Let $f\in C^1(0,+\infty)$, $f(x)> 0$ in $(0,+\infty)$, $f(0+)=+\infty$, and $f$ is decrased in $(0,+\infty)$. Consider $g(x)=x^af(x)$, $(a>1)$. Now one can show that $\liminf_\limits{x\rightarrow0^+}g(x)=\limsup_\limits{x\rightarrow0^+}g(x)$, so we write $\lim_\limits{x\rightarrow0^+}g(x)>0$ to describe both $\lim_\limits{x\rightarrow0^+}g(x)=c>0$ and $\lim_\limits{x\rightarrow0^+}g(x)=+\infty$. Attention, for we only care about what happen near $x=0$,so we can weaken $(0,+\infty)$ to be $(0,\delta)$ and can smoothen $f$ with a smooth function when $x>\delta$.

Problem: Given the $f$ as above, define $g_a(x)=x^af(x)$,$ (a>1)$. If

(1) there exists $\alpha>1$ such that $\lim_\limits{x\rightarrow0^+}g_\alpha(x)>0$,

(2) there exists $\beta>\alpha$ such that $\lim_\limits{x\rightarrow0^+}g_\alpha(x)=0$.

Please show that there exists $\gamma\in[\alpha,\beta)$ such that

(i) $\lim_\limits{x\rightarrow0^+}g_\gamma(x)>0$,

(ii) for any $\epsilon>0$, $\lim_\limits{x\rightarrow0^+}g_{\gamma+\epsilon}(x)=0$.

That is, can one find the "critical index"?

Attempt, example and some information: There are many examples for the proposition:

(a) $f(x)=\frac{1}{x^2}\ln(1+\frac{1}{x})$, then $\lim_\limits{x\rightarrow0^+}g_2(x)>0$ and $\lim_\limits{x\rightarrow0^+}g_3(x)=0$, we can find the "critical index" to be $a=2\in[2,3)$

(b) $f(x)=x^{-2+e^x}\ln(1+\frac{1}{x})$ satisfies our conditions in $(0,\text{small}~\delta)$. One can easily find out the "critical index" is $a=1$.

I think the difficulty of this problem is that "if $a$ make $\lim_\limits{x\rightarrow0^+}g_a(x)=0$, then can one show there exist a small $\epsilon>0$ such that $\lim_\limits{x\rightarrow0^+}g_{a-\epsilon}(x)=0$?". I can't show this can you help me? Or this proposition is wrong and you can find the counterexamples? But I want this proposition to be true, does it need to add any extra conditions for $f(x)$?

If you need you can add the condition $|f'(x)|\leq C \frac{f(x)}{x}$. In fact these conditions for $f$ is given by other propositions and theorem. I just summarized it.

2020/7/26 Addition (you can choose to answer or just have a look): One gives the counterexample in the answer, so if I still want this proposition to be correct I must add some conditions to $f$. In fact, this is relative to this problem: estimate a integral with parameter The problem in the link is one example of a kind of integrals with parameter, if one puts $u=1/r$, then it become the problem here. So the $f$ here should be like $\frac{1}{x^m}(\text{log term, arctan term})$ (the example (a)(b) above), and $\frac{-1}{x^2\log x}$ will not appear. So what can I give the conditions to $f$ such that the proposition can be ture? (I discussed with my friends and we turns out that, it's difficult to describe straight $f$ have such type $\frac{1}{x^m}(\text{log term, arctan term})$, so we want a more general lemma to due with it. )


2 Answers 2


(I don't know if this answers your question, but it can be an hint.)

You can define the function $\lambda\colon [1,+\infty)\to [0,+\infty]$ by $$ \lambda(a) := \limsup_{x\to 0+} x^a f(x). $$ Since $g_a > g_b$ if $a < b$, we clearly have that $\lambda$ is monotone non-increasing. By assumption, $\lambda(\alpha) > 0$ and $\lambda(\beta) = 0$ for some $1 < \alpha < \beta$. Let $$ \gamma := \inf\{a\geq 1:\ \lambda(a) = 0\}. $$ This "critical" value $\gamma$ has the following properties:

(1) $\lambda(a) = +\infty$ for every $a < \gamma$;

(2) $\lambda(a) = 0$ for every $a > \gamma$.

On the other hand, I think that $\lambda(\gamma)$ can be every element of $[0,+\infty]$ (depending on the choice of $f$).

