Does $\int_\cdot f\int_\cdot g$ have the same ordering as $\int_\cdot fg$? Fix an integer $N\geq 2$ and let $f,g:[0,1]\to\mathbb{R}^+$ be continuous. We will divide $[0,1]$ into $N$ subintervals of equal size (disregarding overlapping edges);
$$I_n:=[(n-1)/N,n/N],\qquad n=1,\ldots,N.$$
I'm trying to show that
$$\int_{I_k}f\cdot\int_{I_k}g\leq \int_{I_l}f\cdot\int_{I_l}g\quad\Rightarrow\quad \int_{I_k}fg\leq \int_{I_l}fg.$$
However, I'm not sure whether or not it holds in general. I have managed to break down that if we assume the LHS and additionally assume that there exists a permutation $(n_1,\ldots,n_N)$ of $(1,\ldots,N)$ such that
$$\min_{I_{n_1}}f \:\cdot\: \min_{I_{n_1}}g\leq \max_{I_{n_1}}f \:\cdot\: \max_{I_{n_1}}g\leq\ldots\leq \min_{I_{n_N}}f \:\cdot\: \min_{I_{n_N}}g\leq \max_{I_{n_N}}f \:\cdot\: \max_{I_{n_N}}g,\tag{1}$$
then there exists such a permutation with $i< j$ if $k=n_i$, $l=n_j$ and $k\neq l$ since
$$\min_{I_{n}}f \:\cdot\: \min_{I_{n}}g\leq N^2\int_{I_n}f\cdot \int_{I_n}g\leq \max_{I_{n}}f \:\cdot\: \max_{I_{n}}g.$$
Hence
$$\int_{I_k}fg \leq \max_{I_k}fg\leq \max_{I_{k}}f \:\cdot\: \max_{I_{k}}g\leq \min_{I_{l}}f \:\cdot\: \min_{I_{l}}g\leq\min_{I_l} fg\leq \int_{I_l} fg.$$
But I'm not sure if $(1)$ is a reasonable assumption. I also tried to consider average function values, but it has led me nowhere yet. Perhaps I should look at the integrals as an inner product?
 A: I've discovered a counter example for non-continuous $f$ and $g$. Set $N=2$, $c\in[2/3,1)$ and define
$$f = \tfrac{1}{2}\mathbf{1}_{[0,1/2)} + c\mathbf{1}_{[1/2,3/4)}\quad\text{and}\quad g = \tfrac{1}{2}\mathbf{1}_{[0,3/4)} + \mathbf{1}_{[3/4,1]}.$$
Then 
$$\int_{0}^{1/2}f \cdot\int_{0}^{1/2}g = \frac{1}{16}\quad\text{and}\quad \int_{1/2}^1 f \cdot\int_{1/2}^1g = \frac{c}{4} \cdot \frac{3}{8} = \frac{3c}{32}\geq\frac{1}{16},$$
but
$$\int_{0}^{1/2}fg = \frac{1}{8}\quad\text{and}\quad \int_{1/2}^1fg = \frac{c}{8}<\frac{1}{8}.$$
For arbitrary $N$ we may can
$$f = \tfrac{1}{N}\mathbf{1}_{[0,1/N)} + c\mathbf{1}_{[1/N,1/N+1/N^2)}\quad\text{and}\quad g = \tfrac{1}{N}\mathbf{1}_{[0,1/N+1/N^2)} + \mathbf{1}_{[1/N+1/N^2,1]}$$
and we can derive a similar contradiction. With this idea in mind, it is conceivable that by smoothly approximating $f$ and $g$ above, we can derive a set of continuous functions for which this does not hold. So it is not true and the reverse statement can also be refuted by the same counter example.
