If $\Phi\geq 0$ Is Non-decreasing, Does $\int_1^\infty \frac{\Phi(x)}{x ...

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Learn more about Teams If $\Phi\geq 0$ is non-decreasing, does $\int_1^\infty \frac{\Phi(x)}{x^2}\,dx=\infty$ imply $\int_0^\infty e^{-\Phi(x)}\,dx<\infty$? Ask Question Asked 3 years, 4 months ago Modified 3 years, 4 months ago Viewed 634 times 24 $\begingroup$

Q. Suppose $\Phi : [0, \infty) \to [0, \infty)$ is a non-decreasing function. If $$ \int_{1}^{\infty} \frac{\Phi(x)}{x^2} \, \mathrm{d}x = \infty, $$ then does the inequality $$ \int_{0}^{\infty} e^{-\Phi(x)} \, \mathrm{d}x < \infty $$ hold?

This question is motivated by this posting, but it became a statement of independent interest to me.

I suspect that the statement is true, based on a naive observation that a higher growth rate of $\Phi$ is in favor of both conditions and a slower growth rate is against them.

One obvious observation is that $\Phi(x)$ must tend to $\infty$ as $x \to \infty$. But, to be honest, I have no clue as to where I should start. So I would like to invite you to this strange yet interesting problem!

Edit. The user @Sungjin Kim showed that the answer is false by providing a family of counter-examples. Here is a simplification of his answer:

For each strictly increasing sequence $(b_k)_{k=1}^{\infty}$ in $[1, \infty)$, let

$$ \Phi(t) = \sum_{k \mathop{:} b_k \leq t} b_k = \sum_{k=1}^{\infty} b_k \mathbf{1}_{[b_k, \infty)}(t). $$

Then it follows that

$$ \int_{1}^{\infty} \frac{\Phi(t)}{t^2} \, \mathrm{d}t = \sum_{k=1}^{\infty} b_k \int_{b_k}^{\infty} \frac{\mathrm{d}t}{t^2} = \sum_{k=1}^{\infty} 1 = \infty. $$

On the other hand, since $\Phi(t) = \sum_{l=1}^{k} b_l$ for each $t \in [b_k, b_{k+1})$,

\begin{align*} \int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t \geq \sum_{k=1}^{\infty} \int_{b_k}^{b_{k+1}} e^{-\Phi(t)} \, \mathrm{d}t = \sum_{k=1}^{\infty} (b_{k+1} - b_k) e^{-\sum_{l=1}^{k} b_l}. \end{align*}

So by choosing $(b_{k+1})$ so that it satisfies $(b_{k+1} - b_k) e^{-\sum_{l=1}^{k} b_l} \geq 1$ for each $k$, we have

$$ \int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t = \infty. $$

• integration
• analysis

Share Cite Follow edited Jul 25, 2021 at 15:54 Sangchul Lee asked Jul 22, 2021 at 20:06 Sangchul LeeSangchul Lee 174k1717 gold badges286286 silver badges460460 bronze badges $\endgroup$ 4

• $\begingroup$ As an observation, there seems to be a 'breakeven' point where the first and second integrals are both convergent. This happens for functions like $\Phi(x)=2\ln(x+1)$. Maybe someone can use this to get some bounds on what $\Phi(x)$ can actually be $\endgroup$ – QC_QAOA Commented Jul 22, 2021 at 21:25
• 1 $\begingroup$ Nice simplification! One typo is in $\Phi(t)=\sum_l b_k$. It should be $\sum_l b_l$. $\endgroup$ – Sungjin Kim Commented Jul 25, 2021 at 15:53
• 1 $\begingroup$ But what if we require $\Phi$ to be smooth? Can this counterexample be modified to smooth out the jumps? $\endgroup$ – Simon Parker Commented Jul 27, 2021 at 19:07
• 1 $\begingroup$ @SimonParker, We may mollify $\Phi$ to be smooth. For example, we may mollify $\Phi$ to a smooth function $\tilde{\Phi}$ such that $\Phi(x) \leq \tilde{\Phi}(x) \leq \Phi(x-1)$ and use $\tilde{\Phi}$ instead. A fairly simple comparison will then allow us to directly use our observation for $\Phi$. $\endgroup$ – Sangchul Lee Commented Jul 27, 2021 at 19:13

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1 Answer 1

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 12 $\begingroup$

We may assume $\Phi(1)=0$. Let $\Phi(x)=\int_1^x f(t) dt$ for some nonnegative function $f$. By assumption, we have $$ \int_1^{\infty} \int_1^x \frac{f(t)}{x^2}dtdx =\int_1^{\infty}\int_t^{\infty} \frac{f(t)}{x^2} dxdt = \int_1^{\infty} \frac{f(t)}tdt=\infty. $$ Let $$ a_{\ell} = \int_{2^{\ell-1}}^{2^{\ell}} \frac{f(t)}t dt. $$ We have $\sum_{\ell=1}^{\infty} a_{\ell} = \infty$ and $$ \int_{2^{\ell-1}}^{2^{\ell}} f(t) dt \leq 2^{\ell}\int_{2^{\ell-1}}^{2^{\ell}} \frac{f(t)}t dt=2^{\ell}a_{\ell}. $$ On the other hand, by Cauchy condensation test, the convergence of $\int_1^{\infty} e^{-\Phi(x)} dx$ is equivalent to the convergence of LHS of $$ \sum_{k=1}^{\infty} 2^k e^{-\Phi(2^k)} \geq \sum_{k=1}^{\infty} 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }. $$ Let $a_{\ell}$ be a characteristic function of a very rapidly increasing sequence $\{b_n\}$ of natural numbers.

If $b_m\leq k < b_{m+1}$, we have $$ 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }\geq 2^{b_m} e^{-2^{1+b_m}}. $$ Then the sum over $k$ in this interval is $$ \geq (b_{m+1}-b_m)2^{b_m} e^{-2^{1+b_m}}. $$ For each $m$, we take $b_{m+1}$ large enough that $$ (b_{m+1}-b_m)2^{b_m} e^{-2^{1+b_m}}\geq 1. $$ Then we have the divergence of the series $$ \sum_{k=1}^{\infty} 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }. $$ Hence, we have the divergence of $$ \int_1^{\infty} e^{-\Phi(x)} dx. $$

Share Cite Follow answered Jul 25, 2021 at 14:33 Sungjin KimSungjin Kim 20.6k33 gold badges2929 silver badges7373 bronze badges $\endgroup$ 4

• 1 $\begingroup$ I am not claiming the divergence in all case. I showed a counterexample. $\endgroup$ – Sungjin Kim Commented Jul 25, 2021 at 14:52
• 1 $\begingroup$ Great answer! It is interesting to have a counter-example. :) $\endgroup$ – Sangchul Lee Commented Jul 25, 2021 at 15:03
• $\begingroup$ I tried to simplify your answer by myself and updated my posting. Again, thank you for your elegant idea! $\endgroup$ – Sangchul Lee Commented Jul 25, 2021 at 15:09
• $\begingroup$ @SungjinKim Ahh sry, I misunderstood what you were doing. $\endgroup$ – Maximilian Janisch Commented Jul 25, 2021 at 16:13

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