# How to find the critical index $a$ of $x^af(x)$?

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. )

(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 $$\lambda(\gamma) = 0$$.

• 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 $\lambda(\gamma_0)>0$
– Houa
Jul 26, 2020 at 8:37
• 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 $f$
– Houa
Jul 26, 2020 at 8:39
• I have added some examples. Jul 26, 2020 at 8:50
• 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?
– 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

$$h_a(x)=x^af'(x)\tag{3}$$

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

$$g'_a(x)=ag_{a-1}(x)+h_a(x).\tag{4}$$

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 $$[\alpha,\alpha+1]$$.

• But I think you maybe misunderstood something: you said "since $f'(0+)$ is bounded", but you can see the example $f(x)=x^{-2}\ln(1+\frac{1}{x})$, $|f'(0+)|=+\infty$
– Houa
Jul 26, 2020 at 12:04
• 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 $f\in \mathcal C^1[0,\infty)$? Jul 26, 2020 at 12:35
• 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 Jul 26, 2020 at 12:45
• 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
– Houa
Jul 26, 2020 at 13:31
• @Houa Thanks, it was +20 years ago I took lessons in mathematics :) I will try again. Jul 26, 2020 at 15:39