Let $f(x)=\frac12 x^2$, for $x\in\left (0,1\right]$. It is easily seen that $|f(x)-f(y)|\leq w(|x-y|) \;\;\;(x,y\in \left (0,1\right])$ where $w(t)=t-\frac{1}{2}t^2$, for $t\in\left (0,1\right]$ (is the modulus of continuity of $f$).
W. J. Thron proved (Theorem 3.1, http://eretrandre.org/rb/files/Thron1960_219.pdf) that iterations of every positive function of the form $v(t)=t-bt^{\alpha+1}\;\;\; (t\in\left (0,1\right]),$ where $b>0, \alpha>0$ can be estimate as follows: There exists $n_0\in \mathbb{N}$ such that $v^n(t)\leq \left(\frac12 \alpha b n\right)^{-1/\alpha} \;\;\; (t\in \left (0,1\right],\;n\geqslant n_0).$
Clearly, if $\alpha\in(0,1)$ it follows from the letter inequality that the series $\sum_{n=1}^{\infty} v^n(t)$ is convergent uniformly with respect to $t$. The function $f$, that I have initially defined, is of the form given above but with $\alpha=1$.
My question is: Is it possible to define $f$ such that its modulus of continuity $w$ is of the $w(t)=t-bt^{\alpha+1}$ with $\alpha\in(0,1)$?
I will be really gratefull for any hits. The existence of such a function could solve my recent problem: A series of iterations of a modulus of continuity.