Based on Theorem 3.1, in this section, without loss of generality, we suppose that the sequence \(\{x^{k}\}\) generated by Algorithm A is infinite. In what follows, we will prove that \(\{x^{k}\}\) is bounded under some appropriate conditions, and that any accumulation point \(x^{*}\) of \(\{x^{k}\}\) is either an infeasible stationary point, or a KKT point of NLSDP (1.1). To this end, the following additional assumptions are necessary.
A 3
For any \(c>0\), the level set \(L_{c}:=\{x\in \mathbb{R}^{n}\mid P(x) \leq c\}\) is bounded.
A 4
For any feasible point x of NLSDP (1.1), MFCQ is satisfied at x, that is, there exists \(d\in \mathbb{R}^{n}\) such that
$$ \mathcal{A}(x)+D\mathcal{A}(x)d\prec 0. $$
Lemma 4.1
Suppose that Assumptions A1–A3
hold, then the iterative sequence
\(\{x^{k}\}\)
is bounded.
Proof
One of the following situations occurs:
-
(i)
If there exists an integer \(k_{1}\) such that \(P(x^{k}) \leq \bar{P}\) for any \(k>k_{1}\), then \(x^{k}\in L_{\bar{P}}\) for any \(k>k_{1}\). So \(\{x^{k}\}\) is also bounded because \(L_{\bar{P}}\) is bounded.
-
(ii)
If there exists an integer \(k_{2}\) such that \(P(x^{k})>\bar{P}\) for any \(k>k_{2}\), then it follows from Step 5 that \(x^{k}\in L_{P(x^{k_{2}})}\) for any \(k>k_{2}\). So \(\{x^{k}\}\) is also bounded because \(L_{P(x^{k_{2}})}\) is bounded.
-
(iii)
If both (i) and (ii) do not occur, i.e., \(P(x^{k})\leq \bar{P}\) and \(P(x^{k})>\bar{P}\) occur infinitely, respectively, then there exists an index set \(\{k_{j}\}\) satisfying
$$ P\bigl(x^{k_{j}}\bigr)\leq \bar{P}, \qquad P \bigl(x^{k_{j}+1}\bigr)>\bar{P},\quad \forall j\in \{1,2,\ldots\}. $$
(4.1)
So by arc search strategy, there exists an index set \(\{s_{j}\}\) associated with \(\{k_{j}\}\) such that
$$k_{j}< s_{j}< k_{j+1},\quad P \bigl(x^{s_{j}}\bigr)>\bar{P},\qquad P\bigl(x^{s_{j}+1}\bigr)\leq \bar{P},\quad \forall j\in \{1,2,\ldots\}. $$
For convenience, let \(N:=\{1,2,\ldots\}\), \(N_{j}:=\{k\mid k_{j}< k< k_{j+1} \}\), then we obtain
$$ \{k\in N\mid k>k_{1}\}=\bigcup _{j} \bigl\{ N_{j}\cup \{k_{2},k_{3},k_{4}, \ldots \} \bigr\} . $$
We know from (4.1) that \(\{x^{k_{j}}\}\subseteq L_{\bar{P}}\), so \(\{x^{k_{j}}\}\) is bounded, Hence, \(\{x^{k}\}\) is bounded as long as we can prove that there exists \(\bar{c}>0\) such that \(x^{k}\in L_{ \bar{c}}\), \(\forall j\in N\), \(k\in N_{j}\). Combining with the boundedness of \(\{x^{k_{j}}\}\) and Assumption A1, we get \(\{\triangledown f(x^{k_{j}})\}\) is bounded, i.e., there exists \(M>0\) such that \(\| \triangledown f(x^{k_{j}})\|\leq M\) for any \(j\in N\). In addition, it follows from \(\operatorname{LSDP}(x^{k})\) (2.7) and \(\operatorname{QSDP}(x^{k},H_{k})\) (2.8) that \(\{d^{k_{j}}\}\) is bounded. In view of \(x^{k_{j}+1}=x ^{k_{j}}\), we get \(\{x^{k_{j}+1}\}\) is bounded. Further, one obtains \(\{P(x^{k_{j}+1})\}\) is bounded due to the continuity of \(P(x)\). So there exists \(\bar{c}>0\) such that \(P(x^{k_{j}+1})\leq \bar{c}\).
