\prime \left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)}\right),\underset{k\to \infty }{\lim }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)={{\phi }^{\prime }}_1\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right);\\ \forall j\in 1,\cdots ,K,\forall i\notin {\mathcal{I}}_j\left(\hat{x}\right):\underset{k\to \infty }{\lim \sup }{{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}=m^{\prime }\left(0;d_{\left(j\right)i}\right)=\left|d_{\left(j\right)i}\right|\ge 0,\underset{k\to \infty }{\lim \sup }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)={{\phi }^{\prime }}_1\left(0\right)>0;\\ \forall j\in \mathcal{J}\left(\hat{x}\right):\underset{k\to \infty }{\lim }\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle =m^{\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right),\underset{k\to \infty }{\lim }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)={{\phi }^{\prime }}_2\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right);\\ \forall j\notin \mathcal{J}\left(\hat{x}\right):\underset{k\to \infty }{\lim \sup }\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle =m^{\prime }\left(0;d_{\left(j\right)}\right)=\Vert d_{\left(j\right)}\Vert \ge 0,\underset{k\to \infty }{\lim \sup }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)={{\phi }^{\prime }}_2\left(0\right)>0.\end{array}$

在(20)中利用上述极限，根据定理2.1我们可以得到 $\forall d\in {ℝ}^n$

$\begin{array}{c}0=\underset{k\to \infty }{\lim }\langle \nabla {\tilde{f}}_{{\mu }_k}\left({\hat{x}}^{{\mu }_k}\right),d\rangle \\ =\underset{k\to \infty }{\lim }\langle \nabla \mathcal{L}\left({\hat{x}}^{{\mu }_k}\right),d\rangle +\underset{k\to \infty }{\lim }{\lambda }_1{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +\underset{k\to \infty }{\lim }{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +\underset{k\to \infty }{\lim }{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \\ +\underset{k\to \infty }{\lim }{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \end{array}$

$\begin{array}{c}\le \underset{k\to \infty }{\lim }\langle \nabla \mathcal{L}\left({\hat{x}}^{{\mu }_k}\right),d\rangle +\underset{k\to \infty }{\lim }{\lambda }_1{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right){{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }\underset{k\to \infty }{\lim }{{\tilde{\phi }}^{\prime }}_{1{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)}}\cdot \lim \underset{k\to \infty }{\sup }{{\tilde{m}}^{\prime }}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)i}^{{\mu }_k}\right)d_{\left(j\right)i}\\ +\underset{k\to \infty }{\lim }{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \\ +{\lambda }_2{\displaystyle \underset{j\notin \mathcal{J}\left(\hat{x}\right)}{\sum }\underset{k\to \infty }{\lim }{{\tilde{\phi }}^{\prime }}_{2{\mu }_k}°{\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right)}\cdot \lim \underset{k\to \infty }{\sup }\langle \nabla {\tilde{m}}_{{\mu }_k}\left({\hat{x}}_{\left(j\right)}^{{\mu }_k}\right),d_{\left(j\right)}\rangle \end{array}$

$\begin{array}{c}=\langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{\displaystyle \underset{i\in {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\tilde{\phi }}^{\prime }}_1}}°m\left({\hat{x}}_{\left(j\right)i}\right)\cdot m^{\prime }\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)\\ +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i\notin {\mathcal{I}}_j\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_1}}°m\left({\hat{x}}_{\left(j\right)i}\right)\cdot m^{\prime }\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)+{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_2}°m\left({\hat{x}}_{\left(j\right)}\right)\cdot m^{\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)\\ +{\lambda }_2{\displaystyle \underset{j\in \mathcal{J}\left(\hat{x}\right)}{\sum }{{\phi }^{\prime }}_2}°m\left({\hat{x}}_{\left(j\right)}\right)\cdot m^{\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)\\ =\langle \nabla \mathcal{L}\left(\hat{x}\right),d\rangle +{\lambda }_1{\displaystyle \underset{j=1}{\overset{K}{\sum }}{\displaystyle \underset{i=1}{\overset{n_j}{\sum }}{{\phi }^{\prime }}_1}}\left(\left|{\hat{x}}_{\left(j\right)i}\right|\right)m^{\prime }\left({\hat{x}}_{\left(j\right)i};d_{\left(j\right)i}\right)+{\lambda }_2{\displaystyle \underset{j=1}{\overset{K}{\sum }}{{\phi }^{\prime }}_2}\left(\Vert {\hat{x}}_{\left(j\right)}\Vert \right)m^{\prime }\left({\hat{x}}_{\left(j\right)};d_{\left(j\right)}\right)\\ =f^{\prime }\left(\hat{x};d\right)\end{array}$

因此， $\hat{x}$是问题(2)的方向稳定点。 $\square$

5. 总结

本文研究了稀疏加组稀疏优化问题的非凸松弛模型。给出了非凸松弛模型方向导数的刻画和方向稳定点，分析了方向稳定点的特征及其局部最优性质。进一步构造了松弛模型的光滑化逼近问题，并证明了光滑化问题的稳定点与松弛模型的方向稳定点具有一致性，为后续使用光滑方法计算模型的方向稳定点提供了理论保障。

基金项目

国家自然科学基金项目(11861020, 12261020)、贵州省高层次留学人才创新创业择优资助重点项目([2018]03)、贵州省科技计划项目(ZK[2021]009, [2018]5781)、贵州省青年科技人才成长项目([2018]121)。

NOTES

^*通讯作者。