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

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 $$h>3$$.