At last, by (4.1) and Step 5 in Algorithm A, one has
$$\begin{aligned} \bar{c}\geq P\bigl(x^{k_{j}+1}\bigr)\geq P\bigl(x^{k_{j}+2}\bigr) \geq \cdots\geq P\bigl(x^{s_{j}}\bigr) \geq \bar{P}, \end{aligned}$$
(4.2)
$$\begin{aligned} \bar{P}\geq P\bigl(x^{s_{j}+1}\bigr),\bar{P}\geq P\bigl(x^{s_{j}+2} \bigr),\ldots,\bar{P} \geq P\bigl(x^{k_{j+1}}\bigr). \end{aligned}$$
(4.3)
We can find \(k_{j}\) and \(k_{j+1}\) such that \(k\in (k_{j},k_{j+1})\) for \(k\in N\). So it follows from (4.2) and (4.3) that \({x^{k}}\in L_{\bar{c}}\), i.e., \(\{x^{k}\}\) is bounded. □
Lemma 4.2
Suppose that Assumptions A1–A4
hold. If
\(\alpha _{k}\rightarrow +\infty \), then every accumulation point
\(x^{*}\)
of
\(\{x^{k}\}\)
generated by Algorithm A
is an infeasible stationary point of NLSDP (1.1).
Proof
If \(\alpha _{k}\rightarrow +\infty \), then it follows from (3.1) that the sequence \(\{\frac{\triangledown f(x ^{k})^{\mathrm{T}}d^{k}+(d^{k})^{\mathrm{T}}H_{k} d^{k}}{\lambda _{1}( \mathcal{A}(x^{k}))_{+}-z_{k}}\}\) diverges to +∞.
By (2.15) and (2.18), we have
$$\begin{aligned} \frac{\triangledown f(x^{k})^{\mathrm{T}}d^{k}+(d^{k})^{\mathrm{T}} H _{k} d^{k}}{\lambda _{1}(\mathcal{A}(x^{k}))_{+}-z_{k}} \le &\frac{ \operatorname{Tr}(\varLambda _{k}(\mathcal{A}(x^{k})-z_{k}E_{m}))}{\lambda _{1}(\mathcal{A}(x^{k})-z_{k}E_{m})} \\ \leq& \frac{\operatorname{Tr}(\varLambda _{k})\lambda _{1}(\mathcal{A}(x^{k})-z_{k}E_{m})}{\lambda _{1}( \mathcal{A}(x^{k})-z_{k}E_{m})} \\ =&\operatorname{Tr}(\varLambda _{k}). \end{aligned}$$
(4.4)
If \(x^{*}\) is a feasible point of NLSDP (1.1), then, by Assumption A4, we know that MFCQ is satisfied at \(x^{*}\). Similar to the proof of Theorem 5.1 in [27], we obtain that the set Ω of the KKT Lagrangian multipliers for \(\operatorname{QSDP}(x^{*},H_{*})\) (2.8) is nonempty and bounded. Note that \(\varLambda _{k} \stackrel{ \mathcal{K}}{\to } \varLambda _{*}\) and \(\varLambda _{*} \in \varOmega \), so \(\{\varLambda _{k}\}\) is bounded. Therefore it follows from (4.4) that \(\{\frac{\triangledown f(x^{k})^{\mathrm{T}}d^{k}+(d^{k})^{ \mathrm{T}} H_{k} d^{k}}{\lambda _{1}(\mathcal{A}(x^{k}))_{+}-z_{k}}\}\) is bounded. This is a contradiction. Hence, \(x^{*}\) is an infeasible point, i.e., \(\lambda _{1}(\mathcal{A}(x^{*}))>0\). Further, it is obvious that \((0,\lambda _{1}(\mathcal{A}(x^{*})))\) is a feasible solution of \(\operatorname{LSDP}(x ^{*})\) (2.7), so \(z_{*}\leq \lambda _{1}(\mathcal{A}(x^{*}))\). Let \((d^{*},z_{*})\) be an optimal solution of \(\operatorname{LSDP} (x^{*})\) (2.7), then by the constraint of \(\operatorname{LSDP}(x^{*})\) (2.7), we have
