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.



Integral inequality similar to Hardy's inequality
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. 


