© Wuhan University 2025

0 Introduction

In this paper, we consider the linear complementarity problem (LCP) in the standard form

$\{\begin{array}{l}\mathit{s}=\mathit{M}\mathit{x}+\mathit{q},\\ \mathit{x}\mathit{s}=\mathrm{0},\\ \mathit{x}\ge \mathrm{0},\mathit{s}\ge \mathrm{0},\end{array}(P)$

where $\mathit{M}\in {\mathbb{R}}^{n\times n},\mathit{x},\mathit{s},\mathit{q}\in {\mathbb{R}}^n$ and $\mathit{x}\mathit{s}$ denotes the componentwise product (Hadamard product) of the vectors $\mathit{x}$ and $\mathit{s}$. The LCP has many applications in science, engineering, economics and operations research^[1].

The problem (P) is called $P_{\mathrm{*}}\left(\kappa \right)$-LCP if $\mathit{M}$ is a $P_{\mathrm{*}}\left(\kappa \right)$ matrix, i.e., there exists a constant $\kappa \ge \mathrm{0}$ such that for every $\mathit{x}\in {\mathbb{R}}^n$,

$\left(\mathrm{1}+\mathrm{4}\kappa \right)\sum _{i\in I_{+}(\mathbf{x})}{x}_i{(\mathit{M}\mathit{x})}_i+\sum _{i\in I_{-}(\mathbf{x})}{x}_i{(\mathit{M}\mathit{x})}_i\ge \mathrm{0}$

where $I_{+}(\mathit{x})=\left\{i\in I:x_i{(\mathit{M}\mathit{x})}_i\ge \mathrm{0}\right\}, I_{-}(\mathit{x})=\left\{i\in I:x_i{(\mathit{M}\mathit{x})}_i<\mathrm{0}\right\},I=\left\{\mathrm{1,2},\cdots ,n\right\}.$ In particular, when $\kappa =\mathrm{0},P_{\mathrm{*}}\left(\kappa \right)$-LCP reduces to be monotone LCP (MLCP), which includes linear programming (LP) and convex quadratic programming (CQP) as special cases.

To introduce our method for solving $P_{\mathrm{*}}\left(\kappa \right)$-LCP, we first review the primal-dual interior-point method (IPM), which has proven to be one of the most efficient IPMs. The primal-dual IPM for LP was first introduced by Kojima et al in 1989^[2] and extended to a wider class of problems such as semidefinite programming (SDP)^[3], second order cone programming (SOCP)^[4], and LCP^[5]. It is known that the main feature of the primal-dual IPM is to approximately follow the central path, so the existence of the central path is crucial. Kojima et al^[6] first proved the existence of the central path for $P_{\mathrm{*}}\left(\kappa \right)$-LCP and generalized the primal-dual IPM for LP to $P_{\mathrm{*}}\left(\kappa \right)$-LCP. The algorithm has the currently best known iteration bound for $P_{\mathrm{*}}\left(\kappa \right)$-LCP, namely $O\left(\left(\kappa +\mathrm{1}\right)\sqrt[]{n}L\right)$. After that, the quality of a variant of an IPM is measured by the fact whether it can be generalized to $P_{\mathrm{*}}\left(\kappa \right)$-LCP or not^[7]. Lately, Wang et al^[8] proposed a full-Newton step feasible IPM for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. Zhang et al^[9] introduced a full-Newton step IPM based on the positive-asymptotic kernel function to solve $P_{\mathrm{*}}\left(\kappa \right)$-LCP, enjoying polynomial complexity for the small-update method, both of these two algorithms match the currently best known iteration bound. These two articles are also based on kernel functions to study IPM^[10-11]. In general, the majority of primal-dual IPMs search along a straight line related to derivatives of the central path. However, the central path of most optimization problems are nonlinear, which makes it necessary to replace the center path with another method, namely the arc-search method. To this end, Yang^[12] designed an arc-search IPM for LP, which searches for optimizers along an ellipse that is an approximation of the central path. Shahraki and Delavarkhalafi^[13] also introduced a new arc-search predictor-corrector infeasible IPM(IIPM) for symmetric cone LCP (SCLCP) with Cartesian $P_{\mathrm{*}}\left(\kappa \right)$-property. And more extension of the arc-search IPM see Refs. [14-15].

The majority of primal-dual IPMs are based on the classical logarithmic barrier function. Darvay^[16] presented a new technique for identifying a class of search directions. The search direction of his algorithm which we shall call Darvay's direction in the rest of this paper, was introduced by using an algebraic equivalent transformation (AET) on the nonlinear equation defining the central path, and then applying Newton's method to the new system of equations. Based on this search direction, Darvay^[16] proposed a full-Newton step primal-dual path following IPM for LP. Later on, Achache^[17], Wang et al^[3-4,18] and Mansouri et al^[19] extended Darvay's algorithm for LP to CQP, SDP, SOCP, symmetric cone programming (SCP) and MLCP, respectively. More recently, Asadi et al^[20] presented a polynomial IPM for the $P_{\mathrm{*}}\left(\kappa \right)$-horizontal linear complementarity problem (HLCP). The search direction of this algorithm made use of Darvay's direction, and the algorithm is quadratically convergent with polynomial iteration complexity $O\left(\left(\kappa +\mathrm{1}\right)\sqrt[]{n}\mathrm{l}\mathrm{o}\mathrm{g}\frac{n}{\epsilon })\right)$.

A significant number of researchers employed the AET technique, which is based on the power function $\varphi :(\xi ,+\mathbb{\infty })\to \mathbb{R},\xi \in [\mathrm{0,1}).$ This is a monotonically increasing, continuously differentiable function that is applied to the modified nonlinear equation system, which defines the central path for the $P_{\mathrm{*}}\left(\kappa \right)$-LCP. Illés et al^[21] generalized a whole class of AET functions and presented a unified complexity analysis of the IPM. Kheirfam et al^[22] proposed a power function $\varphi (t)=t^{q/\mathrm{2}},q\ge \mathrm{1}$ to develop a full-Newton short-step IPM for LP based on the AET technique.