$$ \lambda _{1}\bigl(\mathcal{A}\bigl(x^{*}\bigr)+D \mathcal{A}\bigl(x^{*}\bigr)d^{*}\bigr)\preceq \lambda _{1}\bigl(\mathcal{A}\bigl(x^{*}\bigr)\bigr), $$
further,
$$ \max \bigl\{ \lambda _{1}\bigl(\mathcal{A}\bigl(x^{*} \bigr)+D\mathcal{A}\bigl(x^{*}\bigr)d^{*}\bigr),0\bigr\} \leq \lambda _{1}\bigl(\mathcal{A}\bigl(x^{*}\bigr) \bigr). $$
Therefore, we get
$$ \min_{d\in \mathbb{R}^{n}}\max \bigl\{ \lambda _{1}\bigl( \mathcal{A}\bigl(x^{*}\bigr)+D \mathcal{A}\bigl(x^{*} \bigr)d\bigr),0\bigr\} \leq \max \bigl\{ \lambda _{1}\bigl(\mathcal{A} \bigl(x^{*}\bigr)+D \mathcal{A}\bigl(x^{*} \bigr)d^{*}\bigr),0\bigr\} \leq \lambda _{1}\bigl( \mathcal{A}\bigl(x^{*}\bigr)\bigr). $$
Let \(d=0\), then \(\max \{\lambda _{1}(\mathcal{A}(x^{*})+D\mathcal{A}(x ^{*})d),0\}=\lambda _{1}(\mathcal{A}(x^{*}))\), which together with the above inequality implies
$$ \min_{d\in \mathbb{R}^{n}}\max \bigl\{ \lambda _{1}\bigl( \mathcal{A}\bigl(x^{*}\bigr)+D \mathcal{A}\bigl(x^{*} \bigr)d\bigr),0\bigr\} =\lambda _{1}(\mathcal{A}\bigl(x^{*} \bigr), $$
that is, \(x^{*}\) is an infeasible stationary point of NLSDP (1.1). □
In the rest of the paper, we assume \(\alpha _{k}<+\infty \). According to the update rule (3.1), the following conclusion is shown easily.
Lemma 4.3
Suppose that Assumptions A1–A4
hold. Then there exists an integer
\(k_{0}\)
such that
\(\alpha _{k}\equiv \alpha _{k_{0}} \triangleq \alpha >0\)
for any
\(k\geq k_{0}\).
Based on Lemma 4.3, in the rest of the paper, without loss of generality, we assume that \(\alpha _{k}\equiv \alpha \), \(k=1, 2, \ldots \) .
Lemma 4.4
Suppose that Assumptions A1–A2
hold, \(x^{k} {\longrightarrow } x^{*}\), \(H_{k} {\longrightarrow } H_{*}\). Then
\(z_{k} {\longrightarrow } z_{*}\), \(d^{k} {\longrightarrow } d^{*}\), where
\(z_{k}\), \(z_{*}\)
are the optimal solutions of
\(\operatorname{LSDP}(x^{k})\) (2.7) and
\(\operatorname{LSDP}(x^{*})\) (2.7), respectively, and
\(d^{k}\), \(d ^{*}\)
are the optimal solutions of
\(\operatorname{QSDP}(x^{k}, H_{k})\) (2.8) and
\(\operatorname{QSDP}(x^{*}, H_{*})\) (2.8), respectively.
Proof
Since \(z_{k}\) is the optimal value of \(\operatorname{LSDP}(x^{k})\) (2.7), we obtain \(z_{k}<\lambda _{1}(\mathcal{A}(x^{k}))_{+}\) due to the fact that \((0, \lambda _{1}(\mathcal{A}(x^{k}))_{+})\) is a feasible solution of \(\operatorname{LSDP}(x^{k})\) (2.7). By the boundedness of \(\{\lambda _{1}(\mathcal{A}(x^{k}))\}\) and \(z_{k}>0\), it is true that \(\{z_{k}\}\) is bounded. According to the sensitivity theory of semidefinite programming in [28], we know that the first part of the conclusions is true.
