Quasiconvex Semidefinite Minimization Problem

Abstract

We introduce so-called semidefinite quasiconvex minimization problem. We derive new global optimality conditions for the above problem. Based on the global optimality conditions, we construct an algorithm which generates a sequence of local minimizers which converge to a global solution.

1. Introduction

Semidefinite linear programming can be regarded as an extension of linear programming and solves the following problem: min⟨𝐶,𝑋⟩_𝐹,𝐴_𝑗,𝑋_𝐹⩽𝑏_𝑗,𝑗=1,2,…,𝑠,𝑋≽0, (1) where 𝑋∈ℝ^𝑛×𝑛 is a matrix of variables and 𝐴_𝑗∈ℝ^𝑛×𝑛,𝑗=1,2,…,𝑠. 𝑋≽0 is notation for “𝑋 is positive semidefinite”. ⟨⋅,⋅⟩_𝐹 denotes Frobenius norm and ‖𝑋‖_𝐹=⟨𝐴,𝐴⟩_𝐹.

Semidefinite programming finds many applications in engineering and optimization [1]. Most interior-point methods for linear programming have been generalized to semidefinite convex programming [1–3]. There are many works devoted to the semidefinite convex programming problem but less attention so for has been paid to quasiconvex programming semidefinite quasiconvex minimization problem.

The aim of this paper is to develop theory and algorithms for the semidefinite quasiconvex programming. The paper is organized as follows. Section 2 is devoted to formulation of semidefinite quasiconvex programming and its global optimality conditions. In Section 3, we consider an approximation of the level set of the objective function and its properties.

2. Problem Definition and Optimality Conditions

Let 𝑋 be matrices in ℝ^𝑛×𝑛, and define a scalar matrix function as follows: 𝑓∶ℝ^𝑛×𝑛⟶ℝ. (2)

Definition 1. Let 𝑓(𝑋) be a differentiable function of the matrix 𝑋. Then 𝑓^′(𝑋)=𝜕𝑓(𝑋)𝜕𝑥_{𝑖𝑗𝑛×𝑛}. (3)

Introduce the Frobenius scalar product as follows: ⟨𝑋,𝑌⟩_𝐹=^𝑛𝑛_𝑖=1𝑗=1𝑥_𝑖𝑗𝑦_𝑖𝑗,∀𝑋,𝑌∈ℝ^𝑛×𝑛. (4) If 𝑓(⋅) is differentiable, then it can be checked that 𝑓(𝑋+𝐻)−𝑓(𝑋)=𝑓^′(𝑋),𝐻_𝐹+𝑜‖𝐻‖_𝐹. (5)

Definition 2. A set 𝔻⊂ℝ^𝑛×𝑛 is convex if 𝛼𝑋+(1−𝛼)𝑌∈𝔻 for all 𝑋,𝑌∈𝔻 and 𝛼∈[0,1].

Definition 3. The function 𝑓∶𝔻→ℝ is said to be quasiconvex on 𝔻 if 𝑓(()𝑌)𝑓(𝑋)(𝑌[].𝛼𝑋+1−𝛼⩽max,𝑓)∀𝑋,𝑌∈𝔻,𝛼∈0,1 (6)

The well-known property of a convex function [3] can be easily generalized as follows.

Lemma 4. A function 𝑓∶𝑅^𝑛×𝑛→𝑅 is quasiconvex if and only if the set 𝐿_𝑐𝑓=𝑋∈𝑅^𝑛×𝑛∣𝑓(𝑋)⩽𝑐 (7) is convex for all 𝑐∈𝑅.

Proof
Necessity. Suppose that 𝑐∈𝑅 is an arbitrary number and 𝑋,𝑌∈𝐿_𝑐(𝑓). By the definition of quasiconvexity, we have 𝑓(()𝑌)𝑓(𝑋)(𝑌[],𝛼𝑋+1−𝛼⩽max,𝑓)⩽𝑐∀𝛼∈0,1 (8) which means that the set 𝐿_𝑐(𝑓) is convex.
Sufficiency. Let 𝐿_𝑐(𝑓) be a convex set for all 𝑐∈𝑅. For arbitrary 𝑋,𝑌∈𝑅^𝑛, define 𝑐^𝑜=max{𝑓(𝑋),𝑓(𝑌)}. Then 𝑋∈𝐿_𝑐𝑜(𝑓) and 𝑌∈𝐿_𝑐𝑜(𝑓). Consequently, 𝛼𝑋+(1−𝛼𝑌)∈𝐿_𝑐𝑜(𝑓), for any 𝛼∈[0,1]. This completes the proof.

