Take the 2-minute tour ×
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|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
2  
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
add comment

1 Answer 1

up vote 13 down vote accepted

$\int_{0}^{1}[f(g(x))-g(x)]dx=f(g(\xi))-g(\xi)=f(u)-u\leq f(u)-uf(u)=(1-u)f(u) =\int_{u}^{1}f(u)dx\leq\int_{u}^{1}f(x)dx\leq\int_{0}^{1}f(x)dx.$

Over!

Here we use the Mean value theorems for integration in the first $"="$, and let $u=g(\xi)$.

share|improve this answer
add comment

Your Answer

 
discard

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.