Now consider the second part of the conclusions. We first prove \(\{d ^{k}\}\) is bounded. It follows from \(\operatorname{LSDP}(x^{k})\) (2.7) that \(\|\widehat{d}^{k}\|\leq 1\). And obviously, \(\widehat{d}^{k}\) is a feasible solution of \(\operatorname{QSDP}(x^{k},H_{k})\) (2.8), so one obtains
$$\nabla f\bigl(x^{k}\bigr)^{\mathrm{T}}d^{k}+ \frac{1}{2}\bigl(d^{k}\bigr)^{\mathrm{T}}H _{k}d^{k}\leq \nabla f\bigl(x^{k} \bigr)^{\mathrm{T}}\widehat{d}^{k}+ \frac{1}{2}\bigl( \widehat{d}^{k}\bigr)^{\mathrm{T}}H_{k} \widehat{d}^{k}, $$
further, the above inequality gives rise to
$$\nabla f\bigl(x^{k}\bigr)^{\mathrm{T}}d^{k}+ \frac{1}{2}\bigl(d^{k}\bigr)^{\mathrm{T}}H _{k}d^{k}\leq \bigl\Vert \nabla f \bigl(x^{k}\bigr) \bigr\Vert \bigl\Vert \widehat{d}^{k} \bigr\Vert + \frac{1}{2} \bigl\Vert \widehat{d}^{k} \bigr\Vert ^{2}\bar{a}\leq M_{1}+\frac{1}{2} \bar{a}. $$
On the other hand, one has
$$\nabla f\bigl(x^{k}\bigr)^{\mathrm{T}}d^{k}+ \frac{1}{2}\bigl(d^{k}\bigr)^{\mathrm{T}}H _{k}d^{k}\geq - \bigl\Vert \nabla f \bigl(x^{k}\bigr) \bigr\Vert \bigl\Vert d^{k} \bigr\Vert +a \bigl\Vert d^{k} \bigr\Vert ^{2}\geq -M_{1} \bigl\Vert d^{k} \bigr\Vert +a \bigl\Vert d^{k} \bigr\Vert ^{2}. $$
The two inequalities above indicate that \(\{d^{k}\}\) is bounded.
Suppose that \(d^{k} \not \to d^{*}\), then there exists a subsequence \(\{d^{s}\}_{K_{1}}\subseteq \{d^{k}\}\) converging to d̄ (\(\neq d ^{*}\)). For any feasible solution d of \(\operatorname{QSDP}(x^{*}, H_{*})\) (2.8), since \(z_{k} {\rightarrow } z_{*}\), there exists a feasible solution \(d^{m}\) of \(\operatorname{QSDP}(x^{s}, H_{s})\) (2.8) such that
$$d^{m} \stackrel{K_{1}}{\longrightarrow } d. $$
Since \(d^{s}\) is the solution of \(\operatorname{QSDP}(x^{s}, H_{s})\) (2.8), one has
$$\nabla f\bigl(x^{s}\bigr)^{\mathrm{T}}d^{s}+ \frac{1}{2}\bigl(d^{s}\bigr)^{\mathrm{T}}H _{s} d^{s}\leq \nabla f\bigl(x^{s} \bigr)^{\mathrm{T}}d^{m}+\frac{1}{2}\bigl(d^{m} \bigr)^{ \mathrm{T}} H_{s} d^{m}. $$
Let \(s \stackrel{K_{1}}{\longrightarrow } \infty \), \(m \stackrel{K_{1}}{ \longrightarrow } \infty \), one gets
$$\nabla f\bigl(x^{*}\bigr)^{\mathrm{T}}\bar{d}+ \frac{1}{2}\bar{d}^{\mathrm{T}}H _{*}\bar{d} \leq \nabla f\bigl(x^{*}\bigr)^{\mathrm{T}}d+\frac{1}{2}d^{ \mathrm{T}}H_{*}d, $$
which means that d̄ is a solution of \(\operatorname{QSDP}(x^{*}, H_{*})\) (2.8). This contradicts the uniqueness of the solution of \(\operatorname{QSDP}(x ^{*}, B_{*})\) (2.8). □
Lemma 4.5
Suppose that Assumptions A1–A4
hold, \(x^{*}\)
is an accumulation point of the sequence
\(\{x^{k}\}\)
generated by Algorithm A, i.e., \(x^{k}\stackrel{K}{\longrightarrow } x^{*}\). If
\(x^{*}\)
is not an infeasible stationary point of NLSDP (1.1), then
\(d^{k}\stackrel{K}{\longrightarrow }0\).
