Differentiating an integral that grows like log asymptotically












8












$begingroup$


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.)










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$endgroup$












  • $begingroup$
    Am I missing something? Is this not l'Hospital's rule?
    $endgroup$
    – Venkataramana
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 8:46






  • 1




    $begingroup$
    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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 8:54
















8












$begingroup$


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.)










share|cite|improve this question











$endgroup$












  • $begingroup$
    Am I missing something? Is this not l'Hospital's rule?
    $endgroup$
    – Venkataramana
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 8:46






  • 1




    $begingroup$
    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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 8:54














8












8








8


2



$begingroup$


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.)










share|cite|improve this question











$endgroup$




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.)







pr.probability real-analysis ca.classical-analysis-and-odes probability-distributions asymptotics






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 12:40







random_person

















asked Jan 14 at 8:09









random_personrandom_person

955




955












  • $begingroup$
    Am I missing something? Is this not l'Hospital's rule?
    $endgroup$
    – Venkataramana
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 8:46






  • 1




    $begingroup$
    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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 8:54


















  • $begingroup$
    Am I missing something? Is this not l'Hospital's rule?
    $endgroup$
    – Venkataramana
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 8:46






  • 1




    $begingroup$
    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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 8:54
















$begingroup$
Am I missing something? Is this not l'Hospital's rule?
$endgroup$
– Venkataramana
Jan 14 at 8:35




$begingroup$
Am I missing something? Is this not l'Hospital's rule?
$endgroup$
– Venkataramana
Jan 14 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.
$endgroup$
– random_person
Jan 14 at 8:46




$begingroup$
@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.
$endgroup$
– random_person
Jan 14 at 8:46




1




1




$begingroup$
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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 8:54




$begingroup$
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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 8:54










2 Answers
2






active

oldest

votes


















11












$begingroup$

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



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



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – random_person
    Jan 14 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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 10:48










  • $begingroup$
    @MateuszKwaśnicki Thanks, I am looking forward to your answer.
    $endgroup$
    – random_person
    Jan 14 at 12:19






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 12:45



















6












$begingroup$

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 $f(t)asympfrac1t$ (as $toinfty$); as usual, $aasymp b$ 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}simfrac{2j}{t_{j+1}} tag{1}
end{equation}

(as $jtoinfty$), and
begin{equation}
f(x):=c_jquadtext{for}quad xin[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, $0<fle1$, $F(t_j)=c_0+ln t_jsim c_0+ln t_{j+1}=F(t_{j+1})$, whence $F(t)simln t$ (as $ttoinfty$), 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 $b_jin[0,c_0]$, 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 $frac1{t_{j+1}-c_0}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
    $endgroup$
    – random_person
    Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:42








  • 1




    $begingroup$
    @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.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:52











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2 Answers
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active

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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









11












$begingroup$

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



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



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – random_person
    Jan 14 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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 10:48










  • $begingroup$
    @MateuszKwaśnicki Thanks, I am looking forward to your answer.
    $endgroup$
    – random_person
    Jan 14 at 12:19






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 12:45
















11












$begingroup$

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



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



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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?
    $endgroup$
    – random_person
    Jan 14 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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 10:48










  • $begingroup$
    @MateuszKwaśnicki Thanks, I am looking forward to your answer.
    $endgroup$
    – random_person
    Jan 14 at 12:19






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 12:45














11












11








11





$begingroup$

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



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



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.






share|cite|improve this answer











$endgroup$



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



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



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 9:33

























answered Jan 14 at 9:00









RazielRaziel

2,02911425




2,02911425












  • $begingroup$
    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?
    $endgroup$
    – random_person
    Jan 14 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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 10:48










  • $begingroup$
    @MateuszKwaśnicki Thanks, I am looking forward to your answer.
    $endgroup$
    – random_person
    Jan 14 at 12:19






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 12:45


















  • $begingroup$
    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?
    $endgroup$
    – random_person
    Jan 14 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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 10:48










  • $begingroup$
    @MateuszKwaśnicki Thanks, I am looking forward to your answer.
    $endgroup$
    – random_person
    Jan 14 at 12:19






  • 1




    $begingroup$
    @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.
    $endgroup$
    – Mateusz Kwaśnicki
    Jan 14 at 12:45
















$begingroup$
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?
$endgroup$
– random_person
Jan 14 at 9:40




$begingroup$
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?
$endgroup$
– random_person
Jan 14 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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 10:48




$begingroup$
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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 10:48












$begingroup$
@MateuszKwaśnicki Thanks, I am looking forward to your answer.
$endgroup$
– random_person
Jan 14 at 12:19




$begingroup$
@MateuszKwaśnicki Thanks, I am looking forward to your answer.
$endgroup$
– random_person
Jan 14 at 12:19




