Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f(x)$ and $g(x)$ are continuous function from $[0,1]\rightarrow [0,1]$, and $f$ is monotone increasing, then how to prove the following inequality: $$\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx$$

share|cite|improve this question
I'll be the first to say it - users are encouraged to say what they've done on a question so far/what might be a point of confusion so that people can deliver more helpful (and not redundant) advice. –  lamb_da_calculus Jan 13 '13 at 4:20
yes, because it's hard and I have no idea to go any further. I just get a hint: how to estimate $f(x)$, then change $x$ to $g(x)$ –  Larry Eppes Jan 13 '13 at 4:24
Did you consider using the fact that $f$ as a continuous monotone function is differentiable almost everywhere? –  Eckhard Jan 13 '13 at 15:44

1 Answer 1

up vote 24 down vote accepted

By the Mean Value Theorem for Integrals there is a point $\xi$ in $[0,1]$ such that $$ \int_{0}^{1}[f(g(x))-g(x)]dx=f(g(\xi))-g(\xi) $$ Let $u = g(\xi)$. Then $$ f(g(\xi))-g(\xi) = f(u)-u \leq f(u) - uf(u) = (1-u)f(u) $$ (the inequality is due to the fact that $0 \leq f(u) \leq 1$.) Since $f$ is monotone increasing, $f(x) \geq f(u)$ for all $x$ in $[u,1]$. So $$ (1-u)f(u) = \int_u^1 f(u)\,dx \leq \int_u^1 f(x)\,dx \leq \int_0^1 f(x)\,dx $$ Rearranging, $$ \int_0^1 f(g(x)) \,dx \leq \int_0^1 f(x)\,dx + \int_0^1 g(x)\,dx $$ Over!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.