Steffensen's inequality proof

Submitted by MSA on Sat, 12/10/2011 - 17:04.

Johan Frederik Steffensen (1873–1961)

ƒ : [a, b] → R is a non-negative, monotonically decreasing, integrable function and
g:[a, b] →[0, 1] is another integrable function, then:

$\int^b_{b-k} f(x)\,dx \le\int^b_a f(x)g(x)\,dx \le \int^{a+k}_a f(x)\,dx$

where $k=\int^b_ag(x)\,dx$
First we see that $\int_x^bg(t) dt\le b-x$ and $\int_a^x g(t) dt\le x-a$ . We have $x-b \le-\int_x^bg(t) dt$ and

$a+\int_a^xg(t)dt\le a+(x-a)=x=b+(x-b)\le b-\int_x^bg(t) dt$

$f(b-\int_x^bg(t) dt)\le f(x)\le f(a+\int_a^xg(t)dt)$

We have also

$f(b-\int_x^bg(t) dt)g(x)\le f(x)g(x)\le f(a+\int_a^xg(t)dt)g(x)$

$\int_a^bf(b-\int_x^bg(t) dt)g(x)dx\le \int_a^bf(x)g(x)dx\le \int_a^bf(a+\int_a^xg(t)dt)g(x) dx$

take $u=b-\int_x^bg(t) dt$ and $du=g(x) dx$ and $v=a+\int_a^xg(t)dt$ so $dv=g(x)dx$

$\int_a^bf(b-\int_x^bg(t) dt)g(x)dx=\int_{b-\int_a^bg(t)dt}^b f(x)dx$

$\int_a^bf(a+\int_a^xg(t)dt)g(x) dx=\int_a^{a+\int_a^bg(t)dt}f(x) dx$

That proof the inequality.