Let $f,\,g:[a,b]\to \mathbb{R}$ be both monotone increasing or monotone decreasing and also continuous. True or false? If Chebyshev's integral inequality holds as equality , i.e. $$\left(\int\limits_a^bf(x)\,\mathrm{d}x\right)\left(\int\limits_a^bg(x)\,\mathrm{d}x\right)= (b-a)\int\limits_a^bf(x) g(x)\,\mathrm{d}x,$$ then $f$ is constant or $g$ is constant.
Attempt. Since $$\frac{1}{2}\int\limits_a^b\left[\, \int\limits_a^b \big(f(y)-f(x)\big)\big(g(y)-g(x)\big)\,\mathrm{d}x\right]\mathrm{d}y =(b-a)\,\int\limits_a^bf(x) g(x)\,\mathrm{d}x - \left(\int\limits_a^bf(x)\,\mathrm{d}x\right)\, \left(\int\limits_a^bg(x)\,\mathrm{d}x\right),$$ we get the equality if and only if $$\int\limits_a^b\left[\, \int\limits_a^b \big(f(y)-f(x)\big)\big(g(y)-g(x)\big)\,\mathrm{d}x\right]\mathrm{d}y=0.$$ Since $\big(f(y)-f(x)\big)\big(g(y)-g(x)\big)$ is $\geqslant 0$ and continuous we get that for all $x,\,y\in [a,b]$: $$\big(f(y)-f(x)\big)\big(g(y)-g(x)\big)=0$$ i.e. $f(y)=f(x)$ or $g(y)=g(x)$, which doesn't mean that either $f$ or $g$ is constant. May we conclude that the answer is "false"? What would a counterexample be in that case?
Thanks in advance.
At least one function must be constant if $(f(y)-f(x))(g(y)-g(x))=0$ for all $x,y \in [a,b]$ and $f,g$ are both monotone.
Suppose WLOG $f$ and $g$ are both monotone increasing and $f$ is not constant. There exist $a \leqslant x_1 < x_2 \leqslant b$ such that $f(x_2) - f(x_1) > 0$. This implies that $g(x_2) - g(x_1) = 0$ and $g(x_1) = g(x) = g(x_2)$ for all $x \in [x_1,x_2]$ by monotonicity.
However, by monotonicity $f(b) - f(a) > 0$ since $b\geqslant x_2$ and $a \leqslant x_1$. Hence, we must have $g(a) = g(b) = g(x)$ for all $x \in [a,b]$ and $g$ must be constant.