Hello,

I am trying to show that $f,g \in \mathscr{R}$ implies that $(\int fg)^2\leq \int f^2 \int g^2$.

The hint is to use a variant of the Cauchy-Schwartz inequality, $\sum a_i b_i \leq \sum a_i^2 \sum b_i^2$.

We have a theorem which says that the Riemann integrability of $f$ and $g$ imply the integrability of the product fg so we can talk about $\overline{\int}fg$. This number is a lower bound for the upper sums $U(P,fg)$ for any partition $P$ of $[a,b]$ so, $\overline{\int}fg \leq U(P, fg)=\sum s_i t_i \Delta x_i$ (where $s_i$ is the max of $f$ on $[x_{i-1}, x_i]$, and $t_i$ is the max of $g$ on $[x_{i-1}, x_i]$.)

So if I square both sides of the inequality I get $(\overline{\int}fg)^2 \leq (\sum s_i t_i \Delta x_i)^2$.

It's pretty easy I think to argue that $\sum s_i t_i \Delta x_i < \sum s_i \Delta x_i t_i \Delta x_i$, since the $\Delta x_i$ are all positive. And I can write, $$(\overline{\int}fg)^2 \leq (\sum s_i t_i \Delta x_i)^2<(\sum s_i \Delta x_i t_i \Delta x_i)^2\leq (\sum s_i \Delta x_i)^2 (\sum t_i \Delta x_i)^2=U(P,f^2)U(P,g^2)$$

Now I'm not to sure what to do. I have and expression which relates, eventually, the upper sums of a product $(\int fg)^2$ with the upper sums of $f^2$ and $g^2$.