Since $\sigma_k + \tilde{t}_k = \sigma_{k+1} \leq R$, for all $k \geq k(\ell)$, then we conclude that $$\label{eq.boundxk}
        \|x_k - x_{k(\ell)}\| \leq \sum_{j=k(\ell)}^k \|x_{j+1} - x_j\| \leq \sqrt{\frac{1+2 \bar{\gamma}}{1-2 \bar{\lambda}}} \sum_{j=k(\ell)}^k \tilde{t}_j= \sqrt{\frac{1+2 \bar{\gamma}}{1-2 \bar{\lambda}}} \,\, \sigma_{k+1} \leq \sqrt{\frac{1+2 \bar{\gamma}}{1-2 \bar{\lambda}}} \,\, R.$$ Hence, $\{x_k\}$ is bounded. Besides, from the last condition in , the sequence $\{\epsilon_k\}$ is bounded and, by using Proposition \[pr:CompE-subdif\], $\{s_k\}$ is also bounded. Moreover, by we also conclude $\sum_{j=k(\ell)}^{+\infty} \tilde{t}_j<+\infty$, which implies $\lim_{k\to \infty}  \tilde{t}_k = 0$. Thus, due to $\beta_k \in [\underline{\beta}, \bar{\beta}]$, it follows from second equality in that $$\label{eq.boundfk}
        \lim_{k \to \infty} [f(x_k) - f_k^{lev}] = 0.$$ On the other hand, Steps 3 and 5 of Algorithm \[Alg:INPDyn\] yield $$f(x_k) > f_{k(\ell)}^{rec} - \frac{1}{2} \delta_\ell = f_k^{lev} + \delta_\ell - \frac{1}{2} \delta_\ell = f_k^{lev} + \frac{1}{2} \delta_\ell \qquad \quad k= k(\ell), k(\ell)+1,  \ldots,$$ contradicting . Therefore, $\ell$ goes to $+ \infty$. Now, suppose that $ \lim_{\ell\to\infty} \delta_\ell=\delta  > 0$. Then, from Steps 3 and 4 of Algorithm \[Alg:INPDyn\], it follows that for all $\ell$ large enough, we have $\delta_\ell = \delta$ and $
    f_{k(\ell+1)}^{rec}  \leq f_{k(\ell)}^{rec} -\frac{1}{2} \delta,
    $ implying that $\displaystyle\inf_{k \geq 0} f(x_k) = -\infty$, which concludes the claim. If 