In this paper, inspired by Darvay's work^[16], we present a new full-Newton step IPM that generalizes Grimes et al's algorithm^[23] from MLCP to $P_{\mathrm{*}}\left(\kappa \right)$-LCP, and obtain the similar iteration complexity $O\left(\sqrt[]{n}\left(\mathrm{1}+\mathrm{4}\kappa \right)\mathrm{l}\mathrm{o}\mathrm{g}\frac{n}{\epsilon }\right)$, which matches the currently best known iteration bound of IPM for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. As $P_{\mathrm{*}}\left(\kappa \right)$-LCP is a generalization of MLCP, the non-negativity of the inner product of the iterative direction ${\mathit{d}}_x$ and ${\mathit{d}}_s$ is lost, resulting in a distinct analysis in comparison with that of Grimes et al's algorithm^[23].

The paper is organized as follows. In Section 1, we first recall the class of Darvay's directions and some basic tools in the analysis of a new full-Newton step IPM for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. Section 2 consists of the modified search directions for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. And we analyse the feasibility and convergence of the proposed algorithm and derive the complexity results for the algorithm in Section 3. In Section 4, we perform the algorithm on some numerical examples. Finally, the conclusion and remarks are drawn in Section 5.

Notations: For $\mathit{x}=(x_{\mathrm{1}},x_{\mathrm{2}},\cdots ,x_n{)}^{\mathrm{T}}\in {\mathbb{R}}^n$, $x_{\mathrm{m}\mathrm{i}\mathrm{n}}$ denotes the minimum value of the components of $\mathit{x}$, $\mathit{X}$ is the diagonal matrix whose diagonal elements are the coordinates of the vector $\mathit{x}$, i.e., $\mathit{X}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\mathit{x})$. Furthermore, $\mathit{e}$ denotes the all-one vector of length $n$, and $I$ is the index set, i.e., $I=\{\mathrm{1,2},\cdots ,n\}$. As usual, the 1-norm, 2-norm and infinity norm are denoted by $||\cdot |{|}_{\mathrm{1}},||\cdot ||$ and $||\cdot |{|}_{\mathrm{\infty }}$, respectively. Given $\mathit{x},\mathit{y}\in {\mathbb{R}}^n$, $\mathit{x}\mathit{y}=(x_i{y}_i{)}_{\mathrm{1}\le i\le n}$denotes their Hadamard product the same with the vectors $\sqrt[]{\mathit{x}}={(\sqrt[]{x_i})}_{\mathrm{1}\le i\le n}$,$\mathit{x}/\mathit{y}=(x_i/y_i{)}_{\mathrm{1}\le i\le n}$,$y_i\ne \mathrm{0}$, where ${\mathit{x}}^{\mathrm{T}}\mathit{y}={\displaystyle \sum _{i=\mathrm{1}}^n}{x}_i{y}_i$ is their scalar product.

1 Preliminaries and Problem Statement

We assume that (P) satisfies the interior-point condition (IPC), i.e., there exists a pair of vectors $\left({\mathit{x}}_{\mathrm{0}},{\mathit{s}}_{\mathrm{0}}\right)$ such that ${\mathit{s}}_{\mathrm{0}}=\mathit{M}{\mathit{x}}_{\mathrm{0}}+\mathit{q}, {\mathit{s}}_{\mathrm{0}}>\mathrm{0},{\mathit{x}}_{\mathrm{0}}>\mathrm{0}$.

The basic idea of primal-dual IPM is to replace the second equation $\mathit{x}\mathit{s}=\mathrm{0}$ of (P) by the parameterized equation $\mathit{x}\mathit{s}=\mu \mathit{e}$, where $\mu >\mathrm{0}$. Hence, the problem (P) leads to the following equations:

$\{\begin{array}{l}\mathit{s}=\mathit{M}\mathit{x}+\mathit{q},\\ \mathit{x}\mathit{s}=\mu \mathit{e},\\ \mathit{x}\ge \mathrm{0},\mathit{s}\ge \mathrm{0}.\end{array}$(1)

Since $\mathit{M}$ is a $P_{\mathrm{*}}\left(\kappa \right)$ matrix and (P) satisfies IPC, for each $\mu >\mathrm{0}$, the system (1) has a unique solution denoted by $\left(\mathit{x}(\mu ),\mathit{s}(\mu )\right)$, which is called the $\mu$-center of (P)^[6]. The collection of all $\mu$-centers is referred to as the central path of (P). As $\mu$ approaches to zero, $\left(\mathit{x}(\mu ),\mathit{s}(\mu )\right)$ converges to the optimal solution of (P). Using Newton's approach to system (1) with a strictly feasible point $\left(\mathit{x},\mathit{s}\right)$, then the Newton direction $\left(\mathrm{\Delta }\mathit{x},\mathrm{\Delta }\mathit{s}\right)$ at this point is determined as the unique solution of the linear system of equations:

$\{\begin{array}{l}\mathrm{\Delta }\mathit{s}-\mathit{M}\mathrm{\Delta }\mathit{x}=\mathrm{0},\\ \mathit{x}\mathrm{\Delta }\mathit{s}+\mathit{s}\mathrm{\Delta }\mathit{x}=\mu \mathit{e}-\mathit{x}\mathit{s}.\end{array}$(2)

The system (2) gives the classical Newton search directions for $P_{\mathrm{*}}\left(\kappa \right)$-LCP^[24].

