# Integral of composition [duplicate]

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Prove that if $f,g:[0,1]\rightarrow[0,1]$ - continuous functions and f is strictly increasing then $$\int\limits_0^1f(g(x))dx\leq\int\limits_0^1f(x)dx+\int\limits_0^1g(x)dx.$$

I tried to prove that $f(g(x))\leq f(x)+g(x)$ for all $x\in[0,1]$ but realized that it is wrong. For example, $f(x)=\begin{cases}0,0\leq x< \frac{1}{4}\\2x-\frac{1}{2},\frac{1}{4}\leq x < \frac{3}{4} \\1,\frac{3}{4}\leq x\leq 1\end{cases}$, $g(x)=\frac{3}{4}$ but $f(g(0))=1>0+\frac{3}{4}=f(0)+g(0)$.

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## marked as duplicate by Najib Idrissi, Git Gud, gt6989b, AlexR, mauMar 12 '14 at 15:08

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Your $f(x)$ is not strictly increasing. –  gt6989b Mar 12 '14 at 13:12
@gt6989b yes, but we can replace $0$ by somthing like $\frac{x}{100}$ and $1$ by $1-\frac{x}{100}$ and $f(g(0))>f(0)+g(0)$ will be still true –  user122380 Mar 12 '14 at 13:15
Where did you get this question ? It sounds interesting. –  DiffeoR Mar 12 '14 at 13:18
@DiffeoR it was in one student's math olympiad task –  user122380 Mar 12 '14 at 13:21
math.stackexchange.com/q/276722/129458 –  OBDA Mar 12 '14 at 13:34

## 1 Answer

Let $x_0$ the point of $[0,1]$ where $h(x)=f(x)-x$ reach the maximum value.

We have $$\int\limits_0^1f(x)dx \geq \int\limits_{x_0}^{1} f(x)dx \geq (1-x_0)f(x_0)$$ $$=h(x_0)+x_0(1-f(x_0))\geq h(x_0)=max_{[0,1]} h(x)$$ $$\geq \int\limits_{0}^{1} h(g(x))dx=\int\limits_0^1f(g(x))dx -\int\limits_0^1 g(x)dx$$

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@Sabyasachi Yes, Please take a closer look. –  user119228 Mar 12 '14 at 13:33
+1. very smart. –  Sabyasachi Mar 12 '14 at 13:34
I did. I deleted my comment then. –  Sabyasachi Mar 12 '14 at 13:35
@Sabyasachi Ok, and thank you for your comment :) –  user119228 Mar 12 '14 at 13:39
:) you're welcome. –  Sabyasachi Mar 12 '14 at 13:40