a contradiction, which proves that $\tau^{*}_{j_{0}} = 1$.

$\bullet$ Step $2$ : we prove here that $ NS(\varphi^{j_{0}})$ is a sup-critical solution in $\dot{H}^s$.\
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Let us come back to Inequality (\[equation de reference 2\]), which we multiply by the factor $(1 -\tau)^{\frac{2}{\sigma_{s}}}$. As we have shown that $\tau^{*}_{j_{0}} = 1$, hypothesis on $NS(v_{0,n})$ implies that for any $\tau < 1$, $$M^{2}_{c} + \varepsilon_{n}  \geqslant (1 - \tau)^{\frac{2}{\sigma_{s}}} \, \| NS(\varphi^{j_{0}})(\tau) \|^2_{\dot{H}^s}\,  -\,   |\gamma_{n}^{J}(\tau)|.$$ Our aim is to prove that the particular profile $\varphi^{j_{0}}$ generates a sup-critical solution. If not, it means that $$\exists \alpha_{0} >0, \forall \varepsilon >0, \,\, \exists \tau_{\varepsilon},\,\, \hbox{such that} \,\,0<  ( 1 -\tau_{\varepsilon} )^{\frac{2}{\sigma_{s}}}\, < \varepsilon \quad  \hbox{and} \quad  (1 -\tau_{\varepsilon} )^{\frac{2}{\sigma_{s}}}\,  \| NS(u_{0,n})(\tau_{\varepsilon}) \|^{2}_{\dot{H}^s} \geqslant M_{c}^2 + \alpha_{0}.$$ Taking the above inequality at time $\tau_{\varepsilon}$, one has $$M^{2}_{c} + \varepsilon_{n}  \geqslant\,  M_{c}^2 + \alpha_{0}  -\,   |\gamma_{n}^{J}(\tau_{\varepsilon})|.$$

Moreover, assumption on the remaining term $\gamma_{n}^{J}$ implies that $$\forall \eta >0,\,\, \exists \widetilde{J}(\eta) \in \N, \,\, \exists N_{\eta} \in \N\,\, \hbox{such that}\,\, \forall J \geq \widetilde{J}(\eta),\,\, \forall n \geqslant N_{\eta},\,\,  |\gamma_{n}^{J}(\tau_{\varepsilon})|\leqslant \eta.$$

Let $\eta >0$. For any $J \geq \widetilde{J}(\eta)$ and for any $n \geqslant N_{\eta}\,\, $, we get at time $\tau_{\varepsilon}$, $$M^{2}_{c}  \geqslant  M_{c}^2 + \alpha_{0} 