2 New Search Directions for ${\mathit{P}}_{\mathit{*}}\left(\mathit{\kappa }\right)$-LCP

In preparation for the full-Newton step IPM, we first recall the AET technique for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. The AET technique for determining a new search direction for IPM involves transforming the centering equation $\mathit{x}\mathit{s}=\mu \mathit{e}$ of (1) into a new equation $\varphi (\frac{\mathit{x}\mathit{s}}{\mu })=\varphi (\mathit{e})$, $\varphi :{\mathbb{R}}^{+}\to \mathbb{R}$.

Now we give a search direction based on Darvay's method for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. We consider the power function $\varphi :{\mathbb{R}}^{+}\to \mathbb{R},$ which is monotonically increasing, continuously differentiable, and suppose that the invertible function ${\varphi }^{-\mathrm{1}}$ exists. The system (1) can be written equivalently in the following form

$\{\begin{array}{l}\mathit{s}=\mathit{M}\mathit{x}+\mathit{q},\\ \varphi (\frac{\mathit{x}\mathit{s}}{\mu })=\varphi (\mathit{e}),\\ \mathit{x}\ge \mathrm{0},\mathit{s}\ge \mathrm{0}.\end{array}$(3)

The natural way to define the search direction is to follow Newton's approach and linearize the system (3), which leads to the following system

$\{\begin{array}{l}\mathrm{\Delta }\mathit{s}-\mathit{M}\mathrm{\Delta }\mathit{x}=\mathrm{0},\\ \frac{\mathit{s}}{\mu }\varphi \text{'}(\frac{\mathit{x}\mathit{s}}{\mu })\mathrm{\Delta }\mathit{x}+\frac{\mathit{x}}{\mu }\varphi \text{'}(\frac{\mathit{x}\mathit{s}}{\mu })\mathrm{\Delta }\mathit{s}=\varphi (\mathit{e})-\varphi (\frac{\mathit{x}\mathit{s}}{\mu }),\\ \mathit{x}\ge \mathrm{0},\mathit{s}\ge \mathrm{0}.\end{array}$(4)

Since $\mathit{M}$ is a $P_{\mathrm{*}}\left(\kappa \right)$ matrix, this system uniquely defines the search directions $\mathrm{\Delta }\mathit{x}$ and $\mathrm{\Delta }\mathit{s}$. Then the new iterates are given by

${\mathit{x}}^{+}:=\mathit{x}+\mathrm{\Delta }\mathit{x}, {\mathit{s}}^{+}:=\mathit{s}+\mathrm{\Delta }\mathit{s}.$

By introducing the scaled search directions

$\mathit{v}:=\sqrt[]{\frac{\mathit{x}\mathit{s}}{\mu }},{\mathit{d}}_x:=\frac{\mathit{v}\mathrm{\Delta }\mathit{x}}{\mathit{x}},{\mathit{d}}_s:=\frac{\mathit{v}\mathrm{\Delta }\mathit{s}}{\mathit{s}},\mathit{d}:=\sqrt[]{\frac{\mathit{x}}{\mathit{s}}},$(5)

we can rewrite the system (4) as the following scaled Newton-system

$\{\begin{array}{l}{\mathit{d}}_s-\overline{\mathit{M}}{\mathit{d}}_x=\mathrm{0},\\ {\mathit{d}}_x+{\mathit{d}}_s={\mathit{p}}_v,\end{array}$(6)

where ${\mathit{p}}_v:=\frac{\varphi (\mathit{e})-\varphi ({\mathit{v}}^{\mathrm{2}})}{\mathit{v}\varphi \mathrm{\text{'}}({\mathit{v}}^{\mathrm{2}})}$ and $\overline{\mathit{M}}:=\mathit{D}\mathit{M}\mathit{D},\mathit{D}:=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\mathit{d}).$ We observe that $\varphi (t)=t$ yields ${\mathit{p}}_v={\mathit{v}}^{-\mathrm{1}}-\mathit{v}$^[7], which coincides with the classical Newton direction. Moreover, when $\varphi (t)$ equals $\sqrt[]{t}$ and $t-\sqrt[]{t}$, then ${\mathit{p}}_v$ corresponds to $\mathrm{2}(\mathit{e}-\mathit{v})$^[16] and $\frac{\mathrm{2}(\mathit{v}-{\mathit{v}}^{\mathrm{2}})}{\mathrm{2}\mathit{v}-\mathit{e}}$^[25], respectively. Now let's take $\varphi (t)=t^{\mathrm{5}/\mathrm{2}}$^[22-23] based on Darvay's method, and we propose a new full-Newton IPM based on a novel appropriate search direction for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. For the analysis of the algorithm, we define a norm-based proximity measure $\delta (\mathit{v})$ to the central path as follows:

$\delta (\mathit{v}):=\delta (\mathit{x},\mathit{s};\mu ):=\frac{\mathrm{5}}{\mathrm{2}}||{\mathit{p}}_v||=||{\mathit{v}}^{-\mathrm{4}}-\mathit{v}||,$(7)

where

${\mathit{p}}_v=\frac{\mathrm{2}}{\mathrm{5}}({\mathit{v}}^{-\mathrm{4}}-\mathit{v})$(8)

Clearly,

$\delta (\mathit{v})=\mathrm{0}\iff \mathit{v}=\mathit{e}\iff \mathit{x}\mathit{s}=\mu \mathit{e}.$

Let ${\mathit{q}}_v={\mathit{d}}_x-{\mathit{d}}_s,$ it is easy to obtain

${\mathit{d}}_x=\frac{{\mathit{p}}_v+{\mathit{q}}_v}{\mathrm{2}},{\mathit{d}}_s=\frac{{\mathit{p}}_v-{\mathit{q}}_v}{\mathrm{2}},\mathrm{a}\mathrm{n}\mathrm{d} {\mathit{d}}_x{\mathit{d}}_s=\frac{{\mathit{p}}_v^{\mathrm{2}}-{\mathit{q}}_v^{\mathrm{2}}}{\mathrm{4}}.$

It is worthwhile to point out that the scaled search directions ${\mathit{d}}_x$ and ${\mathit{d}}_s$ are not orthogonal and ${{\mathit{d}}_x}^{\mathrm{T}}{\mathit{d}}_x$ is not always non-negative, which results in difficulties in the analysis of the algorithm for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. In Section 3, some new techniques are used to solve this problem.

