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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: $$\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx$$

Martin Sleziak
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Larry Eppes
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    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 04:20
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    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 04:24
  • Did you consider using the fact that $f$ as a continuous monotone function is differentiable almost everywhere? – Eckhard Jan 13 '13 at 15:44

3 Answers3

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By the Mean Value Theorem for Integrals, there is a point $\xi \in [0,1]$ such that $$ \int_{0}^{1}[f(g(x))-g(x)]dx=f(g(\xi))-g(\xi).\tag{1} $$ Let $u = g(\xi)$. Then $$ f(g(\xi))-g(\xi) = f(u)-u \leq f(u) - uf(u) = (1-u)f(u),\tag{2} $$ (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 $$ (1-u)f(u) = \int_u^1 f(u)\,dx \leq \int_u^1 f(x)\,dx \leq \int_0^1 f(x)\,dx.\tag{3} $$ Hence, by $(1),(2),(3)$, we have
$$\int_{0}^{1}[f(g(x))-g(x)]dx\leq\int_0^1 f(x)\,dx.$$ Rearranging, $$ \int_0^1 f(g(x)) \,dx \leq \int_0^1 f(x)\,dx + \int_0^1 g(x)\,dx. $$ Over!

Riemann
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  • I don't see at all how the penultimate line rearranges to the last line. Could someone possibly enlighten me? – Jam Sep 25 '22 at 16:12
  • @Jam Answered my own question. Substitute back $\int_0^1[f(g(x))-g(x)]\leq(1-u)f(u)$ to the left of the penultimate line. That didn't feel obvious. – Jam Sep 25 '22 at 16:18
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This is not an answer, @Rienmann provided one. But it is an interesting corollary of this result:

$$\int_0^1 f(f(x)) dx \leq 2\int_0^1 f(x)dx$$

Also,

$$\int_0^1 f(x) dx + \int_0^1 f^{-1}(x) dx \geq 1$$

and

$$\int_0^1 f(x^2) dx \leq \frac{1}{3}+\int_0^1 f(x) dx.$$ An interesting question is when (for which functions) the last inequality becomes an equality.

Vincent Granville
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    Nice corollaries @ Vincent Granville – Riemann Aug 22 '19 at 10:53
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    wow, interesting question, I just tried the techniques from variation of calculus for your last ineq., after a smooth perturbation, the original function just disappeared... I think this must not a question of variations of calculus. So I can just get the answer from the prove of Riemann, and the monotone continuous $f$ is not exist. – Larry Eppes Sep 01 '19 at 13:05
  • As someone mentioned somewhere else (social media), the second inequality is only defined for f(0)=0, f(1)=1, and then it's an equality (but the OP's inequality only proves 1/2 in the RHS, not 1)? – Clement C. Dec 28 '23 at 12:06
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For equality, we get:

$$\int_{0}^{u} f(x)dx = 0 \Rightarrow f(x) = 0, \forall x \in (0, u)$$ $$\int_{u}^{1} f(u)dx = \int_{u}^{1} f(x)dx \Rightarrow f(x) = f(u), \forall x \in [u, 1)$$

for $f$ to be continuous, we need $u \in \{0, 1\}$, in other words, $f$ is constant. We quickly get that $g$ is constant zero.

As for Vincent's question, there is no equality in the last inequality because $x^2$ isn't constant. Proving a better bound would be more interesting. As pointed out by Clement, the second inequality is wrong and the corollary as a whole seems quite weak.

yawp
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