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Problem statement

Let $f(t) > C_1\cdot e^{-k_1 \cdot t}$ for some $C_1,k_1 > 0$ meeting the following inequality: $$ f(t) \leq C_2 \cdot e^{-k_2 \cdot \int_0^t f(\tau) d\tau}$$ It is also known that $\lim_{t \to \infty} f(t) = 0$ and that $|f(t)| < M$ for some $M \in \mathbb{R}_{>0}$ and also assume that $f$ is continuous.

Question

Can I obtain some estimates on the speed of convergence to zero?

My work:

Assume that $\exists t_0 > 0$ such that $f(t) \geq \frac{2}{k_2\cdot t}$ for all $t \geq t_0$. Then:

$$ \frac{2}{k_2\cdot t} \leq f(t) \leq C_2 \cdot e^{-k_2\cdot \int_{0}^{t_0} f(\tau) d\tau} \cdot e^{-k_2\cdot \int_{t_0}^{t} f(\tau) d\tau} \leq C_3 \cdot e^{-log(t^2)}\cdot t_0^2 \leq C_3 \cdot \frac{1}{t^2}$$ which is a contradiction for a large enough $t$.

However, this only means that $\forall t_0 > 0$ $\exists t> t_0$ such that $f(t) \leq \frac{2}{k_2\cdot t}$, when I would like to say this for all $t > t_0$ ...

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Let $g(t) = \int_{0}^t f(\tau) d\tau$, then $g' = f$ hence the inequality becomes $$ e^{k_2 \cdot g}\cdot g' \leq C_2 \iff \left( e^{k_2\cdot g }\right)' \leq C_2\cdot k_2 \iff g \leq \frac{\log(C_2\cdot k_2\cdot t + 1)}{k_2} $$ which is $$ \int_0^t f(\tau) d\tau \leq \frac{\log(C_2\cdot k_2\cdot t + 1)}{k_2}$$

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