# Upper bound implication for a real, nonnegative and nonincreasing function.

My question is

If there are positive numbers $$a<1$$ and $$\infty>C>0: f(i)\leq Ca^i$$, does it holds that there are finite positive numbers $$C_0,b:f(i)\leq C_0i^{-b}$$, where it is assumed that $$f\geq 0$$, bounded and nonincreasing. Equivalently,

$$\exists C>0,a<1:f(i)\leq Ca^i\implies \exists C_0,b>0:f(i)\leq C_0i^{-b},$$ for any positive $$i$$, where $$f$$ is a real nonnegative, bounded and nonincreasing function?

Observe that for every $$i \geq 0$$ one has $$a^i \leq \frac{C_1}{i}$$, where $$C_1 > 0$$ is un ugly constant depending on $$a$$. Indeed, the function $$ia^i$$ is 0 in 0, tends to 0 at $$+\infty$$ and by differentiating we find its maximum in the point $$i_0$$ such that $$a^{i_0} + i_0 \log a a^{i_0} = 0 \implies i_0 = \frac{-1}{\log a},$$ where the value of the function is $$\frac{-a^{\frac{-1}{\log a}}}{\log a} = C_1 > 0$$. With this definition you can take $$C_0 = CC_1$$ and $$b = 1$$ to get $$f(i) \leq Ca^i \leq CC_0 i^{-1}$$.