I found myself stuck with an "elementary" claim in some article. A simplified version of the problem is:
Let $p : [0,1]\to [0,1]$ be a continuous and non decreasing function such that $p(0)=0$ and $$ \int_0^1 \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u) < + \infty. $$ Show that $$ \lim_{h\to 0} \frac{\displaystyle\int_0^h \left(\log \frac{1}{u}\right)^{\!1/2} \, dp(u)}{p(h) \left(\log \dfrac{1}{h}\right)^{1/2}} $$ exists and is finite.

I think that a good idea is to use integration by parts formula which gives $$ \int_0^h \left(\log \frac{1}{u}\right)^{\!1/2}\, dp(u) = p(h) \left(\log \frac{1}{h}\right)^{\!1/2} + \int_0^h \frac{p(u)}{2u\left(\log \frac{1}{u}\right)^{\!1/2}} du, $$ but I don't see how to conclude...

  • $\begingroup$ What if $p \equiv 0$? You need some additional assumption. $\endgroup$ – Dieter Kadelka Feb 10 at 20:29
  • $\begingroup$ Where did you see such a claim (which is false in general even if $p>0$ on $(0,1]$)? $\endgroup$ – Iosif Pinelis Feb 10 at 20:44
  • $\begingroup$ @IosifPinelis : the reference is doi.org/10.1016/j.spa.2013.04.019, in the proof of Theorem 4.3. I looked at the rest of the paper and I am not sure that they stated additional assumptions. $\endgroup$ – Chev Feb 11 at 11:08

I think you reproduced the "elementary" claim correctly (for the case $n=1$). However, as stated in my earlier comment, the claim is false in general, even if you assume that $p>0$ on $(0,1]$.

Indeed, the claim was that the limit \begin{equation*} \lim_{h\downarrow0}r(h) \tag{1} \end{equation*} exists and is finite, where \begin{equation*} r(h):=\frac{\text{num}(h)}{\text{den}(h)}, \end{equation*} \begin{equation*} \text{num}(h):=\int_0^h \ln^{1/2}\frac1u\;dp(u),\quad \text{den}(h):=p(h)\ln^{1/2}\frac1h. \end{equation*} Note that \begin{equation*} \text{num}(h)\ge\int_0^h \ln^{1/2}\frac1h\;dp(u)=\text{den}(h), \end{equation*} so that $r(h)$ is always $\ge1$.

However, let us show that the limit in (1) can be $\infty$:

Indeed, let \begin{equation*} p(u):=\frac1{\sqrt{\ln\frac1u}\,\ln^2\ln\frac1u} \end{equation*} for $u\in(0,1/3)$, with $p(0):=0$ and $p(u):=\frac1{\sqrt{\ln3}\,\ln^2\ln3}$ for $u\in[1/3,1]$. Then $p$ is a real-valued continuous nondecreasing function on $[0, 1]$. Also, \begin{equation*} p'(u)\sim\frac1{2u\ln^{3/2}\frac1u\,\ln^2\ln\frac1u} \end{equation*} (as $u\downarrow0$) and hence for $h\downarrow0$ \begin{equation*} \text{num}(h) \sim\int_0^h \frac{du}{2u\ln\frac1u\,\ln^2\ln\frac1u} =\frac1{2\ln\ln\frac1h}, \end{equation*} whereas \begin{equation*} \text{den}(h)=\frac1{\ln^2\ln\frac1h}, \end{equation*} so that \begin{equation*} r(h)\sim\frac12\,\ln\ln\frac1h\to\infty. \end{equation*}

On a somewhat positive note, suppose that $p$ is continuous on $[0,1]$ with $p(0)=0$, has a positive derivative $p'$ in a right neighborhood of $0$, and \begin{equation} p(h)=(c+o(1)) hp'(h)\ln\frac1h \tag{2} \end{equation} as $h\downarrow0$ for some $c\in[0,2)$. This condition will hold e.g. if (i) $p(h)=Ch^a\ln^b\frac1h$ for some positive real $C$ and $a$, some real $b$, and all small enough $h>0$ or if (ii) $p(h)=C/\ln^t\frac1h$ for some positive real $C$, some real $t>1/2$, and all small enough $h>0$.

