# Double-integral inequalities for monotone functions

• Jun 18th 2013, 02:15 PM
bkarpuz
Double-integral inequalities for monotone functions
Dear MHF members,

I met somewhere an integral inequality as follows.
$\displaystyle \int_{\tau(a)}^{\tau(b)}f(x)\int_{x}^{b}g(y) \mathrm{d}y \mathrm{d}x\geq\int_{a}^{b}g(y)\int_{\tau(a)}^{ \tau(y)}f(x) \mathrm{d}x \mathrm{d}y,$
where $\displaystyle f,g$ are nonnegative continuous functions
and $\displaystyle \tau$ is a nondecreasing function with $\displaystyle \tau(x)\leq x$.
Have any of you met before with something like this
or can help me how to deduce this?

Thank you very much.
bkarpuz
• Jun 18th 2013, 05:01 PM
chiro
Re: Double-integral inequalities for monotone functions
Hey bkarpuz.

This looks like a change of order: can you try doing a change of order first and then seeing if you can use your constraint (non-decreasing function) to prove the result?
• Jun 18th 2013, 09:44 PM
bkarpuz
Re: Double-integral inequalities for monotone functions
My friend chiro, you are always doing the same.
Please give me somthing more than I have. :)

Okay, when $\displaystyle \tau$ is differentiable, I have the following proof.

Proof. Let
$\displaystyle \varphi(t):=\int_{\tau(a)}^{\tau(t)}f(x)\int_{x}^{ t}g(y) \mathrm{d}y \mathrm{d}x-\int_{a}^{t}g(y)\int_{\tau(a)}^{ \tau(y)}f(x) \mathrm{d}x \mathrm{d}y$,
which yields
$\displaystyle \varphi^{\prime}(t)=\bigg(\int_{\tau(t)}^{t}g(y) \mathrm{d}y\bigg)f(\tau(t))\tau^{\prime}(t)\geq0.$
Hence, for $\displaystyle b\geq a$, we get
$\displaystyle \varphi(b)\geq\varphi(a)=0$,
which completes the proof. $\displaystyle \rule{0.2cm}{0.2cm}$

But what if $\displaystyle \tau$ is not differentiable?
Thank you.
• Jun 19th 2013, 02:34 AM
chiro
Re: Double-integral inequalities for monotone functions
If it is not differentiable you will need to resort to the appropriate Measure Theory results and use the appropriate measure.
• Jun 19th 2013, 10:20 AM
bkarpuz
Re: Double-integral inequalities for monotone functions
Quote:

Originally Posted by chiro
If it is not differentiable you will need to resort to the appropriate Measure Theory results and use the appropriate measure.

What kind of appropriate measure are you talking about chiro?
This is usual integration the functions are continuous.
I will be very glad if you do not spam my posts again.
• Jun 19th 2013, 04:35 PM
chiro
Re: Double-integral inequalities for monotone functions
If something is not differentiable then you use a different measure to integrate.

If something has an analytic smooth anti-derivative you use the Riemann integral and measure.

Other-wise you use another measure that allows you to integrate the function.

An example of a measure that is used for continuous but not differentiable integration is the Brownian Motion stochastic integral.

Take a look at the Lebesgue integral and the appropriate measure theory.