Some examples:

(1) If you take $f(x) = - \log(x) / x^2$, you have that $\lambda(a) = 0$ if $a > 2$ and $\lambda(a) = +\infty$ if $a \leq 2$. In this case, $\gamma = 2$ and $\lambda(\gamma) = +\infty$.

(2) If you take $f(x) = - \frac{1}{x^2 \log x}$, then again $\gamma = 2$, but in this case $\lambda(\gamma) = 0$.

  • $\begingroup$ in fact the difficulty is to show $\lambda(\gamma)>0$. I tried a similar way like you. If $\lambda((\alpha,\beta))=0$, we can claim $\gamma=\alpha$, or we can find $\gamma_1$ s.t. $\lambda([\alpha,\gamma_1])>0$. If $\lambda((\gamma_1,\beta))=0$, we can claim $\gamma=\gamma_1$..... so we get $\{\gamma_n\}$ and its sup is $\gamma_0$. But I can't show that $\lambda(\gamma_0)>0$ $\endgroup$
    – Houa
    Jul 26, 2020 at 8:37
  • $\begingroup$ set $\gamma$ is critical value. If you can find a $f$ s.t. $\lambda(\gamma)=0$, then the proposition is wrong. And I have to add some condition to $f$ $\endgroup$
    – Houa
    Jul 26, 2020 at 8:39
  • $\begingroup$ I have added some examples. $\endgroup$
    – Rigel
    Jul 26, 2020 at 8:50
  • $\begingroup$ your counterexample is good. Even it can satisfy $|f'(x)|\leq C \frac{f(x)}{x}$ near $x=0$! It looks like I have to give more conditions to $f$. In fact this is relative to a problem hanging in the body above, would you like to have a look? $\endgroup$
    – Houa
    Jul 26, 2020 at 12:19

We have 2 conditions, where the first condition is the inequality

$$ |f'(x)|\leq C\frac{f(x)}{x}\tag{1} $$

and the second condition is the limit saying that there exists a $\beta$ such that

$$ g_\beta(0^+)=0.\tag{2} $$

Define the function


then differentiate $g_a(x)$ once and use the definition (3) to get the equation


Using inequality (1) with definition (3) we have

$$ |h_a(x)|\leq Cg_{a-1}(x) $$

which for $a=\beta+1$ yields $h_{\beta+1}(0^+)=0$ which in turn we insert into equation (4) to obtain

$$ g'_{\beta+1}(0^+)=0. $$

The idea is to decrease $a$ by one by one. We start with $a_0=\beta+1$ then we iterate with $a_n=\beta+1-n$ until we find first occurance of $g_{a_n}(0^+)>0$ which yields $\alpha=a_n<\beta$. It follows that $\gamma\in[\alpha,\alpha+1)$. To find $\gamma$ use some iterative method, such as interval halving starting with the interval $[\alpha,\alpha+1]$.

  • $\begingroup$ But I think you maybe misunderstood something: you said "since $f'(0+)$ is bounded", but you can see the example $f(x)=x^{-2}\ln(1+\frac{1}{x})$, $|f'(0+)|=+\infty$ $\endgroup$
    – Houa
    Jul 26, 2020 at 12:04
  • $\begingroup$ Maybe there are many different definitions of $\mathcal C^1$? My thought is if an derivative explodes, then it diverges and therefore does not exists. Maybe it should be $f\in \mathcal C^1[0,\infty)$? $\endgroup$ Jul 26, 2020 at 12:35
  • $\begingroup$ Ah, I found it: "The notion of $\mathcal C^k$ function may be restricted to those whose first k derivatives are bounded functions. " Source: mathworld.wolfram.com/C-kFunction.html $\endgroup$ Jul 26, 2020 at 12:45
  • $\begingroup$ en.wikipedia.org/wiki/Differentiable_function Originally, $C^1(0,+\infty)$ means that $f$ and $f'$ exist and they are continuous in $(0,+\infty)$. And in your link "The notion of Ck function may be restricted to those whose first k derivatives are bounded functions. " This always is used in when we talk about normed space. But generally speaking it just k times differentiable and continuous, not bounded $\endgroup$
    – Houa
    Jul 26, 2020 at 13:31
  • $\begingroup$ @Houa Thanks, it was +20 years ago I took lessons in mathematics :) I will try again. $\endgroup$ Jul 26, 2020 at 15:39

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