Lemma 5. Let 𝑓∶ℝ^𝑛×𝑛→ℝ be a quasiconvex and differentiable function. Then the inequality 𝑓(𝑋)⩽𝑓(𝑌) for 𝑋,𝑌∈ℝ^𝑛×𝑛 implies that 𝑓^′(𝑌),𝑋−𝑌_𝐹⩽0, (9) where 𝑓^′(𝑋)=(𝜕𝑓(𝑋)/𝜕𝑥_𝑖𝑗)_𝑛×𝑛 and ⟨⋅,⋅⟩_𝐹 denotes the Frobenius scalar product of two matrices.

Proof. Since 𝑓 is quasiconvex, 𝑓(𝛼𝑋+(1−𝛼)𝑌)⩽max𝑓(𝑋),𝑓(𝑌)=𝑓(𝑌) (10) for all 𝛼∈[0,1] and 𝑋,𝑌∈𝑅^𝑛×𝑛 such that 𝑓(𝑋)⩽𝑓(𝑌). By Taylor's formula, there is a neighborhood of the point 𝑌 on which: 𝑓(((𝑌)𝑓𝑌+𝛼𝑋−𝑌))−𝑓=𝛼^′(𝑌),𝑋−𝑌_𝐹+𝑜𝛼‖𝑋−𝑌‖_𝐹𝛼⩽0,𝛼>0. (11) From the fact that 𝑜(𝛼‖𝑥−𝑦‖_𝐹)/𝛼−−−−→^𝛼→𝑜0, we obtain ⟨𝑓^′(𝑌),𝑋−𝑌⟩_𝐹⩽0 which completes the proof.

Consider the problem of minimizing a differentiable quasiconvex matrix function subject to constraints min𝑓(𝑋)subjectto𝑔_𝑗(𝑋)⩽𝑏_𝑗,𝑗=1,2,…,𝑠,𝑋≽0, (12)(13)(14) where 𝑔_𝑗∶ℝ^𝑛×𝑛→ℝ,𝑗=1,2,…,𝑠 are scalar functions and 𝑋≽0 are positive semidefinite matrices, 𝑏_𝑗∈ℝ.

We call problem (12)–(14) as the semidefinite quasiconvex minimization problem.

Denote by 𝔻 a constraint set of the problem as follows: 𝔻=𝑋∈ℝ^𝑛×𝑛∣𝑔(𝑋)_𝑗⩽𝑏_𝑗,𝑗=1,2,…,𝑠;𝑋≽0. (15) Then problems (12)–(14) reduce to min_𝑥∈𝔻𝑓(𝑋). (16) In general, the set 𝔻 is nonconvex. Problem (16) is nonconvex and belongs to a class of global optimization problems in Banach space.

We formulate a new global optimality condition for problem (16) in the following. For this purpose,we introduce the level set 𝐸_𝑓(𝑍)(𝑓) of the function 𝑓∶ℝ^𝑛×𝑛→ℝ at a point 𝑍∈ℝ^𝑛×𝑛: 𝐸_𝑓(𝑍)𝑓=𝑌∈ℝ^𝑛×𝑛∣𝑓(𝑌)=𝑓(𝑍). (17) Then global optimality conditions for problem (16) can be formulated as follows.

Theorem 6. Let 𝑍 be a solution of problem (16). Then 𝑓^′(𝑋),𝑋−𝑌_𝐹⩾0∀𝑌∈𝐸_𝑓(𝑍)𝑓,𝑋∈𝐷, (18) where 𝐸_𝑐(𝑓)={𝑌∈𝑅^𝑛×𝑛∣𝑓(𝑌)=𝑐}. If, in addition, lim_{‖𝑋‖→∞}𝑓(𝑋)=+∞,𝑓^′𝑋+𝛼𝑓^′(𝑋)≠0 (19) holds for all 𝑋∈𝐷 and 𝛼≥0, then condition (18) becomes sufficient.

