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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.

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1 Answer 1

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If $f(x)$ is increasing and concave on $[0,1]$, the maximum of $f(x+t) - f(x)$ for $0 \le x \le 1-t$ is at $x=0$, so the modulus of continuity is $w(t) = f(t) - f(0)$. Taking for convenience $f(0)=0$, you want $f(t) = t - b t^{\alpha+1}$, which is indeed increasing and concave if $0 \le b \le 1/(1+a)$.

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So, it is enough to take $f(t)=t−bt^{α+1}$ with α<1 and the modulus of continuity of $f$ is just equal to f? Thank you very much. –  dawid Apr 2 '12 at 19:26
    
Thanks to you, finally I have an example I need in my work :) –  dawid Apr 2 '12 at 20:09

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