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I have the follwing inequality, which I am not sure if it is correct or not. $$\int_{0}^{h} \int_{0}^{h} \max(u,v) f(u) f(v) du dv \geq \int_{0}^{h} \int_{0}^{h} \min(u,v) du dv \int_{0}^{h}\int_{0}^{h} f(u)f(v) du dv, $$ where $f$ is an $L^2$ function , $f$ is the a.e derivative of $F$, and $F$ is also in $L^2$, $0 \leq u, v \leq h $

The final thing that I am trying to prove is the following: $$\int_{0}^{h} \int_{0}^{h} \max(u,v) f(u) f(v) du dv \geq \int_{0}^{h} \int_{0}^{h} \min(u,v) du dv \int_{0}^{h} f^2 du .$$

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  • $\begingroup$ I guess you mean $0\le u,v\le h$ $\endgroup$
    – Fedor Petrov
    Commented Dec 2, 2018 at 14:05
  • $\begingroup$ right, you are correct here $\endgroup$
    – user2714795
    Commented Dec 2, 2018 at 14:07
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    $\begingroup$ You certainly are in trouble if $f$ is concentrated near $0$, say $f=1$ on $[0,\delta]$ and $0$ on $(\delta,h]$ with $\delta\ll h$. $\endgroup$
    – fedja
    Commented Dec 2, 2018 at 14:22
  • $\begingroup$ The question is unclearly formulated. In particular, could $h$ be $\infty$? $\endgroup$
    – user64494
    Commented Dec 2, 2018 at 18:58
  • $\begingroup$ No h is finite. I am pasting a link to the place from where the basic problem has arisen. statistics.stanford.edu/sites/default/files/EFS%20NSF%20159.pdf. On pg 9 of this the authors are trying to prove that $ \frac{1}{h^2}\int_0^h (f -f_h)^2 - \frac{1}{12} \int_{0}^{h}f^{'}^{2}$ can be small. The $1/12$ fraction only comes if we are able to get some kind of inequality as above. Since the author has not explicity mentioned hence I was trying to derive it. The $1/12 $ fraction appears to be coming from the min value that the expression can have. $\endgroup$
    – user2714795
    Commented Dec 2, 2018 at 19:09

2 Answers 2

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Your inequalities are false in general, in view of homogeneity considerations. Indeed, note first that \begin{equation} \int_0^h\int_0^h \min(u,v) du dv=\int_0^h dv\int_0^v du\, u+\int_0^h du\int_0^u dv\, v=\frac{h^3}3. \end{equation} Let now $f=1$ on $[0,h]$. Then the left side of your displayed inequalities is \begin{equation} \int_0^h\int_0^h \max(u,v) f(u) f(v)\, du\, dv \le\int_0^h\int_0^h h \, du\, dv=h^3, \end{equation} whereas their right sides are, respectively, $\frac{h^3}3\,h^2$ and $\frac{h^3}3\,h$, which are greater than $h^3$ if $h>3$.

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  • $\begingroup$ Can you kindly elaborate your answer? In particular, why "Then the left side of your displayed inequalities is $\leq h^3$"? $\endgroup$
    – user64494
    Commented Dec 3, 2018 at 1:45
  • $\begingroup$ I have provided details. $\endgroup$
    – Iosif Pinelis
    Commented Dec 3, 2018 at 2:53
  • $\begingroup$ I woud state this simpler: if $f$ is dimensionless, and $u,v,h$ are measured in feet, then the left hand side has dimension $ft^3$ while the RHS has $ft^5$ and $ft^4$. And we have the same inconsistency whatever the dimension of $f$ is. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 3, 2018 at 14:44
  • $\begingroup$ @AlexandreEremenko : This is how I was actually thinking (except that it was centimeters rather than feet :-); I certainly prefer the simple and natural Gauss metric unit CGS system (en.wikipedia.org/wiki/Gaussian_units) to any other), and this is what I meant by the "homogeneity considerations". $\endgroup$
    – Iosif Pinelis
    Commented Dec 3, 2018 at 18:18
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    $\begingroup$ @AlexandreEremenko : I know that, and I think it is a great pity. Instead of the simple and natural formula $F=q_1q_2/r^2$ in CGS for the Coulomb law, in SI they have this terrible extra factor $1/(4\pi\epsilon_0)$. Instead of only three basic CGS units, in SI they have 7 basic units (if my count is correct), including (yes) the basic unit candela; and then they have about 3 million derived SI units, mostly named after persons. Looks quite crazy to me! When I was in high school (late 60s), we only had to deal with CGS, thankfully. $\endgroup$
    – Iosif Pinelis
    Commented Dec 3, 2018 at 22:20
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Both your inequalities are wrong, because the LHS can be negative. Indeed, take $h=3,\; f(x)=\delta(x-1)-\delta(x-2)$ then the left hand side will be $-1$. Now approximate the deltas with smooth functions.

On the other hand, in your second inequalty (the final thing) the right hand side is always non-negative, while in the first inequality it is zero for the example above.

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  • $\begingroup$ Thanks a lot for pointing this out. $\endgroup$
    – user2714795
    Commented Dec 2, 2018 at 19:10
  • $\begingroup$ Can you kindly elaborate your answer, giving us details? $\endgroup$
    – user64494
    Commented Dec 3, 2018 at 1:40

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