Proof
Necessity. Assume that 𝑧 is a solution of problem (16). Let 𝑋∈𝐷 and 𝑌∈𝐸_𝑓(𝑍)(𝑓). Then we have 0⩾𝑓(𝑍)−𝑓(𝑋)=𝑓(𝑌)−𝑓(𝑋), and Lemma 5 implies ⟨𝑓^′(𝑋),𝑋−𝑌⟩_𝐹⩾0.
Sufficiency. Suppose, on the contrary, that 𝑍 is not a solution of (16). Then there exists a 𝑈∈𝐷 such that 𝑓(𝑈)<𝑓(𝑍). Construct a ray 𝑌_𝛼 for 𝛼>0 defined by 𝑌_𝛼=𝑈+𝛼𝑓^′(𝑈). (20)
We claim that 𝑓(𝑌_𝛼)>𝑓(𝑈) holds for all positive 𝛼. By Taylor's formula, we have 𝑓𝑈+𝛼𝑓^′(𝑈)−𝑓(𝑈)=𝛼𝑓^′(𝑈)²+𝑜𝛼𝑓^′(𝑈)𝛼 (21) for small 𝛼>0, where lim_𝛼→0+𝑜(𝛼‖𝑓^′(𝑈)‖)/𝛼=0. Therefore, there exists 𝛼_𝑜>0 such that 𝑓(𝑌_𝛼)−𝑓(𝑈)>0 holds for all 𝛼∈(0,𝛼_𝑜). Hence, by Lemma 5, we have ⟨𝑓^′(𝑈+𝛼_𝑜𝑓^′(𝑈)),𝑓^′(𝑈)⟩_𝐹⩾0 since 𝑓^′(𝑈)≠0 and 𝑓^′(𝑈+𝛼_𝑜𝑓^′(𝑈))≠0 by the assumption. Note that for all 𝛾>1, we also have 𝑓(𝑈+𝛾𝛼_𝑜𝑓^′(𝑈))>𝑓(𝑈+𝛼_𝑜𝑓^′(𝑈)); for otherwise, we would have 𝑓(𝑈+𝛾𝛼_𝑜𝑓^′(𝑈))⩽𝑓(𝑈+𝛼_𝑜𝑓^′(𝑈)), and consequently, by Lemma 5, ⟨𝑓^′(𝑈+𝛼_𝑜𝑓^′(𝑈)),𝛼_𝑜(𝛾−1)𝑓^′(𝑈)⟩_𝐹⩾0, which would imply 𝛾⩽1 which is contradicting to the assumption that 𝛾>1. Moreover, we can show that 𝑓(𝑈+𝛾𝛼_𝑜𝑓^′(𝑈)) is increasing in 𝛾>0. If 𝑓(𝑈+𝛾^′𝛼_𝑜𝑓^′(𝑈))<𝑓(𝑈+𝛾𝛼_𝑜𝑓^′(𝑈)) holds for some 𝛾^′>𝛾, then 𝛼_𝑜(𝛾^′−𝛾)⟨𝑓^′(𝑈+𝛾𝛼_𝑜𝑓^′(𝑈)),𝑓^′(𝑈)⟩_𝐹⩽0, which would contradict the fact that 𝛾^′>𝛾. These prove our claim 𝑓(𝑌_𝛼)>𝑓(𝑈) for all 𝛼>0.
Now it is obvious that the function 𝜑∶𝑅⁺→𝑅 defined as 𝜑(𝛼)=𝑓𝑌_𝛼 (22) is continuous on [0,∞). Also, with assumption (19) implies lim_𝛼→∞𝜑(𝛼)=+∞, and therefore, there exists an 𝛼 such that 𝜑(𝛼)>𝑓(𝑍). Using the continuity of 𝜑(𝛼) and the inequalities 𝜑(𝛼)>𝑓(𝑍)>𝑓(𝑈), there exists an 𝛼 such that 𝑓𝑌+𝛼𝑓^′(𝑈)=𝑓(𝑍), (23) which means that 𝑌_𝛼∈𝐸_𝑓(𝑍)(𝑓). On the other hand, we have 𝑓^′(𝑈)=(1/𝛼)(𝑌_𝛼−𝑈). Thus we get 𝑓^′(𝑈),𝑈−𝑌_𝛼𝐹=1𝛼𝑌_𝛼−𝑈,𝑈−𝑌_𝛼𝐹1=−𝛼𝑌_𝛼−𝑌²_𝐹<0, (24) which contradicts (18). This means that 𝑍 must be a solution of (16).