Moreover, by using (5), (6), (8), system (4) becomes

$\{\begin{array}{l}\mathrm{\Delta }\mathit{s}-\mathit{M}\mathrm{\Delta }\mathit{x}=\mathrm{0},\\ \mathit{s}\mathrm{\Delta }\mathit{x}+\mathit{x}\mathrm{\Delta }\mathit{s}=\frac{\mathrm{2}\mu }{\mathrm{5}}({\mathit{v}}^{-\mathrm{3}}-{\mathit{v}}^{\mathrm{2}}).\end{array}$(9)

Furthermore, we define the $\tau$-neighborhood of the central path as follows:

$N(\tau ,\mu )=\left\{(\mathit{x},\mathit{s}):\mathit{x}>\mathrm{0},\mathit{s}=\mathit{M}\mathit{x}+\mathit{q}>\mathrm{0}:\delta (\mathit{x},\mathit{s};\mu )\le \tau \right\}$(10)

where $\tau$ is a threshold and $\mu >\mathrm{0}$.

Now we establish the full-Newton step IPM for $P_{\mathrm{*}}\left(\kappa \right)$-LCP. Let's assume that the algorithm starts with a strictly feasible pair $\left({\mathit{x}}_{\mathrm{0}},{\mathit{s}}_{\mathrm{0}}\right)$ and ${\mu }_{\mathrm{0}}=\frac{{{\mathit{x}}_{\mathrm{0}}}^{\mathrm{T}}{\mathit{s}}_{\mathrm{0}}}{n}$ such that $\delta ({\mathit{x}}_{\mathrm{0}},{\mathit{s}}_{\mathrm{0}};{\mu }_{\mathrm{0}})\le \tau .$ By using Newton's method, we can construct a new pair $\left(\mathit{x},\mathit{s}\right)$ which is 'close' to $\left(\mathit{x}\left(\mu \right),\mathit{s}\left(\mu \right)\right)$. Then $\mu$ is reduced by the factor $\left(\mathrm{1}-\theta \right)$ with $\mathrm{0}<\theta <\mathrm{1}$. This process is repeated until $\mu$ is small enough, i.e., $n\mu \le \epsilon$. Finally, an $\epsilon$-approximate solution of (P) can be found through the following Algorithm 1.

Algorithm 1: A full-Newton step IPM for $P_{\mathrm{*}}\left(\kappa \right)$-LCP based on a new AET technique

Input:  A threshold parameter $0<\tau <1$;  an accuracy parameter $\epsilon >0$;  barrier update parameter $0<\theta <1$;  a strictly feasible $\left({\mathit{x}}_0,{\mathit{s}}_0\right)$ and ${\mu }_0=\frac{{{\mathit{x}}_0}^{\mathrm{T}}{\mathit{s}}_0}{n}$ such that $\delta ({\mathit{x}}_0,{\mathit{s}}_0;{\mu }_0)\le \tau .$ begin  $\mathit{x}:={\mathit{x}}_0,\mathit{s}:={\mathit{s}}_0;\mu :={\mu }_0;$  while $n\mu \ge \epsilon$ do  begin  $\begin{array}{l}\left(\mathit{x},\mathit{s}\right):=\left(\mathit{x},\mathit{s}\right)+(\mathrm{\Delta }\mathit{x},\mathrm{\Delta }\mathit{s});\\ \mu :=\left(1-\theta \right)\mu \end{array}$  end end

3 Analysis of the Algorithm

In this section, we first examine the feasibility of Algorithm 1 by establishing several technical lemmas. Subsequently, we provide a proof for the local quadratic convergence of a full-Newton step to the central path. Finally, we determine the polynomial iteration complexity of the algorithm.

3.1 Feasibility Analysis of the Algorithm

Lemma 1   The solution $\left({\mathit{d}}_x,{\mathit{d}}_s\right)$ of linear system (6) satisfies

$\{\begin{array}{l}||{\mathit{d}}_x{\mathit{d}}_s|{|}_{\mathrm{1}}\le \frac{\mathrm{1}}{\mathrm{25}}\left(\mathrm{2}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}},\\ ||{\mathit{d}}_x{\mathit{d}}_s||\le \frac{\mathrm{1}}{\mathrm{25}}\sqrt[]{\left(\mathrm{2}+\mathrm{4}\kappa \right)\left(\mathrm{1}+\mathrm{4}\kappa \right)}{\delta }^{\mathrm{2}},\\ ||{\mathit{d}}_x{\mathit{d}}_s|{|}_{\mathrm{\infty }}\le \frac{\mathrm{1}}{\mathrm{25}}\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}},\\ {{\mathit{d}}_x}^{\mathrm{T}}{\mathit{d}}_s\le \frac{\mathrm{2}}{\mathrm{25}}{\delta }^{\mathrm{2}},\end{array}$(11)

where $\delta =\delta (\mathit{v}):=\delta (\mathit{x},\mathit{s};\mu ),\mu >\mathrm{0}.$

Proof   From Ref. [26], the solution $\left(\mathrm{\Delta }\mathit{x},\mathrm{\Delta }\mathit{s}\right)$ of linear system $\{\begin{array}{l}\mathrm{\Delta }\mathit{s}-\mathit{M}\mathrm{\Delta }\mathit{x}=\mathrm{0}\\ \mathit{s}\mathrm{\Delta }\mathit{x}+\mathit{x}\mathrm{\Delta }\mathit{s}=\frac{\mathrm{2}\mu }{\mathrm{5}}({\mathit{v}}^{-\mathrm{3}}-{\mathit{v}}^{\mathrm{2}})\end{array}$ satisfies the following relations:

