Suppose that for a certain function $b \colon \mathbb R \to \mathbb R$ I can show that there exists a local modulus of continuity $\omega_R$ and $C>0$ (both independent of $\epsilon$) such that $$|b(x)-b(y)|\leq \frac{|x-y|}{\epsilon} \omega_R(|x-y|) $$ for every $\epsilon>0$.
Can I say then that $$|b(x)-b(y)|\leq \gamma_R \left (\frac{|x-y|^2}{\epsilon} \right ) $$ for some local modulus of continuity $\gamma_R$ (independent of $\epsilon$)?
Indeed if $\omega_R=C_R$ some constant depending of $R$ then this is true. In general I would like to write $\epsilon=\sqrt \epsilon \sqrt \epsilon$ but then I cannot say that $\omega_R(|x-y|)/ \sqrt \epsilon \leq \omega_R(|x-y|/\epsilon)$. If this was true then I would have $\gamma_R$. Any ideas?
If you divide by $|x-y|$ you get $$ \left\vert \frac{b(x)-b(y)}{x-y}\right\vert \leq\frac{1}{\varepsilon} \omega_{R}(|x-y|). $$ Letting $y\rightarrow x$ you get $$ 0\leq\limsup_{y\rightarrow x}\left\vert \frac{b(x)-b(y)}{x-y}\right\vert \leq\lim_{y\rightarrow x}\frac{1}{\varepsilon}\omega_{R}(|x-y|)=0 $$ since $\omega_{R}$ is a modulus of continuity. Hence, $$ \lim_{y\rightarrow x}\left\vert \frac{b(x)-b(y)}{x-y}\right\vert =0, $$ which implies that $$ b^{\prime}(x)=\lim_{y\rightarrow x}\frac{b(x)-b(y)}{x-y}=0. $$ Thus, $b^{\prime}(x)=0$ for every $x$, and so $b$ is a constant function.