1




1




$begingroup$
@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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 12:45




$begingroup$
@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.
$endgroup$
– Mateusz Kwaśnicki
Jan 14 at 12:45











6












$begingroup$

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 $f(t)asympfrac1t$ (as $toinfty$); as usual, $aasymp b$ 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}simfrac{2j}{t_{j+1}} tag{1}
end{equation}

(as $jtoinfty$), and
begin{equation}
f(x):=c_jquadtext{for}quad xin[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, $0<fle1$, $F(t_j)=c_0+ln t_jsim c_0+ln t_{j+1}=F(t_{j+1})$, whence $F(t)simln t$ (as $ttoinfty$), 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 $b_jin[0,c_0]$, 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 $frac1{t_{j+1}-c_0}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
    $endgroup$
    – random_person
    Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:42








  • 1




    $begingroup$
    @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.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:52
















6












$begingroup$

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 $f(t)asympfrac1t$ (as $toinfty$); as usual, $aasymp b$ 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}simfrac{2j}{t_{j+1}} tag{1}
end{equation}

(as $jtoinfty$), and
begin{equation}
f(x):=c_jquadtext{for}quad xin[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, $0<fle1$, $F(t_j)=c_0+ln t_jsim c_0+ln t_{j+1}=F(t_{j+1})$, whence $F(t)simln t$ (as $ttoinfty$), 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 $b_jin[0,c_0]$, 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 $frac1{t_{j+1}-c_0}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
    $endgroup$
    – random_person
    Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:42








  • 1




    $begingroup$
    @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.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:52














6












6








6





$begingroup$

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 $f(t)asympfrac1t$ (as $toinfty$); as usual, $aasymp b$ 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}simfrac{2j}{t_{j+1}} tag{1}
end{equation}

(as $jtoinfty$), and
begin{equation}
f(x):=c_jquadtext{for}quad xin[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, $0<fle1$, $F(t_j)=c_0+ln t_jsim c_0+ln t_{j+1}=F(t_{j+1})$, whence $F(t)simln t$ (as $ttoinfty$), 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 $b_jin[0,c_0]$, 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 $frac1{t_{j+1}-c_0}$.






share|cite|improve this answer











$endgroup$



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 $f(t)asympfrac1t$ (as $toinfty$); as usual, $aasymp b$ 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}simfrac{2j}{t_{j+1}} tag{1}
end{equation}

(as $jtoinfty$), and
begin{equation}
f(x):=c_jquadtext{for}quad xin[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, $0<fle1$, $F(t_j)=c_0+ln t_jsim c_0+ln t_{j+1}=F(t_{j+1})$, whence $F(t)simln t$ (as $ttoinfty$), 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 $b_jin[0,c_0]$, 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 $frac1{t_{j+1}-c_0}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 13:58

























answered Jan 14 at 12:26









Iosif PinelisIosif Pinelis

18.7k22159




18.7k22159












  • $begingroup$
    Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
    $endgroup$
    – random_person
    Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:42








  • 1




    $begingroup$
    @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.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:52


















  • $begingroup$
    Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
    $endgroup$
    – random_person
    Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:42








  • 1




    $begingroup$
    @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.
    $endgroup$
    – Iosif Pinelis
    Jan 14 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.
    $endgroup$
    – random_person
    Jan 14 at 12:52
















$begingroup$
Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
$endgroup$
– random_person
Jan 14 at 12:38




$begingroup$
Thanks for the nice counter-example. Would it still be possible to establish the upper bound $f(t) le frac{C}{t}$ though?
$endgroup$
– random_person
Jan 14 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 $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
$endgroup$
– Iosif Pinelis
Jan 14 at 12:41






$begingroup$
@random_person : This very example shows that the upper bound $frac Ct$ on $f(t)$ is impossible in general, as we have $f(t_{j+1}-)/frac1{t_{j+1}}toinfty$.
$endgroup$
– Iosif Pinelis
Jan 14 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.
$endgroup$
– random_person
Jan 14 at 12:42






$begingroup$
Oh I have asked a dumb question. I actually want to ask if a lower bound $f(t) ge frac{1}{Ct}$ is possible.
$endgroup$
– random_person
Jan 14 at 12:42






1




1




$begingroup$
@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.
$endgroup$
– Iosif Pinelis
Jan 14 at 12:48






$begingroup$
@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.
$endgroup$
– Iosif Pinelis
Jan 14 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.
$endgroup$
– random_person
Jan 14 at 12:52




$begingroup$
I am feeling so embarrassed that I have missed this observation...thank you so much for your patience and again your counter-example.
$endgroup$
– random_person
Jan 14 at 12:52


















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