# Asymptotic of an improper integral

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...

• What if $p \equiv 0$? You need some additional assumption. – Dieter Kadelka Feb 10 at 20:29
• Where did you see such a claim (which is false in general even if $p>0$ on $(0,1]$)? – Iosif Pinelis Feb 10 at 20:44
• @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. – 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 $$$$p(h)=(c+o(1)) hp'(h)\ln\frac1h \tag{2}$$$$ as $$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 $$h\downarrow0$$.

• 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? – Chev Feb 12 at 9:54
• @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. – Iosif Pinelis Feb 12 at 14:52
• 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 !! – Chev Feb 12 at 16:00
• @Chev : All right. So, are you satisfied with the answer overall? – Iosif Pinelis Feb 12 at 18:09
• Yeah. I am new to these posts and forgot to "accept" your answer. – 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.

• 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. – Chev Feb 11 at 11:03
• @Chev They are differentiable in $(0,1)$. Actually the limit should be taken as $h\rightarrow0^+$. Why do you say that they are not ? – Daniel Castro Feb 11 at 12:46
• $p$ is continuous but may not be differentiable – Chev Feb 11 at 14:08
• @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. – Iosif Pinelis Feb 11 at 15:44
• @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. – Daniel Castro Feb 11 at 16:14