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**

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?

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.

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.

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.

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.