## 4 Global Convergence
We first consider case (i), in which two subcases may occur: \(t_{\infty}:=\lim_{k}t_{k}=\infty\) and \(t_{\infty}<\infty\). The first subcase of \(t_{\infty}=\infty\) is analyzed in the following lemma. **Lemma 4**. : _Suppose that Algorithm 1 generates finitely many descent steps, that is, there exists an index \(\bar{k}\) such that only null steps occur for all \(k\geq\bar{k}\), and that \(t_{\infty}=\infty\). Denote \(\mathcal{K}:=\{k\geq\bar{k}:t_{k+1}>t_{k}\}\), then \(V_{k}\to 0\) as \(k\in\mathcal{K},\ k\to\infty\)._  
Proof. : For the last time \(t_{k}\) increases before Step 5 for \(k\in\mathcal{K}\), one has  
\[V_{k}<\left(\frac{2\varepsilon_{x}^{\bar{k}}}{t_{k}}\right)^{1/2}\left(1+\left\| \bar{x}^{\bar{k}}\right\|\right),\]  
which along with \(t_{k}\to\infty\) shows the lemma. The following lemma analyzes the second subcase of \(t_{\infty}<\infty\). **Lemma 5**. : _Suppose that there exists \(\bar{k}\) such that \(\hat{x}^{k}=\hat{x}^{\bar{k}}\) and \(t_{\min}\leq t_{k+1}\leq t_{k}\) for all \(k\geq\bar{k}\). If the descent criterion (16) fails for all \(k\geq\bar{k}\), then \(V_{k}\to 0\)._  
Proof. : In view of the facts that \(t_{\min}\leq t_{k+1}\leq t_{k}\) and \(\hat{x}^{k}=\hat{x}^{\bar{k}}\) for all \(k\geq\bar{k}\), we know that only null steps occur and \(t_{k}\) does not increase at Step 5. By [PERSON]'s expansion, Cauchy-Schwarz inequality, and the properties of subproblems (11) and (13), we can conclude that \(v_{k}\to 0\), so the conclusion holds from (27). For more details, one can refer to [11], Lemma 3.2. By combining Lemmas 4 and 5, we have the following lemma. **Lemma 6**. : _Suppose that there exists \(\bar{k}\) such that only null steps occur for all \(k\geq\bar{k}\). Let \(\mathcal{K}:=\{k\geq\bar{k}:t_{k+1}>t_{k}\}\) if \(t_{k}\to\infty\); \(\mathcal{K}:=\{k:k\geq\bar{k}\}\) otherwise. Then \(V_{k}\overset{\mathcal{K}}{\to}0\)._  
Now, we can present the main convergence result for the case where the algorithm generates finitely many descent steps. **Theorem 1**. : _Suppose that Algorithm 1 generates finitely many descent steps, and that \(\hat{x}^{\bar{k}}\) is the last stability center. Then, \(\hat{x}^{\bar{k}}\) is an optimal solution to problem (1) for instances Ex and PI; an \(\varepsilon\)-optimal solution for IE; and an \(\varepsilon_{x}^{\bar{k}(l)}\)-optimal solution for AE and PAE._  
Proof. : Under the stated assumption, we know that \(\hat{x}^{k}\equiv\hat{x}^{\bar{k}}\) and \(f^{k}_{\bar{x}}\equiv f^{\bar{k}}_{\bar{x}}\) for all \(k\geq\bar{k}\). This together with (24) and Lemma 6 shows that  
\[F^{\bar{k}}_{\bar{x}}\leq\inf\{F(x):\ x\in\mathbb{R}^{n}\}=F_{*}.\]  
Hence, similar to the proof of Lemma 3, we obtain the results of the theorem. Next, we consider the second case where the algorithm generates infinitely many descent steps. **Lemma 7**. : _Suppose that Algorithm 1 generates infinitely many descent steps, and that \(F^{\infty}_{\bar{x}}:=\underset{k}{\lim}\,F^{k}_{\bar{x}}>-\infty\). Let \(K:=\{k\ :F^{k+1}_{\bar{x}}<F^{k}_{\bar{x}}\}\). Then \(v_{k}\overset{K}{\to}0\) and \(\underset{k\in K}{\lim}V_{k}=0\). Moreover, if \(\{\hat{x}^{k}\}\) is bounded, then \(V_{k}\overset{K}{\to}0\)._  
Proof. : From the descent test condition (16), we may first prove that \(v_{k}\overset{K}{\to}0\), and therefore \(\epsilon_{k},t_{k}\|_{p}k\|^{2}\overset{K}{\to}0\) from (26) and \(\|p_{k}\|\overset{K}{\to}0\) from the fact that \(t_{k}\geq t_{\min}\). It can be further proved that \(\epsilon_{k},\|p_{k}\|\overset{K}{\to}0\), so it follows \(\underset{k\in K}{\lim}V_{k}=0\) from the definition of \(V_{k}\). Moreover, under the condition that \(\{\hat{x}^{k}\}\) is bounded, we have \(V_{k}\overset{K}{\to}0\) by (v) of Lemma 1. For more details, one can refer to [11], Lemma 3.4. Finally, we present the convergence results for the second case. **Theorem 2**. : _Suppose that Algorithm 1 generates infinitely many descent steps, \(F^{\infty}_{\bar{x}}>-\infty\), and that the index set \(K\) is defined in Lemma 7. Then_  
1.