# Inequality involving three functions

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: $$\int_{0}^{h} \int_{0}^{h} \max(u,v) f(u) f(v) du dv \geq \int_{0}^{h} \int_{0}^{h} \min(u,v) du dv \int_{0}^{h} f^2 du .$$

• I guess you mean $0\le u,v\le h$ – Fedor Petrov Dec 2 '18 at 14:05
• right, you are correct here – user2714795 Dec 2 '18 at 14:07
• You certainly are in trouble if $f$ is concentrated near $0$, say $f=1$ on $[0,\delta]$ and $0$ on $(\delta,h]$ with $\delta\ll h$. – fedja Dec 2 '18 at 14:22
• The question is unclearly formulated. In particular, could $h$ be $\infty$? – user64494 Dec 2 '18 at 18:58
• 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. – user2714795 Dec 2 '18 at 19:09

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 $$h>3$$.

• Can you kindly elaborate your answer? In particular, why "Then the left side of your displayed inequalities is $\leq h^3$"? – user64494 Dec 3 '18 at 1:45
• I have provided details. – Iosif Pinelis Dec 3 '18 at 2:53
• 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. – Alexandre Eremenko Dec 3 '18 at 14:44
• @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". – Iosif Pinelis Dec 3 '18 at 18:18
• @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. – Iosif Pinelis Dec 3 '18 at 22:20

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.

• Thanks a lot for pointing this out. – user2714795 Dec 2 '18 at 19:10
• Can you kindly elaborate your answer, giving us details? – user64494 Dec 3 '18 at 1:40