$\lim_{\ell\to\infty} \delta_\ell > 0$ then, according to above claim, we have $\inf_{k \geq 0} f(x_k) = -\infty$, obtain the desired result. Now, we assume by contradiction that $\lim_{\ell\to\infty} \delta_\ell = 0$ and $\inf_{k \geq 0} f(x_k) > f^*$. Thus, it follows from Remark \[eq:bfkrec\] that $\inf_{k \geq 0} f_{k}^{rec} =\inf_{k \geq 0} f(x_k) $. Hence, we conclude that $\inf_{k \geq 0} f_{k}^{rec}> f^*$. In this case, by using the definition of $\{f_k^{lev}\}$ in Step 5 and taking into account that $\lim_{\ell\to\infty} \delta_\ell = 0$, we conclude that $$\displaystyle\inf_{k \geq 0} f_k^{lev} =\displaystyle\inf_{\ell \geq 0} (f_{k(\ell)}^{rec} - \delta_\ell) = \displaystyle\inf_{\ell \geq 0}  f_{k(\ell)}^{rec} > f^*.$$ Therefore, there exist $\bar{\delta}>0$, ${\bar x}\in  C$ and $\bar{k} \in \mathbb{N}$ such that $$\label{eq:ahc}
        f_k^{lev} - f({\bar x}) \geq \bar{\delta}, \qquad  k =\bar{k}, \bar{k}+1, \ldots.$$ Hence, by using the definition of $\tilde{t}_k$ in , it follows from that $$\label{eq:limtildtk}
        \tilde{t}_k = \beta_k\frac{f(x_k)-f_{k}^{lev}}{\left\|s_k\right\|} < \bar{\beta} \frac{f(x_k)-f({\bar x})-\bar{\delta}}{\left\|s_k\right\|}, \qquad  k =\bar{k}, \bar{k}+1, \ldots.$$ Now, applying Lemma \[Le:FejerConv\] with $x={\bar x}$ and then using and , we obtain $$\|x_{k+1} - {\bar x}\|^2  \leq \|x_k-{\bar x}\|^2+ \tilde{t}_k \left(\nu \tilde{t}_k - \frac{2}{\|s_k\|} \left[f(x_k)-f({\bar x})-\epsilon_k\right] \right), \qquad  k =\bar{k}, \bar{k}+1, \ldots.$$ Thus, the combination of the last inequality with , , and 