$\begin{array}{l}||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}|{|}_{\mathrm{1}}\le \left(\frac{\mathrm{1}}{\mathrm{2}}+\kappa \right)||\frac{a}{\sqrt[]{\mathit{x}\mathit{s}}}|{|}^{\mathrm{2}},\\ ||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}||\le \sqrt[]{\left(\frac{\mathrm{1}}{\mathrm{4}}+\kappa \right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\kappa \right)}||\frac{a}{\sqrt[]{\mathit{x}\mathit{s}}}|{|}^{\mathrm{2}},\\ ||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}|{|}_{\mathrm{\infty }}\le \left(\frac{\mathrm{1}}{\mathrm{4}}+\kappa \right)||\frac{a}{\sqrt[]{\mathit{x}\mathit{s}}}|{|}^{\mathrm{2}},\end{array}$

where $a=\frac{\mathrm{2}\mu }{\mathrm{5}}({\mathit{v}}^{-\mathrm{3}}-{\mathit{v}}^{\mathrm{2}})$, therefore, by

$||\frac{a}{\sqrt[]{\mathit{x}\mathit{s}}}|{|}^{\mathrm{2}}=\frac{\mathrm{1}}{\mu }{\left(\frac{\mathrm{2}\mu }{\mathrm{5}}\right)}^{\mathrm{2}}||{\mathit{v}}^{-\mathrm{4}}-\mathit{v}|{|}^{\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{25}}\mu ||{\mathit{v}}^{-\mathrm{4}}-\mathit{v}|{|}^{\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{25}}\mu {\delta }^{\mathrm{2}},$

we can deduce that

$\begin{array}{l}||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}|{|}_{\mathrm{1}}\le \frac{\mathrm{4}}{\mathrm{25}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\kappa \right)\mu {\delta }^{\mathrm{2}},\\ ||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}||\le \frac{\mathrm{4}}{\mathrm{25}}\sqrt[]{\left(\frac{\mathrm{1}}{\mathrm{4}}+\kappa \right)\left(\frac{\mathrm{1}}{\mathrm{2}}+\kappa \right)}\mu {\delta }^{\mathrm{2}},\\ ||\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}|{|}_{\mathrm{\infty }}\le \frac{\mathrm{4}}{\mathrm{25}}\left(\frac{\mathrm{1}}{\mathrm{4}}+\kappa \right)\mu {\delta }^{\mathrm{2}}.\end{array}$

By $\frac{{\mathit{v}}^{\mathrm{2}}\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}}{\mathit{x}\mathit{s}}=\frac{\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}}{\mu }={\mathit{d}}_x{\mathit{d}}_s$, we get the first three inequalities of (11). From (6), one get

$\begin{array}{l}{\Vert {\mathit{p}}_v\Vert }^{\mathrm{2}}={\left({\mathit{d}}_x+{\mathit{d}}_s\right)}^{\mathrm{T}}\left({\mathit{d}}_x+{\mathit{d}}_s\right)\\ ={\Vert {\mathit{d}}_x\Vert }^{\mathrm{2}}+{\Vert {\mathit{d}}_s\Vert }^{\mathrm{2}}+\mathrm{2}{\mathit{d}}_x^{\mathrm{T}}{\mathit{d}}_s\ge \mathrm{2}{\mathit{d}}_x^{\mathrm{T}}{\mathit{d}}_s\end{array}$

and according to (7), we have $||{\mathit{p}}_v|{|}^{\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{25}}{\delta }^{\mathrm{2}}$, hence, we obtain the last inequality.

Lemma 2   The full-Newton step is strictly feasible, i.e. ${\mathit{x}}^{+}=\mathit{x}+\mathrm{\Delta }\mathit{x}>\mathrm{0}$ and ${\mathit{s}}^{+}=\mathit{s}+\mathrm{\Delta }\mathit{s}>\mathrm{0}$, provided that $\delta <\frac{\mathrm{5}}{\sqrt[]{\mathrm{1}+\mathrm{4}\kappa }}$.

Proof   Let $\alpha \in [\mathrm{0,1}]$ and $\left(\mathit{x},\mathit{s}\right)$ be a strictly feasible point for (P), then using (5), we get

$\begin{array}{l}\mathit{x}\left(\alpha \right)\mathit{s}\left(\alpha \right)=\left(\mathit{x}+\alpha \mathrm{\Delta }\mathit{x}\right)\left(\mathit{s}+\alpha \mathrm{\Delta }\mathit{s}\right)\\ =\mathit{x}\mathit{s}+\alpha \left(\mathit{x}\mathrm{\Delta }\mathit{s}+\mathit{s}\mathrm{\Delta }\mathit{x}\right)+{\alpha }^{\mathrm{2}}\mathrm{\Delta }\mathit{x}\mathrm{\Delta }\mathit{s}\\ =\mu \left(\left(\mathrm{1}-\alpha \right){\mathit{v}}^{\mathrm{2}}+\alpha \left({\mathit{v}}^{\mathrm{2}}+\mathit{v}{\mathit{p}}_v+\alpha {\mathit{d}}_x{\mathit{d}}_s\right)\right) .\end{array}$(12)

Due to (11) in Lemma 1 and (7) with $\delta <\frac{\mathrm{5}}{\sqrt[]{\mathrm{1}+\mathrm{4}\kappa }}$, it follows that

