Home › Forums › Math questions › Analysis questions

Steffensen's inequality proof

Posted December 10th, 2011 by MSA
in

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

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

so

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

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.

‹ Mathematical induction : integral evaluation ›

Comments

Answer

On December 14th, 2011 Structure says:

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

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

so

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

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.

Back to top