Suppose we have the iterative inequality $\gamma\_{k+1} \leq \gamma\_k(1 - c \gamma\_k^\alpha)$ with $c, \alpha \in (0, 1)$ and $(1 - c \gamma\_k^\alpha)>0$ for all non-negative terms $\gamma\_k$.


-- How could we show that $\gamma\_k$ is polynomially decay, i.e., $\gamma\_k \leq C (k+1)^{-\frac{1}{\alpha}}$ for some constant $C$?


-- Could we show that the above constant $C$ is uniformly bounded with respect to $\alpha$? If not, what is a good form to describe the decay rate of $\gamma\_k$.



How to find a proper decay rate from an iterative inequality 
From the iterative inequality (assuming $1-c\gamma\_k^\alpha >0$ as said)
$$\gamma\_{k+1}^{-\alpha}\ge \gamma\_k^{-\alpha}\big(1- c\, \gamma\_k^\alpha\big)^{-\alpha}\ge \gamma\_k^{-\alpha}+\alpha c\, ,
$$
the last inequality just coming from the convexity inequality $(1-x)^{-\alpha}\ge 1+\alpha x $, for $0 < x < 1$.


Therefore $\gamma\_k^{-\alpha}\ge \gamma\_0^{-\alpha}+ k\alpha c\ge(k+1) \alpha c$, and your polynomial bound follows with $C:=(\alpha c ) ^{-1/\alpha}$.


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As to the other question, there is no uniform constant for all $0<\alpha<1$, and the reason is that the above constant is indeed optimal. Precisely, for any iteration $\gamma\_{k+1}= \gamma\_k(1- c \gamma\_k^\alpha)$ with $1-c\gamma\_0^\alpha >0$ we have 
$$\gamma\_k=Ck^{-1/\alpha}\big(1+o(1)\big)\, ,\quad\mathrm{for}\, k\to\infty \, .$$


Indeed, since $\gamma\_k\to 0$, we have 
 $$
\gamma\_{k+1}^{-\alpha}= \gamma\_k^{-\alpha}\big(1- c\, \gamma\_k^\alpha\big)^{-\alpha}= \gamma\_k^{-\alpha}+\alpha c+o(1)\, ,\quad\mathrm{for}\, k\to\infty \, ,
$$
because $(1-x)^{-\alpha}=1+\alpha x +o(x)$ for $x\to0$. Thus $\gamma\_{k^{-\alpha}}=k\alpha c+o(k)=k\alpha c\big(1+o(1)\big)$, and the above asymptotic follows. 