$\begin{array}{l}{\mathit{v}}^{\mathrm{2}}+v{\mathit{p}}_v+\alpha {\mathit{d}}_x{\mathit{d}}_s\ge {\mathit{v}}^{\mathrm{2}}+v{\mathit{p}}_v-\alpha ||{\mathit{d}}_x{\mathit{d}}_s|{|}_{\mathrm{\infty }}\mathit{e}\\ \ge {\mathit{v}}^{\mathrm{2}}+v{\mathit{p}}_v-\frac{\mathrm{1}}{\mathrm{25}}\alpha (\mathrm{1}+\mathrm{4}\kappa ){\delta }^{\mathrm{2}}\mathit{e}\\ ={\mathit{v}}^{\mathrm{2}}+\mathit{v}\cdot \frac{\mathrm{2}}{\mathrm{5}}({\mathit{v}}^{-\mathrm{4}}-\mathit{v})-\frac{\mathrm{1}}{\mathrm{25}}\alpha (\mathrm{1}+\mathrm{4}\kappa ){\delta }^{\mathrm{2}}\mathit{e}\\ \ge \frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}-\mathit{e}.\end{array}$

Now following from (12), for all $\mathrm{0}\le \alpha \le \mathrm{1}$, we have $\mathit{x}(\alpha )\mathit{s}(\alpha )>\mathrm{0}$ if $\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}-\mathit{e}\ge \mathrm{0}$. Here, we consider the function $h(t)=\frac{\mathrm{3}}{\mathrm{5}}{t}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{t}^{-\mathrm{3}}-\mathrm{1} , t>\mathrm{0}$. It is easy to see that $h(t)$ obtains a minimum at $t=\mathrm{1}$. Therefore, for all $t>\mathrm{0}$, $h(t)\ge h(\mathrm{1})=\mathrm{0}$, hence

$\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}-\mathit{e}\ge \mathrm{0}.$(13)

Thus $\mathit{x}(\alpha )\mathit{s}(\alpha )>\mathrm{0}$ for $\alpha \in [\mathrm{0,1}]$. Since $\mathit{x}>\mathrm{0}$ and $\mathit{s}>\mathrm{0}$, we obtain $\mathit{x}(\alpha )>\mathrm{0}$ and $\mathit{s}(\alpha )>\mathrm{0}$ for all $\alpha \in [\mathrm{0,1}]$. Then, by continuity the vectors $\mathit{x}(\mathrm{1})={\mathit{x}}^{+}>\mathrm{0}$ and $\mathit{s}(\mathrm{1})={\mathit{s}}^{+}>\mathrm{0}$, i.e., the full-Newton step is strictly feasible. This completes the proof.

For convenience, we denote ${\mathit{v}}^{+}=\sqrt[]{\frac{{\mathit{x}}^{+}{\mathit{s}}^{+}}{\mu }}.$

3.2 Convergence Analysis of the Algorithm

Lemma 3   If $\delta <\frac{\mathrm{5}}{\sqrt[]{\mathrm{1}+\mathrm{4}\kappa }}$, then $v_{\mathrm{m}\mathrm{i}\mathrm{n}}^{+}\ge \frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}$.

Proof   By setting $\alpha =\mathrm{1}$ in (12), (13) and Lemma 1, one has

$\begin{array}{l}{{\mathit{v}}_{+}}^{\mathrm{2}}={\mathit{v}}^{\mathrm{2}}+\mathit{v}{\mathit{p}}_v+{\mathit{d}}_x{\mathit{d}}_s\\ =\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}+{\mathit{d}}_x{\mathit{d}}_s\\ \ge \mathit{e}+{\mathit{d}}_x{\mathit{d}}_s\\ \ge \mathit{e}-||{\mathit{d}}_x{\mathit{d}}_s|{|}_{\mathrm{\infty }}\mathit{e}\end{array}$

$\begin{array}{l}\ge \left(\mathrm{1}-\frac{\mathrm{1}+\mathrm{4}\kappa }{\mathrm{25}}{\delta }^{\mathrm{2}}\right)\mathit{e}\\ =\frac{\mathrm{1}}{\mathrm{25}}\left(\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}\right)\mathit{e},\end{array}$

hence $v_{\mathrm{m}\mathrm{i}\mathrm{n}}^{+}\ge \frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}$. This completes the proof.

Next, we prove that the iteration of the proximity measure is locally quadratically convergent.

Lemma 4   Let $\delta <\frac{\mathrm{5}}{\sqrt[]{\mathrm{1}+\mathrm{4}\kappa }}$ and $\left(\mathit{x},\mathit{s}\right)$ be a strictly feasible point for (P). Then

$\begin{array}{l}{\delta }^{+}:=\delta ({\mathit{v}}^{+}):=\delta ({\mathit{x}}^{+},{\mathit{s}}^{+};\mu )\\ \le (\frac{\mathrm{25}}{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}+\frac{\mathrm{625}}{{\left(\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}\right)}^{\mathrm{2}}}\\ +\frac{\mathrm{5}}{\mathrm{5}+\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}})\gamma \left(\kappa ,\delta \right),\end{array}$

where $\gamma \left(\kappa ,\delta \right)=\left(\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{25}}\sqrt[]{\left(\mathrm{1}+\mathrm{4}\kappa \right)\left(\mathrm{2}+\mathrm{4}\kappa \right)}\right){\delta }^{\mathrm{2}}$.

Furthermore, if $\delta <\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$, then ${\delta }^{+}\le \frac{\mathrm{32}}{\mathrm{319}}\left(\mathrm{15}+\sqrt[]{\left(\mathrm{1}+\mathrm{4}\kappa \right)\left(\mathrm{2}+\mathrm{4}\kappa \right)}\right){\delta }^{\mathrm{2}}$, which shows the quadratic convergence of the proximity measure.

