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I got this question in a prelims exam. $f$ is a positive, continuously differentiable function on $(0,\infty)$ such that $f’ >0$. If $f(x) \leq Cx^2$ for some $C>0$ then $\int \frac{1}{f’(x)}$ diverges.

There is a hint in the problem which asks to establish that if $\int \frac{1}{f’(x)}< \infty$ then $\int\limits_{a}^{\infty} \frac{1}{f’(x)} \to 0$ as $a\to \infty$. I can establish the claim made in the hint by writing $\int f = \sum_{n} \int_{n}^{n+1} f$ and then observing that the above sum converges absolutely and therefore the tail of the series (which is the required integral) goes to 0. I believe that I am now supposed to show that in our case $\int_{a}^{\infty} \frac{1}{f’(x)}$ does not go to zero as $a\to \infty$. But, I am unable to make use of the fact that $f(x)\leq Cx^2$. I do not know if it is useful to note that $\frac{1}{f’(x)}=(f^{-1})’(f(x))$.

Any help would be highly appreciated. I would like some hint on how to use this condition.

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  • $\begingroup$ By a change of variables $x=f^{-1}(y)$, and setting $g(y) = \sqrt C f^{-1}(y)$, your last remark can be converted into the following equivalent form: if $g$ is a positive increasing differentiable function with $g(y) \ge \sqrt y$ then $\int^\infty (g'(y))^2\,dy$ diverges. (But I don't know how to solve the new form either.) $\endgroup$ – Greg Martin Jul 17 '19 at 7:22
  • $\begingroup$ Yeah! I was thinking of this. But, we need to have some estimate which says that $g’$ is reasonably large. I am not sure how to get that. $\endgroup$ – WhoKnowsWho Jul 17 '19 at 7:41
  • $\begingroup$ By applying MVT in intervals $(a,2a)$, it can be shown that $\dfrac1{f'(x)}>\dfrac1{2Cx}$ for infinitely many $x\in(0,\infty)$. $\endgroup$ – ajotatxe Jul 17 '19 at 7:44
  • $\begingroup$ @ajotatxe Yep! This seems reasonable. We can use the continuity to get that it is reasonably big on a small neighbourhood of these infinitely many points. I think doing some epsilon-delta would yield it. I will try after a few hours (right now I am travelling). $\endgroup$ – WhoKnowsWho Jul 17 '19 at 7:49
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Note that by Cauchy-Schwarz inequality it holds $$ \int_A^{2A} \frac {dx} {f'(x)}\cdot \int_A^{2A} f'(x)dx \ge \left(\int_A^{2A} 1 \ dx\right)^2 = A^2. $$ We can also see that $$ 4CA^2 \ge f(2A) > f(2A) - f(A) = \int_A^{2A} f'(x) dx $$ by the fundamental theorem of calculus. This gives $$ \int_A^{2A} \frac {dx}{f'(x)} \ge \frac 1{4C},\quad \forall A>0, $$ which implies \begin{align*} \int_0^\infty \frac {dx}{f'(x)} =& \sum_{j=-\infty}^\infty \int_{2^j}^{2^{j+1}} \frac {dx}{f'(x)} \\ \ge & \sum_{j=-\infty}^\infty \frac 1{4C} = \infty. \end{align*}

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