Throughout the third part, $f \in \mathcal{C}_{0}$ satisfies property $(\mathcal{P}_{1})$: there exist $x_{0} \in [0,1]$, $s \in ]0,1[$ and $c_{1} \in ]0, +\infty[$, such that for all $(j, k) \in \mathcal{I}$, $$|c_{j,k}(f)| \leq c_{1} (2^{-j} + |k 2^{-j} - x_{0}|)^{s}$$ We fix $x_{0}$, $s$, $c_{1}$ and $x \in [0,1] \backslash \{x_{0}\}$. Let $n_{0}$ be the unique natural integer such that $2^{-n_{0}-1} < |x - x_{0}| \leq 2^{-n_{0}}$. We suppose that $\|f\|_{\infty} = 1$. Show that there exists a unique $n_{1} \in \mathbf{N}$ such that $\omega_{f}(2^{-n_{1}-1}) < 2^{-n_{0} s} \leq \omega_{f}(2^{-n_{1}})$.
Throughout the third part, $f \in \mathcal{C}_{0}$ satisfies property $(\mathcal{P}_{1})$: there exist $x_{0} \in [0,1]$, $s \in ]0,1[$ and $c_{1} \in ]0, +\infty[$, such that for all $(j, k) \in \mathcal{I}$,
$$|c_{j,k}(f)| \leq c_{1} (2^{-j} + |k 2^{-j} - x_{0}|)^{s}$$
We fix $x_{0}$, $s$, $c_{1}$ and $x \in [0,1] \backslash \{x_{0}\}$. Let $n_{0}$ be the unique natural integer such that $2^{-n_{0}-1} < |x - x_{0}| \leq 2^{-n_{0}}$. We suppose that $\|f\|_{\infty} = 1$.
Show that there exists a unique $n_{1} \in \mathbf{N}$ such that $\omega_{f}(2^{-n_{1}-1}) < 2^{-n_{0} s} \leq \omega_{f}(2^{-n_{1}})$.