Proof
By contradiction, suppose that
, then there exist a constant \(b>0\) and an index subset \(K'\subseteq K\) such that
$$ \bigl\Vert d^{k} \bigr\Vert \geq b>0 $$
(4.5)
for any \(k\in K'\). The following proof is divided into two steps.
Step A. We first prove \(\underline{t}:=\inf \{t_{k}, k \in K'\}>0\).
By Taylor expansion and the boundedness of the sequence \(\{d^{k}\}\), one has
$$\begin{aligned} f\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr) =&f\bigl(x^{k}\bigr)+t\nabla f \bigl(x^{k}\bigr)^{ \mathrm{T}}d^{k}+o(t), \\ {\mathcal{A}}\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr) =&\mathcal{A}\bigl(x ^{k} \bigr)+t\sum_{i=1}^{n} d_{i}^{k}\frac{\partial {\mathcal{A}}(x^{k})}{ \partial x_{i}}+o(t). \end{aligned}$$
(4.6)
In view of \(t\leq 1\), combining with the convexity of \(\lambda _{1}( \cdot )_{+}\) and \(\operatorname{QSDP}(x^{k},H_{k})\) (2.8), one obtains
$$\begin{aligned}& \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr)\bigr)_{+} \\& \quad \leq (1-t)\lambda _{1}\bigl(\mathcal{A}\bigl(x^{k} \bigr)\bigr)_{+}+t\lambda _{1} \Biggl( \mathcal{A} \bigl(x^{k}\bigr)+\sum_{i=1}^{n} d_{k_{i}}\frac{\partial {\mathcal{A}}(x ^{k})}{\partial x_{i}} \Biggr)_{+}+o(t) \\& \quad \leq (1-t)\lambda _{1}\bigl(\mathcal{A}\bigl(x^{k} \bigr)\bigr)_{+}+t\lambda _{1}(z_{k}E _{m})_{+}+o(t), \end{aligned}$$
(4.7)
which together with (2.12), (4.6), and (4.7) gives
$$\begin{aligned}& \theta _{\alpha }\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr) \\& \quad \leq f\bigl(x^{k}\bigr)+t\nabla f\bigl(x^{k} \bigr)^{\mathrm{T}}d^{k}+o(t)+\alpha \bigl[(1-t) \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}\bigr) \bigr)_{+}+t\lambda _{1}(z_{k}E_{m})_{+}+o(t) \bigr] \\& \quad =f\bigl(x^{k}\bigr)+\alpha \lambda _{1}\bigl( \mathcal{A}\bigl(x^{k}\bigr)\bigr)_{+}+t\bigl(\nabla f \bigl(x ^{k}\bigr)^{\mathrm{T}}d^{k}-\alpha \bigl( \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}\bigr) \bigr)_{+}- \lambda _{1}(z_{k} E_{m})_{+}\bigr)\bigr)+o(t) \\& \quad =\theta _{\alpha }\bigl(x^{k}\bigr)+t\Delta \bigl(x^{k},\alpha \bigr)+o(t), \end{aligned}$$
so we obtain
$$\begin{aligned} \theta _{\alpha }\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr)-\theta _{\alpha }\bigl(x^{k} \bigr)-\beta t\triangle \bigl(x^{k},\alpha \bigr)\leq (1-\beta )t \triangle \bigl(x ^{k},\alpha \bigr)+o(t). \end{aligned}$$
(4.8)
By Lemma 4.4 and (3.12), we get
