# Delayed Star Subgradient Methods for Constrained Nondifferentiable Quasi-Convex Optimization
## 3 Algorithms and Convergence Results
### Delayed Star Subgradient Method I
This yields that  
\[f(\mathbf{x}_{k-\tau_{k}})>f^{*}+\epsilon\qquad\text{ for all }k\geq k_{1}+\tau. \tag{6}\]  
On the other hand, since \(\lim_{k\to\infty}\alpha_{k}=0\) and \(\lim_{k\to\infty}\|\mathbf{x}_{k}-\mathbf{x}_{k-\tau_{k}}\|=0\) (from Lemma 4), there exists \(k_{2}\in\mathbb{N}_{0}\) such that  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]0  
Let \(\mathbf{x}^{*}\in X^{*}\) and let \(k\in\mathbb{N}_{0}\) be such that \(k\geq N_{1}:=\max\{k_{1}+\tau,k_{2}\}\). Then, we obtain from Lemma 3 together with the relations (6) and (7) that for any \(k\geq N_{1}\),  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]1 \[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]2  
which yields  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]3  
Since \(\sum_{k=0}^{\infty}\alpha_{k}=\infty\), we have \(\left(\frac{\epsilon}{L}\right)^{\frac{1}{p}}\sum_{k=N_{1}}^{\infty}\alpha_{k}=\infty\), and it follows that there exists \(N_{2}\in\mathbb{N}_{0}\) such that  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]4  
Thus, for any \(k\geq N:=\max\{N_{1},N_{2}\}\), we obtain from (8) and (9) that  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]5which is a contradiction. Therefore, we can conclude that \(\liminf_{k\to\infty}f(\mathbf{x}_{k})\leq f^{*}\), and hence \(\liminf_{k\to\infty}f(\mathbf{x}_{k})=f^{*}\), as desired. Corollary 1: _Suppose that Assumptions 1 and 2 hold. Let \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) be a sequence generated by DSSM-I, and let \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) be a nonincreasing sequence with \(\lim_{k\to\infty}\alpha_{k}=0\) and \(\sum_{k=0}^{\infty}\alpha_{k}=\infty\). If the sequence \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) is bounded, then there exists a subsequence of \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) that converges to an optimal solution in \(X^{*}\)._  
Proof: We first note from the fact that \(\liminf_{k\to\infty}f(\mathbf{x}_{k})=f^{*}\) in Theorem 1 together with the boundedness of the sequence \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) and the continuity of \(f\) that there exists a subsequence \(\{\mathbf{x}_{k_{j}}\}_{j=0}^{\infty}\) of \(\{\mathbf{x}_{k}\}_{k=0}^{\infty}\) such that  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]6  
Since the sequence \(\{\mathbf{x}_{k_{i}}\}_{i=0}^{\infty}\) is also bounded, there exists a subsequence \(\{\mathbf{x}_{k_{i_{j}}}\}_{j=0}^{\infty}\) of \(\{\mathbf{x}_{k_{i}}\}_{i=0}^{\infty}\) such that \(\lim_{j\to\infty}\mathbf{x}_{k_{i_{j}}}=\bar{\mathbf{x}}\in\mathbb{R}^{n}\). Again, the continuity of \(f\) and using (10) yield  
\[f(\mathbf{x}_{k})-f^{*}\leq L\langle\mathbf{g}_{k},\mathbf{x}_{k}-\mathbf{x}^ {*}\rangle^{p}\qquad\text{ for all }\mathbf{g}_{k}\in\partial^{*}f(\mathbf{x}_{k})\cap \mathcal{S}.\]7  
Note that since \(\{\mathbf{x}_{k_{i_{j}}}\}_{j=0}^{\infty}\subset X\), the closedness of \(X\) yields \(\bar{\mathbf{x}}\in X\), which therefore implies that \(\bar{\mathbf{x}}\) is an element in \(X^{*}\).