3
$\begingroup$

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:

\begin{equation} \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. \end{equation}

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:

\begin{equation} \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. \end{equation}

$\bf{EDIT}$: 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?

$\endgroup$
2
  • 1
    $\begingroup$ 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 $\endgroup$ May 6, 2013 at 14:52
  • $\begingroup$ Ah, I stated it wrong! Thank you David, I will make the edits now. $\endgroup$
    – Zamoura
    May 6, 2013 at 16:03

1 Answer 1

3
$\begingroup$

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 $x=\frac{1}{2}$).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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