$$ \Delta \bigl(x^{k},\alpha \bigr)\leq -\bigl(d^{k} \bigr)^{\mathrm{T}}H_{k}d^{k}\longrightarrow - \bigl(d^{*}\bigr)^{\mathrm{T}}H_{*}d^{*}< 0 \quad \mbox{as } k\ \bigl(\in K'\bigr)\longrightarrow \infty , $$
so we have
$$ \triangle \bigl(x^{k},\alpha \bigr)\leq -0.5\bigl(d^{*} \bigr)^{\mathrm{T}}H_{*}d^{*} $$
for k (\(\in K'\)) sufficiently large. Substituting the above inequality into (4.8) gives
$$ \theta _{\alpha }\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr)-\theta _{\alpha }\bigl(x^{k} \bigr)-\beta t\triangle \bigl(x^{k},\alpha \bigr)\leq -0.5(1-\beta )t \bigl(d^{*}\bigr)^{ \mathrm{T}}H_{*}d^{*}+o(t), $$
(4.9)
which means that, for k (\(\in K'\)) sufficiently large and t sufficiently small, inequality (3.3) or the first inequality of (3.4) holds.
In what follows, we consider the second inequality of (3.4).
Note that \(P(x^{k})>\bar{P}>0\), so \(P(x^{*})=\lim_{K'} P(x^{k}) \ge \bar{P}>0\), which means \(x^{*}\) is an infeasible solution of NLSDP (1.1). Since \(x^{*}\) is not an infeasible stationary point of NLSDP (1.1), it follows that \(z_{*}-\lambda _{1}(\mathcal{A}(x ^{*}))_{+}<0\). Further, we have
$$ \lambda _{1}(z_{k}E_{m})_{+}- \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}\bigr) \bigr)_{+} \longrightarrow z_{*}-\lambda _{1}\bigl(\mathcal{A}\bigl(x^{*}\bigr) \bigr)_{+}< 0\quad \mbox{as }k\bigl(\in K'\bigr) \rightarrow \infty , $$
so it follows that, for k (\(\in K'\)) sufficiently large,
$$ \lambda _{1}(z_{k}E_{m})_{+}- \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}\bigr) \bigr)_{+} < 0.5\bigl(z _{*}-\lambda _{1} \bigl(\mathcal{A}\bigl(x^{*}\bigr)\bigr)_{+}\bigr). $$
By (4.7), (2.11), and the above inequality, one has
$$\begin{aligned} P\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr) \leq &P\bigl(x^{k}\bigr)+t \bigl(\lambda _{1}(z _{k}E_{m})_{+}- \lambda _{1}\bigl(\mathcal{A}\bigl(x^{k}\bigr) \bigr)_{+}\bigr)+o(t) \\ \leq &P\bigl(x^{k}\bigr)+0.5t\bigl(z_{*}-\lambda _{1}\bigl(\mathcal{A}\bigl(x^{*}\bigr) \bigr)_{+}\bigr)+o(t), \end{aligned}$$
equivalently,
$$\begin{aligned} P\bigl(x^{k}+td^{k}+t^{2} \widetilde{d}^{k}\bigr)-P\bigl(x^{k}\bigr)\leq 0.5t \bigl(z_{*}-\lambda _{1}\bigl(\mathcal{A} \bigl(x^{*}\bigr)\bigr)_{+}\bigr)+o(t), \end{aligned}$$
(4.10)
which implies that, for k (\(\in K'\)) sufficiently large and t sufficiently small, the second inequality in (3.4) holds.
Summarizing the analysis above, we can conclude \(\underline{t}:= \inf \{t_{k}, k\in K'\}>0\).
Step B. Based on \(\underline{t}= \inf \{t_{k}, k\in K'\}>0 \), we prove a contradiction will occur.
