Suppose I have a continuous function $f(x)$ that is non-increasing and always stays between $0$ and $1$, and it is known that

$$ \int_0^t f(x) dx = \log t + o(\log t), \qquad t \to \infty.$$

Unfortunately one has no control over the error term $o(\log t)$ (other than what is implied by the above asymptotic behaviour and the properties of $f$). My question is whether it is possible to conclude that

$$ f(t) = \frac{1}{t} + o(t^{-1}), \qquad t \to \infty.$$

Update: thanks to Raziel's response, the claim above does not hold in general and the problem is related to de Haan theory (which I am not very familiar with). I would therefore like to ask if it is still possible to find some $C > 0$ such that for $t$ sufficiently large,

$$ \frac{1}{Ct} \le f(t) \le \frac{C}{t}.$$

---------Old follow-up question below; please ignore---------

If this is possible, a follow-up question is whether the claim can be extended to $f$ that has countably many jumps (and continuous otherwise). Of course I will still be assuming that $f(x) \in [0,1]$ and the function is non-increasing, which in particular means that the jumps are negative and the size of the jump at $x_i$ (if any) is bounded by $f(x_i)$.

(One may start by proposing a solution to the toy problem with $f$ being differentiable if it simplifies the problem and offers any useful insights.)

Suppose I have a continuous function $f(x)$ that is non-increasing and always stays between $0$ and $1$, and it is known that

$$ \int_0^t f(x) dx = \log t + o(\log t), \qquad t \to \infty.$$

Unfortunately one has no control over the error term $o(\log t)$ (other than what is implied by the above asymptotic behaviour and the properties of $f$). My question is whether it is possible to conclude that

$$ f(t) = \frac{1}{t} + o(t^{-1}), \qquad t \to \infty.$$

Update: thanks to Raziel's response, the claim above does not hold in general and the problem is related to de Haan theory (which I am not very familiar with). I would therefore like to ask if it is still possible to find some $C > 0$ such that for $t$ sufficiently large,

$$ \frac{1}{Ct} \le f(t) \le \frac{C}{t}.$$

---------Old follow-up question below; please ignore---------

If this is possible, a follow-up question is whether the claim can be extended to $f$ that has countably many jumps (and continuous otherwise). Of course I will still be assuming that $f(x) \in [0,1]$ and the function is non-increasing, which in particular means that the jumps are negative and the size of the jump at $x_i$ (if any) is bounded by $f(x_i)$.

(One may start by proposing a solution to the toy problem with $f$ being differentiable if it simplifies the problem and offers any useful insights.)

Suppose I have a continuous function $f(x)$ that is non-increasing and always stays between $0$ and $1$, and it is known that

$$ \int_0^t f(x) dx = \log t + o(\log t), \qquad t \to \infty.$$

Unfortunately one has no control over the error term $o(\log t)$ (other than what is implied by the above asymptotic behaviour and the properties of $f$). My question is whether it is possible to conclude that

$$ f(t) = \frac{1}{t} + o(t^{-1}), \qquad t \to \infty.$$

If this is possible, a follow-up question is whether the claim can be extended to $f$ that has countably many jumps (and continuous otherwise). Of course I will still be assuming that $f(x) \in [0,1]$ and the function is non-increasing, which in particular means that the jumps are negative and the size of the jump at $x_i$ (if any) is bounded by $f(x_i)$.

(One may start by proposing a solution to the toy problem with $f$ being differentiable if it simplifies the problem and offers any useful insights (which I doubt would be the case), even though as one can see in the follow-up question I really don't want to assume differentiability.)

Suppose I have a continuous function $f(x)$ that is non-increasing and always stays between $0$ and $1$, and it is known that

$$ \int_0^t f(x) dx = \log t + o(\log t), \qquad t \to \infty.$$

Unfortunately one has no control over the error term $o(\log t)$ (other than what is implied by the above asymptotic behaviour and the properties of $f$). My question is whether it is possible to conclude that

$$ f(t) = \frac{1}{t} + o(t^{-1}), \qquad t \to \infty.$$

Update: thanks to Raziel's response, the claim above does not hold in general and the problem is related to de Haan theory (which I am not very familiar with). I would therefore like to ask if it is still possible to find some $C > 0$ such that for $t$ sufficiently large,

$$ \frac{1}{Ct} \le f(t) \le \frac{C}{t}.$$

---------Old follow-up question below; please ignore---------

If this is possible, a follow-up question is whether the claim can be extended to $f$ that has countably many jumps (and continuous otherwise). Of course I will still be assuming that $f(x) \in [0,1]$ and the function is non-increasing, which in particular means that the jumps are negative and the size of the jump at $x_i$ (if any) is bounded by $f(x_i)$.

(One may start by proposing a solution to the toy problem with $f$ being differentiable if it simplifies the problem and offers any useful insights.)