Example 7. Consider the following problem: min_𝑋∈𝐷𝑓(𝑋)=‖𝑋‖²_𝐹,subjectto𝔻=𝑋∈ℝ^𝑛×𝑛∣𝑋=2341⩽𝑋⩽𝑋=.7589,𝑋≽0 (25)(26)

Example 8. Consider the fractional programming problem min_𝑋∈𝐷𝑓(𝑥)=𝑓₁(𝑋)𝑓₂(𝑋), (27) where 𝑓₁ is convex and differentiable on 𝑅^𝑛×𝑛 and 𝑓₂ is concave and differentiable on 𝑅^𝑛×𝑛. Suppose that 𝑓₁ and 𝑓₂ are defined positively on a ball 𝐵 containing a subset 𝐷⊂𝑅^𝑛×𝑛; that is, 𝑓₁(𝑋)>0,𝑓₂(𝑋)>0∀𝑋∈𝐷⊂𝐵. (28) We will call this problem as the mixed fractional minimization problem. By Lemma 4, we can easily show that 𝑓(𝑋) is quasiconvex. Hence, the optimality condition (13) at a solution 𝑍 of (27) is as follows: ^𝑛𝑛_𝑖=1𝑗=1𝜕𝑓₁(𝑋)𝜕𝑥_𝑖𝑗𝑓₂(𝑋)−𝜕𝑓₂(𝑋)𝜕𝑥_𝑖𝑗𝑓₁(𝑋)𝑥_𝑖𝑗−𝑦_𝑖𝑗𝑓²₂(𝑋)≥0∀𝑌∈𝐸_𝑓(𝑍)𝑓,𝑋∈𝐷. (29)

3. An Algorithm for the Convex Minimization Problem

We consider the quasiconvex minimization problem as a special case of problem (16): min_𝑋∈𝐷𝑓(𝑋), (30) where 𝑓∶𝑅^𝑛×𝑛→𝑅 is strongly convex and continuously differentiable and 𝐷 is an arbitrary compact set in 𝑅^𝑛×𝑛. In this case, then we can weaken condition (19) as shown in the next theorem.

Theorem 9. Let 𝑍 be a solution of problem (30). Then 𝑓^′(𝑋),𝑋−𝑌_𝐹⩾0∀𝑌∈𝐸_𝑓(𝑍)𝑓,𝑋∈𝐷. (31) If, in addition, min_𝑋∈𝐷𝑓^′(𝑋)_𝐹>0 (32) holds, then condition (31) is also sufficient.