Proof   We have ${\delta }^{+}=\Vert {{\mathit{v}}^{+}}^{-\mathrm{4}}-{\mathit{v}}^{+}\Vert =\Vert \frac{\mathit{e}-{{\mathit{v}}^{+}}^{\mathrm{5}}}{{{\mathit{v}}^{+}}^{\mathrm{4}}}\Vert =\Vert \left(\mathit{e}-{{\mathit{v}}^{+}}^{\mathrm{2}}\right)\left(\frac{\mathit{e}+{\mathit{v}}^{+}+{{\mathit{v}}^{+}}^{\mathrm{2}}+{{\mathit{v}}^{+}}^{\mathrm{3}}+{{\mathit{v}}^{+}}^{\mathrm{4}}}{(\mathit{e}+{\mathit{v}}^{+}){{\mathit{v}}^{+}}^{\mathrm{4}}}\right)\Vert .$

Consider the function

$f(t)=\frac{\mathrm{1}+t+t^{\mathrm{2}}+t^{\mathrm{3}}+t^{\mathrm{4}}}{(\mathrm{1}+t)t^{\mathrm{4}}}=\frac{\mathrm{1}}{t^{\mathrm{2}}}+\frac{\mathrm{1}}{t^{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{1}+t},t>\mathrm{0}.$

Therefore, we can get

${\delta }^{+}=\Vert f({\mathit{v}}^{+})(\mathit{e}-{{\mathit{v}}^{+}}^{\mathrm{2}})\Vert \le {\Vert f({\mathit{v}}^{+})\Vert }_{\mathrm{\infty }}\Vert \mathit{e}-{{\mathit{v}}^{+}}^{\mathrm{2}}\Vert .$

There is no difficulty to check that the function $f(t)$ is continuous, monotone decreasing and positive on $(\mathrm{0},+\mathrm{\infty })$. Hence,

$\mathrm{0}<\left|f(({\mathit{v}}^{+}{)}_i)\right|=f(({\mathit{v}}^{+}{)}_i)\le f(v_{\mathrm{m}\mathrm{i}\mathrm{n}}^{+})\le f\left(\frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-(\mathrm{1}+\mathrm{4}\kappa ){\delta }^{\mathrm{2}}}\right)$

Consequently

$\begin{array}{l}{\Vert f\left({\mathit{v}}^{+}\right)\Vert }_{\mathrm{\infty }}=f\left(\frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}\right)\\ =\frac{\mathrm{25}}{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}+\frac{\mathrm{625}}{{\left(\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}\right)}^{\mathrm{2}}}\\ +\frac{\mathrm{5}}{\mathrm{5}+\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}}\cdot \end{array}$

Next, setting $\alpha =\mathrm{1}$ in (12), and from (8), we have

$\begin{array}{l}\Vert \mathit{e}-{{\mathit{v}}_{+}}^{\mathrm{2}}\Vert =\Vert \mathit{e}-({\mathit{v}}^{\mathrm{2}}+\mathit{v}{\mathit{p}}_v+{\mathit{d}}_x{\mathit{d}}_s)\Vert \\ =\Vert \mathit{e}-\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}-\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}-{\mathit{d}}_x{\mathit{d}}_s\Vert \\ \le \Vert \mathit{e}-\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}-\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}\Vert +||{\mathit{d}}_x{\mathit{d}}_s||.\end{array}$

Recalling the proof of Lemma 4 in Ref. [23], we have

$\Vert \mathit{e}-\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}-\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}\Vert \le \frac{\mathrm{3}}{\mathrm{5}}{\delta }^{\mathrm{2}}$

and together with (11), we obtain

$\begin{array}{l}\Vert \mathit{e}-{{\mathit{v}}_{+}}^{\mathrm{2}}\Vert \le \Vert \mathit{e}-\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}-\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}\Vert +\Vert {\mathit{d}}_x{\mathit{d}}_s\Vert \\ \le \left(\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{25}}\sqrt[]{\left(\mathrm{2}+\mathrm{4}\kappa \right)\left(\mathrm{1}+\mathrm{4}\kappa \right)}\right){\delta }^{\mathrm{2}}.\end{array}$

This implies that

${\delta }^{+}\le \left(\frac{\mathrm{25}}{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}+\frac{\mathrm{625}}{{\left(\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}\right)}^{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{5}+\sqrt[]{\mathrm{25}-\left(\mathrm{1}+\mathrm{4}\kappa \right){\delta }^{\mathrm{2}}}}\right)\gamma \left(\kappa ,\delta \right).$

Letting $\delta <\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$, by Lemma 3, $v_{\mathrm{m}\mathrm{i}\mathrm{n}}^{+}\ge \frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\frac{\mathrm{1}}{\mathrm{16}\left(\mathrm{1}+\mathrm{4}\kappa \right)}}\ge \frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\frac{\mathrm{1}}{\mathrm{16}}}$, we have

${\Vert f({\mathit{v}}_{+})\Vert }_{\mathrm{\infty }}\le f\left(\frac{\mathrm{1}}{\mathrm{5}}\sqrt[]{\mathrm{25}-\frac{\mathrm{1}}{\mathrm{16}}}\right)=\frac{\mathrm{800}}{\mathrm{319}}$

So we can get

${\delta }^{+}\le \frac{\mathrm{32}}{\mathrm{319}}\left(\mathrm{15}+\sqrt[]{\left(\mathrm{1}+\mathrm{4}\kappa \right)\left(\mathrm{2}+\mathrm{4}\kappa \right)}\right){\delta }^{\mathrm{2}}$

which shows the quadratic convergence of the proximity measure. This completes the proof.

The next lemma shows the influence of a full-Newton step on the duality gap and gives an upper bound for it.

