to $\Lambda$, the inequality $$\label{bout1}
\max_{0 \leq \eta \leq \xi}\;\min_{||x||=\eta, \; x \in \Lambda}||\nabla P_{I,\Lambda}(x)||>C'\xi^{m_0-1}$$ holds true for all $0 < \xi \leq\delta'$. Assume $r\leq \delta'$, then for $x \in \Lambda$ such that $||x||\leq \xi \leq \delta=r$ and for any $I \in B_{2r}$, we have $I+x\in B_{3r} \subseteq B_{\bar{r}}$ hence by Taylor’s formula (applied to $\nabla h$ at order $m_0$) we obtain $$||\nabla h(I+x)-\nabla h(I)-\nabla P_I(x)||\leq Mm_0!||x||^{m_0}.$$ Therefore, if we assume $r$ sufficiently small with respect to $M$, $m_0$ and $C'$, we get $$||\nabla h(I+x)-\nabla h(I)-\nabla P_I(x)||\leq (C'/2)||x||^{m_0-1}\leq (C'/2)\xi^{m_0-1}$$ and hence, projecting the above vector onto $\Lambda$, we have $$\label{bout2}
||\nabla h_{\lambda}(I+x)-\nabla h_{\lambda}(I)-\nabla P_{I,\Lambda}(x)||\leq (C'/2)\xi^{m_0-1}$$ where $\lambda=I+\Lambda$. Letting $I'=I+x$, the inequalities  and  yield, for any $I \in B_{2r}$, the inequality $$\max_{0 \leq \eta \leq \xi}\;\min_{||I-I'||=\eta, \; I' \in \lambda}||\nabla h_{\lambda}(I')-\nabla h_{\lambda}(I)||> (C'/2)\xi^{m_0-1}=C\xi^{m_0-1}$$ for all $0 < \xi\leq \delta=r$. This concludes the proof.

We will also need a version of Proposition \[propsteep1\], in which $I=0 \in {\mathbb{R}}^n$ is replaced by a compact set $K \subseteq {\mathbb{R}}^n$.

\[propsteep2\] Let $K \subseteq {\mathbb{R}}^n$ be a compact set, $h : V_{\bar{r}}K \rightarrow {\mathbb{R}}$ be a function of class $C^{m_0+1}$ such that $||\nabla h(I)||\geq\varpi>0$ and $T_I^{m_0}h$ is $(\rho',C',\delta')$-stably steep for any $I \in K$. Then 