It follows from (3.3) or (3.4) that \(\{\theta _{ \alpha }(x^{k})\}\) is nonincreasing and
$$ \theta _{\alpha }\bigl(x^{k+1}\bigr)\leq \theta _{\alpha }\bigl(x^{k}\bigr)-0.5ab^{2}\beta \underline{t} $$
(4.11)
for any \(k\in K'\), where b is defined in (4.5). And one obtains from (2.12) that
$$ \theta _{\alpha }\bigl(x^{k}\bigr)=f\bigl(x^{k} \bigr)+\alpha \bigl(\lambda _{1}\bigl(\mathcal{A}\bigl(x ^{k}\bigr)\bigr)_{+}\bigr)\geq f\bigl(x^{k} \bigr), $$
combining with the boundedness of \(\{f(x^{k})\}\), we conclude that \(\{\theta _{\alpha }(x^{k})\}\) is convergent. Taking \(k\stackrel{ K'}{ \longrightarrow }\infty \) in (4.11), we obtain \(-0.5ab^{2} \beta \underline{t}\geq 0\). This is a contradiction. So \(\lim_{K}d^{k}=0\). □
Based on the above results, we are now in a position to present the global convergence of Algorithm A.
Theorem 4.1
Suppose that Assumptions A1–A4
hold, \(x^{*}\)
is an accumulation point of the sequence
\(\{x^{k}\}\)
generated by Algorithm A. Then either
\(x^{*}\)
is an infeasible stationary point, or a KKT point of NLSDP (1.1).
Proof
Without loss of generality, we suppose that \(x^{*}\) is not an infeasible stationary point of NLSDP (1.1). In what follows, we show that \(x^{*}\) is a KKT point of NLSDP (1.1).
By Lemmas 4.4–4.5, we know that \(d^{*}=0\) is an optimal solution of \(\operatorname{QSDP}(x^{*},H_{*})\) (2.8), so it follows from Lemma 2.5 that there exists \(\varLambda _{*}\in \mathbb{S}^{m}_{+}\) such that
$$\begin{aligned}& \nabla f\bigl(x^{*}\bigr)+D\mathcal{A}\bigl(x^{*} \bigr)^{\ast }\varLambda _{*} = 0, \end{aligned}$$
(4.12a)
$$\begin{aligned}& \mathcal{A}\bigl(x^{*}\bigr) \preceq z_{*} E_{m}, \end{aligned}$$
(4.12b)
$$\begin{aligned}& \operatorname{Tr}\bigl(\varLambda _{*}\bigl(\mathcal{A} \bigl(x^{*}\bigr)-z_{*}E_{m}\bigr)\bigr)=0. \end{aligned}$$
(4.12c)
Now we prove \(z_{*}=0\). By contradiction, if \(z_{*}\neq 0\), then \(z_{*}>0\). Since (0, \(\lambda _{1}(A(x^{*}))_{+}\)) is a feasible solution of \(\operatorname{LSDP}(x^{*})\) (2.7), we get \(\lambda _{1}(A(x^{*}))_{+}\geq z _{*}>0\), which implies \(x^{*}\) is an infeasible point of NLSDP (1.1). On the other hand, we get \(z_{*}\geq \lambda _{1}(A(x^{*}))_{+}>0\) by (4.12b), so \(z_{*}=\lambda _{1}(A(x^{*}))_{+}\). Obviously, \((0, z_{*}=\lambda _{1}(A(x^{*}))_{+})\) is an optimal solution of \(\operatorname{LSDP}(x^{*})\) (2.7). In a manner similar to the proof of Theorem 3.1, we can conclude that \(x^{*}\) is an infeasible stationary point of NLSDP (1.1), this is a contradiction.
Substituting \(z_{*}=0\) into (4.12a)–(4.12c), one obtains
$$\begin{aligned}& \nabla f\bigl(x^{*}\bigr)+D\mathcal{A}\bigl(x^{*} \bigr)^{\ast }\varLambda _{*} =0, \\& \mathcal{A}\bigl(x^{*}\bigr)\preceq 0, \qquad \operatorname{Tr}\bigl( \varLambda _{*}{\mathcal{A}}\bigl(x^{*}\bigr)\bigr) =0, \end{aligned}$$
which means that \(x^{*}\) is a KKT point of NLSDP (1.1). □