Prove $(b-a) \cdot \int_a^b \alpha(x) \beta(x) dx \ge \int_a^b \alpha(x) dx \cdot \int_a^b \beta(x) dx$.

Question

Suppose $\alpha: [a, b] \to (-\infty, \infty)$ and $\beta:[a,b] \to (-\infty, \infty)$ are both nondecreasing. Then, $$(b-a) \cdot \int_a^b \alpha(x) \beta(x) dx \ge \int_a^b \alpha(x) dx \cdot \int_a^b \beta(x) dx.$$

This is Lemma A.3 from Robson, A.J., 1992. Status, the distribution of wealth, private and social attitudes to risk. Econometrica: Journal of the Econometric Society, pp.837-857.

The author omit this proof since it is straightforward, but I am struggling to prove this. I appreciate if you give some help, and please tell me if more context is needed.

Answer 1

Proof. Let $D = [a,b]^2$. Then $$ \int_a^b 1 \int_a^b \alpha\cdot \beta- \int_a^b \alpha \int_a^b \beta\stackrel{(1)}= \iint_D( \alpha(y)\beta(y)-\alpha(x)\beta(y) )\,\mathrm dx \,\mathrm dy \stackrel{(2)}= \iint_D (\alpha (x)\beta(x)-\alpha(y)\beta(x))\,\mathrm dx\,\mathrm dy = \frac 12 \iint_D (\alpha(x)\beta(x)+\alpha(y)\beta(y)-\alpha (x)\beta(y)-\alpha(y)\beta(x)) \,\mathrm dx \,\mathrm dy = \frac 12 \iint_D (\alpha(x)-\alpha(y))(\beta(x)-\beta(y))\,\mathrm dx\,\mathrm dy \geqslant 0, $$ since both $\alpha, \beta$ are nondecreasing, i.e. $(\alpha(x) - \alpha (y)), (\beta(x)- \beta(y))$ have the same sign, equivalently $(\alpha(x)-\alpha(y))(\beta(x)-\beta(y)) \geqslant 0$ for all $(x,y)\in [a,b]^2$.

EXPLANATION

1. For two functions $f,g \colon [a,b] \to \mathbb R$, $$ \int_a^b f \int_a^b g = \int_a^b f(x)\,\mathrm dx \int_a^b g(x)\,\mathrm dx = \int_a^b f(x)\,\mathrm dx \int_a^b g(y)\,\mathrm dy \;[\text{change the integration parameter}] = \iint_D f(x)g(y)\,\mathrm dx \,\mathrm dy \;[\text{transfer to a double integral}]. $$ Hence the $(1)$.
2. Use a different parameter, we have $$ \int_a^b f \int_a^b g = \int_a^b f(y)\,\mathrm dy \int_a^b g(x)\, \mathrm dx = \iint_D f(y)g(x)\,\mathrm dx \,\mathrm dy. $$ Go back to the question, we have $$ (b-a)\int_a^b \alpha\cdot \beta - \int_a^b \alpha \int_a^b \beta = \iint_D (\alpha(y)\beta(y) - \alpha(x)\beta(y)), $$ also $$ (b-a)\int_a^b \alpha\cdot \beta - \int_a^b \alpha \int_a^b \beta = \iint_D (\alpha(x)\beta(x) - \alpha(y)\beta(x)), $$ add these two equations we get the $(2)$.