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.)

  • $\begingroup$ Am I missing something? Is this not l'Hospital's rule? $\endgroup$ Jan 14 '19 at 8:35
  • $\begingroup$ @Venkataramana I could be wrong, but aren't you suggesting the converse of L'Hospital's rule (if the ratio of two functions has a limit, then the ratio of the derivatives has the same limit), which does not seem to be true in general? I have no control over $o(\log t)$ and just can't say much about its derivative, and I am not sure how L'Hospital may be applied. $\endgroup$ Jan 14 '19 at 8:46
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    $\begingroup$ This belongs to Karamata's Tauberian theory, it is a kind of monotone density theorem for de Hahn classes, I believe. I'll look into Bingham-Goldie-Teugels book later today and write more. $\endgroup$ Jan 14 '19 at 8:54

The answer is no, even in the smooth case. Take for example:

$$ f(x) = \frac{2}{x} + \frac{\cos(\log(x))}{x} $$

Alter it on a small neighborhood of $0$ in such a way that there is no singularity there, preserving smoothness (this will be irrelevant for the asymptotics). This function is decreasing and, for $t$ sufficiently large, we have

$$ \int_0^t f(x) dx = C + 2\log(t) + \sin(\log(t)) = 2\log(t) + o(\log(t)) $$

The monotone density theorem mentioned in the comments does not work in general if your r.h.s. is simply a slowly varying function (as any function asymptotic to $\log(t)$). You want your r.h.s. to be a de Haan function. The specific result you may want to use is Theorem 3.6.8 here:

Bingham, N. H.; Goldie, C. M.; Teugels, J. L., Regular variation., Encyclopedia of Mathematics and its Applications, 27. Cambridge etc.: Cambridge University Press. 512 p. £ 20.00/pbk (1989). ZBL0667.26003.

  • $\begingroup$ What about bounds? Is it possible to show that there exists some $C > 0$ such that $\frac{1}{Ct} \le f(t) \le \frac{C}{t}$ for $t$ sufficiently large? $\endgroup$ Jan 14 '19 at 9:40
  • $\begingroup$ Original asymptotic equation is equivalent to $f$ being in the right de Haan class. Two-sided estimate would correspond to de Haan's analogue of $O$-regular variation, I think, and it does not follow automatically. The counter-example is less explicit, though, I will type it later today on a computer, if you like. $\endgroup$ Jan 14 '19 at 10:48
  • $\begingroup$ @MateuszKwaśnicki Thanks, I am looking forward to your answer. $\endgroup$ Jan 14 '19 at 12:19
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    $\begingroup$ @random_person: I just started to type when Iosif Pinelis gave esentially the same construction. I can oly add that what I meant in my first comment was Theorem 3.6.8 in the BGT book, which gives a necessary and sufficient condition for the desired asymptotics of $f$ in terms of its primitive function. See also Sections 3.7.1–3.7.2 therein. $\endgroup$ Jan 14 '19 at 12:45

The post by Raziel shows that the answer to the original question is no. The OP then asked, in a comment to that post, if one one still conclude that $f(t)\asymp\frac1t$ (as $\to\infty$); as usual, $a\asymp b$ means here that $\limsup|\frac ab+\frac ba|<\infty$.

Let us show that the answer is still no. E.g., for $j=0,1,\dots$ let $t_j:=e^{j^2}$, \begin{equation} c_j:=\frac{\ln t_{j+1}-\ln t_j}{t_{j+1}-t_j}\sim\frac{2j}{t_{j+1}} \tag{1} \end{equation} (as $j\to\infty$), and \begin{equation} f(x):=c_j\quad\text{for}\quad x\in[t_j,t_{j+1}), \end{equation} with $f:=c_0=\frac1{e-1}$ on $[0,t_0)$. Let also $F(t):=\int_0^t f(x)\,dx$.

Then $f$ is nonincreasing, $0<f\le1$, $F(t_j)=c_0+\ln t_j\sim c_0+\ln t_{j+1}=F(t_{j+1})$, whence $F(t)\sim\ln t$ (as $t\to\infty$), whereas $f(t_{j+1}-)=c_j$ is much greater than $\frac1{t_{j+1}}$, by (1). We also see that $f(t_j)=c_j$ is much less than $\frac1{t_j}$, again by (1).

The only condition missed here is the continuity of $f$, as $f$ is not left-continuous at $t_{j+1}$ for $j=0,1,\dots$. This omission is quite easy, but tedious, to fix by approximation. For instance, one can replace the above $f$ on every interval $[t_{j+1}-c_02^{-j},t_{j+1}]$ by the linear interpolation of $f$ on the same interval. Then instead of the value $c_0+\ln t_{j+1}$ of $F(t_{j+1})$ we will have $b_j+\ln t_{j+1}\sim c_0+\ln t_j=F(t_j)$ for some $b_j\in[0,c_0]$, and instead of $f(t_{j+1}-)=c_j$ being much greater than $\frac1{t_{j+1}}$, we will have that $f(t_{j+1}-c_0)=c_j$ is much greater than $\frac1{t_{j+1}-c_0}$.

  • $\begingroup$ Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) \le \frac{C}{t}$ though? $\endgroup$ Jan 14 '19 at 12:38
  • $\begingroup$ @random_person : This very example shows that the upper bound $\frac Ct$ on $f(t)$ is impossible in general, as we have $f(t_{j+1}-)/\frac1{t_{j+1}}\to\infty$. $\endgroup$ Jan 14 '19 at 12:41
  • $\begingroup$ Oh I have asked a dumb question. I actually want to ask if a lower bound $f(t) \ge \frac{1}{Ct}$ is possible. $\endgroup$ Jan 14 '19 at 12:42
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    $\begingroup$ @random_person : I have now added a sentence showing that, in the same example, the lower bound $\frac1{Ct}$ on $f(t)$ is impossible either. $\endgroup$ Jan 14 '19 at 12:48
  • $\begingroup$ I am feeling so embarrassed that I have missed this observation...thank you so much for your patience and again your counter-example. $\endgroup$ Jan 14 '19 at 12:52

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