# Integral inequality for convex function

Let $u(x)$ be a smooth function from $\mathbb{R}$ to $\mathbb{R}$. Suppose that for some real numbers $a,b$ with $a < b$ the following equality is true:

$$\frac{1}{b-a} \int_a^b u(x) \ \rm{d} x = \frac{1}{u(a)-u(b)} \int_{u(b)}^{u(a)} x \ \rm{d} x.$$

I would like to prove that, if $f$ is a non-negative convex function from $\mathbb{R}$ to $\mathbb{R}$, then the following inequality holds:

$$\frac{1}{b-a} \int_a^b f(u(x)) \ \rm{d} x \geq \frac{1}{u(a)-u(b)} \int_{u(b)}^{u(a)}f(x) \ \rm{d} x.$$

$\bf{EDIT}$: It is clear now that, as written, the above claim is false. A related question is for what class of functions $f$ is the stated inequality true?

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I think this is false. For example pick u(x) = 0, c=-1, d=1 (it doesn't matter what a and b are). Then the claim is that $f(0) \geq \frac{1}{2} \int_{-1}^{1} f(x) dx$ which $f(x) = x^2$ shows is false – David Benson-Putnins May 6 '13 at 14:52
Ah, I stated it wrong! Thank you David, I will make the edits now. – Zamoura May 6 '13 at 16:03

The inequality is false in general. To see this, take $a=0,$ $b=1,$ $f(x)=x^2.$ Let $u(0)=0$ and $u(1)=1.$ Then our problem can be reformulated as follows: given that $\int_{0}^1u(x)dx=\frac{1}{2}$ show that $\int_{0}^1u^2(x)dx\ge\frac{1}{3}.$ Now take $u(x)=x+th(x)$ where $h(0)=h(1)=0$ and $\int_{0}^1h(x)dx=0$ to end up with the inequality $$t^2\int_{0}^1h^2(x)dx+2t\int_0^1xh(x)dx\ge 0$$ for all $t\in\mathbb{R}.$ Taking $t$ sufficiently small we get $\int_0^1xh(x)dx\ge 0$ for all appropriately chosen $h(x).$ It is now easy to choose $h$ in way that last inequality is false (just take something antisymmetric with respect to $x=\frac{1}{2}$).