7
$\begingroup$

I am not very sure if that's research level, I hope you don't find it too elementary for this place.

I am trying to solve the following puzzle:

We are given a real function $f$, where $f(x) \geq 0$ and $F := \int_0^x f(t) dt$ and some real $p>1$.

Does $\int_0^\infty f(x)^p e^{-x}dx < \infty$ imply $\int_0^\infty F(x)^pe^{-x}dx < \infty$ ?

My current approach was modifying proofs of Hardy's theorem. This approach has not been very successful. I keep running into hard to compute integrals that I would need to bound. I hope somebody could maybe provide me with some direction in which I should look. If somebody was interested I could provide details of calculations so far.

$\endgroup$
1
  • $\begingroup$ Which Hardy's inequality? $\endgroup$ Jul 13, 2015 at 9:36

2 Answers 2

4
$\begingroup$

Here is an idea and I will leave details to you (it might be that I made a stupid mistake somewhere therefore these computations must be checked carefully). All functions below should be sufficiently nice so that all formulas make sense. First I will formulate a lemma.

Lemma: Let $p>1$. If $\psi(0)=0, \psi' >0$ and $\psi(\infty)=\infty$ then for any $f\geq 0$ and any (nice) measures $d\mu(t)$ we have $$ \int_{0}^{\infty}\left(\int_{0}^{t}f(s)ds\right)^{p}d\mu(t)\leq \int_{0}^{\infty}f(y)^{p} \left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]dy \quad (1) $$

Remarks:
1) We have a freedom of choosing $\psi$. One can optimize the quantity $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]$ in the right hand side of (1) over all $\psi$ and find the best bounds (This becomes some ODE problem which should not be a difficult problem so that I will leave details to the readers).

However, for our purposes we do not need to know the best $\psi$. One just needs to guess the right one to find some bound. Let me show you how it works on your example. In your case $d\mu(t)=e^{-t}dt$. Instead of $\psi$ take something like $\psi(t)=te^{\frac{t}{10(p-1)}}$. Then $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right] \leq C e^{-y}$. So the claim follows.

Proof of Lemma:

Let $\psi$ be a convex function, and let $w$ be some positive function (we will choose them later). Then Jensen's inequality together with Fubini's theorem implies:

\begin{align*} &\int_{0}^{\infty} \varphi\left( \frac{1}{x}\int_{0}^{x} g(s)ds\right) w(x)dx\leq \int_{0}^{\infty}\frac{w(x)}{x}\int_{0}^{x}\varphi(g(s))dsdx=\\ &\int_{0}^{\infty} \varphi(g(s))\int_{s}^{\infty}\frac{w(x)}{x} dxds \end{align*}

Let $\varphi=u^{p}$ and lets make a change of variables $s=\psi(y)$ and $x=\psi(t)$. Then the above inequality takes the form:

\begin{align*} \int_{0}^{\infty} \left(\int_{0}^{t} g(\psi(y))\psi'(y) ds\right)^{p} \frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt\leq \int_{0}^{\infty} [g(\psi(y))\psi'(y)]^{p}\int_{y}^{\infty}\frac{w(\psi(t)) \psi'(t)}{\psi(t) (\psi'(y))^{p-1}} dtdy \end{align*}

Now choose $g$ so that $g(\psi(y))\psi'(y)=f(y)$. And choose $\frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt = d\mu(t)$ (for example take $w(\psi(t))dt =\frac{\psi^{p}(t)d\mu(t)}{\psi'(t)}$ ). Then the lemma follows.

$\endgroup$
1
  • $\begingroup$ Dear Paata, thank you very much for your insight. It solves the riddle. $\endgroup$
    – zen-dev
    Jul 14, 2015 at 21:51
2
$\begingroup$

The question has been already solved by Paata Ivanisvili. I just want to share my insight.

I think I have found the solution. I will use generalized Hardy's inequality, which unfortunately is a bit like killing a fly with a cannon. As a source I will cite http://www.encyclopediaofmath.org/index.php/Hardy_inequality where you can find references to the original source.

Lemma - generalized Hardy's (cited from Encyclopedia of Math). $$ \int_0^\infty \left| \phi(x) \int_0^x f(t) dt \right|^p dx \leq C \int_0^\infty \left| \psi(x) f(x) \right|^p dx $$ $\iff$ $$ \sup_{x > 0} \left[ \int_x^\infty | \phi(t) |^p dt \right]^{\frac{1}{p}} \left[ \int_0^x | \psi(t) |^{-q} dt \right]^{\frac{1}{q}} < + \infty, $$ where $\frac{1}{p} + \frac{1}{q} = 1 $.

After using this inequality, the task comes down to verification of the assumption, which is pretty easy. Let me verify.

Given a function $f \geq 0$ and $F(x) = \int_0^{x} f(t) dt$ I claim $\int_0^\infty f(x)^p e^{-x}dx < \infty$ implies $\int_0^\infty F(x)^pe^{-x}dx < \infty$.

Proof: Assume $\psi(t) = \phi(t) = e^{- \frac{t}{p} }$. When $\frac{1}{p} + \frac{1}{q} = 1 $, then $\frac{1}{q} = \frac{p-1}{p}$ and $-q = \frac{p}{1-p}$. We investigate $$ \sup_{x > 0} \left[ \int_x^\infty e^{-t} dt \right]^{\frac{1}{p}} \left[ \int_0^x e^{-t \frac{1}{1-p}} dt \right]^{\frac{p-1}{p}} = \sup_{x > 0} e^{- \frac{x}{p}} \left( e^{\frac{x}{p-1}} - 1 \right)^{\frac{p-1}{p}} = $$ $$ \sup_{x > 0} \left( e^{- \frac{x}{p-1}} e^{\frac{x}{p-1}} - e^{- \frac{x}{p-1}} \right)^{\frac{p-1}{p}} = \sup_{x > 0} \left( 1 - e^{- \frac{x}{p-1}} \right)^{\frac{p-1}{p}}, $$ which is bounded as $p>1$. Now we apply Hardy and our theorem is verified.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.