Question:

Inequality involving three functions

Ella: 2 days ago

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

Answer:
Jameson: 2 days ago

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 $h>3$.