Lemma 5   If $\delta :=\delta (\mathit{x},\mathit{s};\mu )<\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$, then after a full-Newton step, the duality gap satisfies

$({\mathit{x}}^{+}{)}^{\mathrm{T}}{\mathit{s}}^{+}\le \mathrm{2}\mu n$(14)

Proof   Due to (5) and (7), we have

${{\mathit{v}}^{+}}^{\mathrm{2}}={\mathit{v}}^{\mathrm{2}}+\mathit{v}{\mathit{p}}_v+{\mathit{d}}_x{\mathit{d}}_s=\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}+{\mathit{d}}_x{\mathit{d}}_s$

Then one gets

$\begin{array}{l}{\mathit{x}}^{+}{\mathit{s}}^{+}=\mu \left({\mathit{v}}^{\mathrm{2}}+\mathit{v}{\mathit{p}}_v+{\mathit{d}}_x{\mathit{d}}_s\right)\\ =\mu \left(\mathit{e}+\frac{\mathrm{25}}{\mathrm{4}}{{\mathit{p}}_v}^{\mathrm{2}}\cdot \frac{\frac{\mathrm{3}}{\mathrm{5}}{\mathit{v}}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{\mathit{v}}^{-\mathrm{3}}-\mathit{e}}{{\left({\mathit{v}}^{-\mathrm{4}}-\mathit{v}\right)}^{\mathrm{2}}}+{\mathit{d}}_x{\mathit{d}}_s\right)\\ \le \mu \left(\mathit{e}+\frac{\mathrm{25}}{\mathrm{4}}{{\mathit{p}}_v}^{\mathrm{2}}+{\mathit{d}}_x{\mathit{d}}_s\right).\end{array}$

Since, after some reductions, $\frac{\frac{\mathrm{3}}{\mathrm{5}}{v_i}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{v_i}^{-\mathrm{3}}-\mathrm{1}}{{\left({v_i}^{-\mathrm{4}}-v_i\right)}^{\mathrm{2}}}<\frac{\mathrm{3}}{\mathrm{5}}<\mathrm{1}$ for all $i$, see details in Lemma 4 of Ref. [23]. Then from the last inequality of (11) and (7), we get $({\mathit{x}}^{+}{)}^{\mathrm{T}}{\mathit{s}}^{+}={\mathit{e}}^{\mathrm{T}}\left({\mathit{x}}^{+}{\mathit{s}}^{+}\right)\le \mu \left(n+\frac{\mathrm{25}}{\mathrm{4}}||{\mathit{p}}_v|{|}^{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{25}}{\delta }^{\mathrm{2}}\right)\le \mu \left(n+\mathrm{2}{\delta }^{\mathrm{2}}\right)$.

Using $\delta <\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}<\mathrm{1}$, and $n+\mathrm{2}\le \mathrm{2}n,n\ge \mathrm{2}$, we may obtain ${\left({\mathit{x}}^{+}\right)}^{\mathrm{T}}{\mathit{s}}^{+}\le \mathrm{2}\mu n.$ This completes the proof.

In the next theorem, we analyse the effect of a full-Newton step on the proximity measure by updating the parameter $\mu$ by a factor $\left(\mathrm{1}-\theta \right)$.

Theorem 1   Let $\delta <\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$ and ${\mu }^{+}=\left(\mathrm{1}-\theta \right)\mu$, where $\mathrm{0}<\theta <\mathrm{1}$, then $\delta \left({\mathit{x}}^{+},{\mathit{s}}^{+};{\mu }^{+}\right)<\frac{\mathrm{5}\sqrt[]{\mathrm{2}n}\theta }{\sqrt[]{\mathrm{1}-\theta }}+{\delta }^{+}$.

Furthermore, if $\theta =\frac{\mathrm{1}}{\mathrm{36}\sqrt[]{\mathrm{2}n}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$ and $n\ge \mathrm{2}$, then we have $\delta \left({\mathit{x}}^{+},{\mathit{s}}^{+};{\mu }^{+}\right)<\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+\mathrm{4}\kappa \right)}$.

Proof   By $\sqrt[]{\frac{{\mathit{x}}^{+}{\mathit{s}}^{+}}{{\mu }^{+}}}=\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}{\mathit{v}}^{+}$ and ${\delta }^{+}=\Vert {{\mathit{v}}^{+}}^{-\mathrm{4}}-{\mathit{v}}^{+}\Vert$, one get

$\begin{array}{l}\delta \left({\mathit{x}}^{+},{\mathit{s}}^{+};{\mu }^{+}\right)=\Vert {\left(\sqrt[]{\frac{{\mathit{x}}^{+}{\mathit{s}}^{+}}{\left(\mathrm{1}-\theta \right)\mu }}\right)}^{-\mathrm{4}}-\sqrt[]{\frac{{\mathit{x}}^{+}{\mathit{s}}^{+}}{\left(\mathrm{1}-\theta \right)\mu }}\Vert \\ =\Vert {\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{{\mathit{v}}^{+}}^{-\mathrm{4}}-\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}{\mathit{v}}^{+}\Vert \\ =\Vert {\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{{\mathit{v}}^{+}}^{-\mathrm{4}}-{\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{\mathit{v}}^{+}+{\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{\mathit{v}}^{+}-\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}{\mathit{v}}^{+}\Vert .\end{array}$

Using the triangular inequality, we get

$\begin{array}{l}\delta \left({\mathit{x}}^{+},{\mathit{s}}^{+};{\mu }^{+}\right)\le \Vert {\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{{\mathit{v}}^{+}}^{-\mathrm{4}}-{\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{\mathit{v}}^{+}\Vert \\ +\Vert {\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{\mathit{v}}^{+}-\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}{\mathit{v}}^{+}\Vert \\ ={\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}{\delta }^{+}+\left|{\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}-\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}\right|||{\mathit{v}}^{+}||\\ \le {\delta }^{+}+\left|{\left(\mathrm{1}-\theta \right)}^{\mathrm{2}}-\frac{\mathrm{1}}{\sqrt[]{\mathrm{1}-\theta }}\right|||{\mathit{v}}^{+}||\\ ={\delta }^{+}+\frac{\mathrm{1}}{\mathrm{1}-\theta }\left|{\left(\mathrm{1}-\theta \right)}^{\mathrm{3}}-\sqrt[]{\mathrm{1}-\theta }\right|||{\mathit{v}}^{+}||\end{array}$