Proof
Necessity. Assume that 𝑧 is a solution of problem (30). Consider 𝑋∈𝐷 and 𝑌∈𝐸_𝑓(𝑍)(𝑓). Then by the convexity of 𝑓, we have 0⩾𝑓(𝑍)−𝑓(𝑋)=𝑓(𝑌)−𝑓(𝑋)⩾𝑓^′(𝑋),𝑌−𝑋_𝐹. (33)
Sufficiency. Let us prove the assertion by contradiction. Assume that (31) holds and there exists a point 𝑈∈𝐷 such that 𝑓(𝑈)<𝑓(𝑍). (34) Clearly, 𝑓^′(𝑈)≠0 by assumption (32). Now define 𝑈_𝛼 as follows for 𝛼>0: 𝑈_𝛼=𝑈+𝛼𝑓^′(𝑈). (35) Then, by the convexity of 𝑓, we have 𝑓𝑈_𝛼−𝑓(𝑈)⩾𝑓^′(𝑈),𝑈_𝛼−𝑈_𝐹=𝛼𝑓^′(𝑈)²_𝐹, (36) which implies 𝑓𝑈_𝛼⩾𝑓(𝑈)+𝛼𝑓^′(𝑈)²_𝐹>𝑓(𝑈). (37) Then find 𝛼=𝛼 such that 𝑓(𝑈)+𝛼𝑓^′(𝑈)²_𝐹=𝑓(𝑍); (38) that is, 𝑓(𝑍)(𝑈)𝛼=−𝑓𝑓^′(𝑈)²_𝐹>0. (39) Thus we get 𝑓𝑈_𝛼⩾𝑓(𝑈)+𝛼𝑓^′(𝑈)²_𝐹=𝑓(𝑍)>𝑓(𝑈). (40) Define a function ℎ∶𝑅⁺→𝑅 as ℎ(𝛼)=𝑓𝑈+𝛼𝑓^′(𝑈)−𝑓(𝑍), (41) where 𝑅⁺={𝛼∈𝑅∣𝛼⩾0}. It is clear that ℎ is continuous on [0,+∞). Note that ℎ(𝛼)⩾0 and ℎ(0)<0. There are two cases with respect to the values of ℎ(𝛼) which we should consider.
Case a.  ℎ(𝛼)=0 (or 𝑓(𝑈+𝛼𝑓^′(𝑈))=𝑓(𝑍)), then 𝑓^′(𝑈),𝑈−𝑈_𝛼𝐹=−𝑓^′(𝑈),𝛼𝑓^′(𝑈)_𝐹=−𝛼𝑓^′(𝑈)²_𝐹<0, (42) contradicting condition (31).
Case b.  ℎ(𝛼)>0 and ℎ(0)<0. Since ℎ is continuous, there exists a point 𝛼_𝑜∈(0,𝛼) such that ℎ(𝛼_𝑜)=0 (or 𝑓(𝑈+𝛼_𝑜𝑓^′(𝑈))=𝑓(𝑍)). Then we have 𝑓^′(𝑈),𝑈−𝑈_𝛼𝑜𝐹=−𝛼_𝑜𝑓^′(𝑈)²_𝐹<0, (43) again contradicting (31).
Thus, in both cases, we find contradictions, proving the theorem.

Now using the function 𝑃(𝑌)=min_𝑋∈𝐷⟨𝑓^′(𝑋),𝑋−𝑌⟩_𝐹,𝑌∈𝑅^𝑛×𝑛, we reformulate Theorem 9 in terms of function 𝜓(𝑧) defined as follows: 𝜓(𝑍)=min_{𝑌∈𝐸𝑓(𝑍)𝑓}𝑃(𝑌),𝑍∈𝐷. (44)

Theorem 10. Assume that 𝑓∶𝑅^𝑛×𝑛→𝑅 is strongly convex and continuously differentiable and 𝐷 is a compact set in 𝑅^𝑛×𝑛. Let min_𝑋∈𝐷‖𝑓^′(𝑋)‖>0. If 𝜓(𝑍)=0, then the point 𝑍 is a solution to problem (30).

Proof. This is an obvious consequence of the following relations: 0=𝜓(𝑍)⩽𝑃(𝑌)⩽𝑓^′(𝑋),𝑋−𝑌_𝐹, (45) which are fulfilled for all 𝑌∈𝐸_𝑓(𝑍)(𝑓) and 𝑋∈𝐷.
Now we are ready to present an algorithm for solving problem (30). We also suppose that one can efficiently solve the problem of computing min_𝑋∈𝐷⟨𝑓^′(𝑋),𝑋−𝑌⟩_𝐹 for any given 𝑌∈𝑅^𝑛×𝑛.
Algorithm MIN
Input. A strongly quasiconvex function 𝑓 and a compact set 𝐷.
Output. A solution 𝑋 to the minimization problem (30).

Step 1. Choose a feasible point 𝑋⁰∈𝐷. Set 𝑘∶=0.

Step 2. Solve the following problem: min_{𝑌∈𝐸𝑘)𝑓(𝑋𝑓}𝑃(𝑌). (46) Let 𝑌_𝑘 be a solution of this problem (i.e., 𝑃(𝑌_𝑘)=min_𝑋∈𝐷⟨𝑓^′(𝑋),𝑋−𝑌_𝑘⟩_𝐹=min_{𝑌∈𝐸𝑘)𝑓(𝑋(𝑓)}𝑃(𝑌)), and let 𝑋_𝑘+1 realizes 𝑃(𝑌_𝑘) (i.e., 𝜓(𝑋_𝑘)=𝑃(𝑌_𝑘)=⟨𝑓^′(𝑋_𝑘+1),𝑋_𝑘+1−𝑌_𝑘⟩_𝐹).

