# consequences of a functional inequality

## 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 $$t > t_0$$ ...

## 1 Answer

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 $$\int_0^t f(\tau) d\tau \leq \frac{\log(C_2\cdot k_2\cdot t + 1)}{k_2}$$