\infty}f(x_{k_j})= f^*.$ Without lose of generality, we can assume that $\{f(x_k)\}$ is decreasing, is monotonous, and converges to $f^*$. Being bounded, the sequence $\{x_k\}$ has a convergent subsequence $\{x_{k_\ell}\}$. Let us say that $\lim_{\ell\rightarrow\infty}x_{k_\ell}=x_*,$ which by the continuity of $f$ implies $f(x_*)=\lim_{\ell\rightarrow\infty}f(x_{k_\ell})=f^*,$ and then $x_*\in\Omega^*$. Hence, $\{x_k\}$ has an cluster point $x_*\in\Omega$, and due to $\{x_k\}$ be quasi-Féjer convergent to $\Omega$, Theorem \[teo.qf\] implies that $\{x_k\}$ converges to $x_*$.

Next theorem presents an iteration-complexity bound; similar bound can be found in [@Nesterov2004 Theorem 3.2.2].

\[teo:complrule1\] Assume that the sequence $\{x_k\}$ converges to a point $x_*\in\Omega^*$. Then, for every $N \in \mathbb{N}$, the following inequality holds $$\min \{f(x_k) - f^{*}:~ \, k = 0, 1,\ldots, N \} \leq \varGamma \frac{\|x_0 - x_{*}\|^2 + \rho\sum_{k=0}^{N}\alpha_k^{2}}{2\sum_{k=0}^{N}\alpha_k}.$$

Since $\left\lbrace \epsilon_k \right\rbrace $ and $\left\lbrace x_k\right\rbrace $ are bounded sequences, then using Proposition \[pr:CompE-subdif\], it follows that $\left\lbrace s_k\right\rbrace$ is also bounded, i.e. there exists $c > 0$ such that $\|s_k\| \leq c$, for all $k \geq 0$. Therefore, using the definition of $\eta_k$ in , we have $\eta_k = \max\left\{1, \| s_k\|\right\} \leq \max\left\{1, c \right\} =: \varGamma$. Now, applying Lemma \[Le:FejerConvExog\] with $x = x_*$ and due to $f^* = f(x_*)$, we obtain 