Given the additional condition (2), you will have $r(0+)=\frac1{1-c/2}\in(0,\infty)$, by l'Hospital's rule. Indeed, then for the "derivative ratio" we have \begin{equation*} \frac{\text{num}'(h)}{\text{den}'(h)} =\frac{p'(h)\ln^{1/2}\frac1h}{p'(h)\ln^{1/2}\frac1h-\frac{p(h)}{2h}\ln^{-1/2}\frac1h} =\frac1{1-\dfrac{p(h)}{2hp'(h)\ln\frac1h}}\to\frac1{1-c/2} \end{equation*} as $h\downarrow0$.

  • $\begingroup$ I completely agree with your counter-example (except that the multiplicative factor in the equivalent of $p'(u)$ should be 1/2 instead of 2). Thank you very much !!! If I understand correctly, the problem comes from the fact that the two terms in $den'(h)$ cancels in terms of equivalent. I tried to find such counter-example without the composition $\ln \ln$ and failed to find it. Do you think that the initial claim could be proved with additional assumptions? $\endgroup$ – Chev Feb 12 at 9:54
  • $\begingroup$ @Chev : I have replaced the factor $2$ by $1/2$ and added a "positive note" with an additional condition to make the initial claim hold. $\endgroup$ – Iosif Pinelis Feb 12 at 14:52
  • $\begingroup$ I did the same computations on my side and I agree. I was hoping for some "simplest" condition but in fact it is quite satisfying like this. For instance, it is easy to check for $p(h) = h^p$ for any $p>0$. Given that the typical example we have in mind for application to Brownian motion is $p(h) = \sqrt{h}$, I would say that your answer is perfect for what I need. Thanks !! $\endgroup$ – Chev Feb 12 at 16:00
  • $\begingroup$ @Chev : All right. So, are you satisfied with the answer overall? $\endgroup$ – Iosif Pinelis Feb 12 at 18:09
  • $\begingroup$ Yeah. I am new to these posts and forgot to "accept" your answer. $\endgroup$ – Chev Feb 13 at 8:56

Note that $\displaystyle\int_0^h \left(\log \frac{1}{u}\right)^{\!1/2} \, \mathrm{d}p(u)\rightarrow 0$ as $h\rightarrow 0$, and that $p(h) \left(\log \dfrac{1}{h}\right)^{1/2}\rightarrow 0$ as $h\rightarrow 0$. Moreover, $ \mathrm{d} \left(p(h) \left(\log \dfrac{1}{h}\right)^{1/2} \right)\bigg/\mathrm{d}h\neq 0$ for $h$ sufficiently close to $0$, therefore L'Hôpital's theorem ensures that the limit exists. Since $\mathrm{d}p(u)/\mathrm{d}u$ is positive you can verify that the limit is non-zero.

  • $\begingroup$ Could you give a reference to L'Hôpital's theorem you are using? The only one I found assumes that the functions are differentiable which is not the case here. $\endgroup$ – Chev Feb 11 at 11:03
  • $\begingroup$ @Chev They are differentiable in $(0,1)$. Actually the limit should be taken as $h\rightarrow0^+$. Why do you say that they are not ? $\endgroup$ – Daniel Castro Feb 11 at 12:46
  • $\begingroup$ $p$ is continuous but may not be differentiable $\endgroup$ – Chev Feb 11 at 14:08
  • $\begingroup$ @DanielCastro : I do not see any reasons for your claims "that $p(h) \left(\log \dfrac{1}{h}\right)^{1/2}\rightarrow 0$ as $h\rightarrow 0$. Moreover, $ \mathrm{d} \left(p(h) \left(\log \dfrac{1}{h}\right)^{1/2} \right)\bigg/\mathrm{d}h\neq 0$ for $h$ sufficiently close to $0$", or for the applicability of l'Hospital's rule. $\endgroup$ – Iosif Pinelis Feb 11 at 15:44
  • $\begingroup$ @IosifPinelis (Assuming differentiability) Taylor expand $p(h)$, then since $p(0)=0$ we have $(\mathrm{d} p(0)/\mathrm{d} h)h\left(\log \dfrac{1}{h}\right)^{1/2}$ and this function vanishes for $h\rightarrow 0^+$. For a proper applicability of the rule we would need to extend the function to negative $h$, that is, to define the limit for $h\rightarrow 0^-$, I guess that's the only missing piece. $\endgroup$ – Daniel Castro Feb 11 at 16:14

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