Step 3. If 𝜓(𝑋_𝑘)=0 then output 𝑋=𝑋_𝑘 and terminate. Otherwise, let 𝑘∶=𝑘+1 and return Step 2.
The convergence of this algorithm is based on the following theorem.

Theorem 11. Assume that 𝑓∶𝑅^𝑛×𝑛→𝑅 is strongly convex and continuously differentiable and 𝐷 is a compact set in 𝑅^𝑛×𝑛. Let min_𝑋∈𝐷‖𝑓^′(𝑋)‖_𝐹>0. Then the sequence {𝑋_𝑘,𝑘=0,1,…} generated by Algorithm MIN is a minimizing sequence for problem (30); that is, lim_𝑘→∞𝑓𝑋_𝑘=min_𝑋∈𝐷𝑓(𝑋), (47) and every accumulation point of the sequence {𝑋_𝑘} is a global minimizer of (30).

Proof. From the construction of {𝑋_𝑘}, we have 𝑋_𝑘∈𝐷 and 𝑓(𝑋_𝑘)≥𝑓_∗ for all 𝑘, where 𝑓_∗=𝑓(𝑋_∗)=min_𝑋∈𝐷𝑓(𝑋). Clearly, 𝑓^′(𝑋_∗)≠0 by assumption. Also, note that for all 𝑌∈𝐸_{𝑓(𝑋𝑘)}(𝑓) and 𝑋∈𝐷, we have 𝜓𝑋_𝑘=min_{𝑌∈𝐸𝑘)𝑓(𝑋𝑓}min_𝑋∈𝐷𝑓^′(𝑋),𝑋−𝑌_𝐹⩽𝑓^′(𝑋),𝑋−𝑌_𝐹⩽0. (48) If there exists a 𝑘 such that 𝜓(𝑋_𝑘)=0 then, by Theorem 11, 𝑋^𝑘 is a solution to problem (30) and in this case the proof is complete. Therefore, without loss of generality, we can assume that ignored 𝜓(𝑋_𝑘)<0 for all 𝑘 and prove the theorem by contradiction. If the assertion is false; that is, 𝑋_𝑘 is not a minimizing sequence for problem (30), the following inequality holds: lim_𝑘→∞inf𝑓𝑋_𝑘>𝑓_∗. (49) By the definition of 𝜓(𝑋_𝑘) and Algorithm MIN, we have 𝑝𝑌_𝑘=𝜓𝑋_𝑘=min_{𝑌∈𝐸𝑘)𝑓(𝑋𝑓}min_𝑋∈𝐷𝑓^′(𝑋),𝑋−𝑌_𝐹=𝑓^′𝑋_𝑘+1,𝑋_𝑘+1−𝑌_𝑘𝐹 (50) and 𝑓(𝑌_𝑘)=𝑓(𝑋_𝑘). The convexity of 𝑓 implies that 𝑓𝑋_𝑘−𝑓𝑋_𝑘+1=𝑓𝑌_𝑘−𝑓𝑋_𝑘+1⩽𝑓^′𝑋_𝑘+1,𝑌_𝑘−𝑋_𝑘+1𝐹=−𝜓𝑋_𝑘>0. (51) Hence, we obtain 𝑓(𝑋_𝑘+1)<𝑓(𝑋_𝑘) for all 𝑘, and the sequence {𝑓(𝑋_𝑘)} is strictly decreasing. Since the sequence is bounded from below by 𝑓_∗, it has a limit and satisfies lim_𝑘→∞𝑓𝑋_𝑘+1−𝑓𝑋_𝑘=0. (52) Then, from (49) and (50), we obtain lim_𝑘→∞𝜓𝑋_𝑘=0. (53) From (51) we have 𝑓(𝑋_𝑘)>𝑓(𝑋_∗) for all 𝑘. Now define 𝑣_𝛼 as follows: 𝑉_𝛼=𝑋_∗+𝛼𝑓^′𝑋_∗,𝛼>0. (54) Then, by the convexity of 𝑓, we have 𝑓𝑉_𝛼−𝑓𝑋_∗⩾𝑓^′𝑋_∗,𝑉_𝛼−𝑋_∗𝐹=𝛼𝑓^′𝑋_∗², (55) which implies 𝑓𝑉_𝛼≥𝑓𝑋_∗+𝛼𝑓^′𝑋_∗²>𝑓𝑋_∗,𝛼>0. (56) Choose 𝛼=𝛼_𝑘 such that 𝑓𝑋_∗+𝛼_𝑘𝑓^′𝑋_∗²>𝑓𝑋_𝑘; (57) that is, 𝛼_𝑘>𝑓𝑋_𝑘−𝑓𝑋_∗𝑓^′𝑋_∗²>0. (58) Define a function ℎ_𝑘∶𝑅⁺→𝑅 as ℎ_𝑘(𝛼)=𝑓𝑋_∗+𝛼𝑓^′𝑋_∗−𝑓𝑋_𝑘, (59) where 𝑅⁺={𝛼∈𝑅∣𝛼≥0}. It is clear that ℎ_𝑘 is continuous on [0,+∞). Note that ℎ_𝑘(𝛼_𝑘)>0 and ℎ_𝑘(0)<0. Since ℎ_𝑘 is continuous, there exists a point 𝛼_𝑘∈(0,𝛼_𝑘) such that ℎ_𝑘(𝛼_𝑘)=0; that is, 𝑓(𝑣_𝛼𝑘)=𝑓(𝑥^𝑘) and 𝑣_𝛼𝑘=𝑥_∗+𝛼_𝑘𝑓^′(𝑥_∗). Also, note that 𝜓𝑋_𝑘=min_{𝑌∈𝐸𝑘)𝑓(𝑋𝑓}min_𝑋∈𝐷𝑓^′(𝑋),𝑋−𝑌_𝐹⩽𝑓^′𝑋_∗,𝑋_∗−𝑉_𝛼𝑘𝐹. (60) Taking into account 𝑉_𝛼𝑘=𝑋_∗+𝛼_𝑘𝑓^′(𝑋_∗), we have −𝜓𝑋_𝑘⩾𝑓^′𝑋_∗,𝑉_𝛼𝑘−𝑋_∗𝐹=𝑓^′𝑋_∗𝑉_𝛼𝑘−𝑋_∗⩾min_𝑋∈𝐷𝑓^′(𝑋)𝑣_𝛼𝑘−𝑋_∗>0. (61) Since lim_𝑘→∞𝜓(𝑋_𝑘)=0, this implies lim_𝑘→∞𝑉_𝛼𝑘=𝑋_∗. (62) The continuity of 𝑓 on 𝑅^𝑛×𝑛 yields lim_𝑘→∞𝑓𝑋_𝑘=lim_𝑘→∞𝑓𝑉_𝛼𝑘=𝑓𝑋_∗, (63) which is a contradiction to (49).
Consequently, {𝑋_𝑘} is a minimizing sequence for problem (30). Since 𝐷 is compact, we can always select the convergent subsequences {𝑋_𝑘𝑙} from {𝑋_𝑘} such that lim_𝑙→∞𝑋_𝑘𝑙=𝑋∈𝐷. (64) Then together with (63), we obtain lim_𝑙→∞𝑓𝑋_𝑘𝑙=𝑓𝑋=𝑓_∗, (65) which completes the proof.

4. Numerical Experiments

The proposed algorithm has been tested on the following numerical examples.

Problem 12. min_𝑋∈𝔻‖𝐴𝑋−𝐵𝑋‖²_𝐹,𝐴=,2334,𝐵=1324 (66) where 𝐴𝔻=¹𝑋,𝑋_𝐹+𝐴₂,𝑋_𝐹𝐴⩽0₂𝑋,𝑋_𝐹+𝐴₁,𝑋_𝐹𝐴⩽0₁=−1234,𝐴₂=2314𝑋≽0. (67) The global solution is 𝑋^∗=4425. (68)

Problem 13. min_𝑋∈𝐷‖𝑋‖²_𝐹subjectto𝔻=𝑋∈ℝ^𝑛×𝑛∣𝑋=2341⩽𝑋⩽=𝑋.7589,𝑋≽0 (69)(70) The global solution is 𝑋^∗=4389. (71)