Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove the following inequality: (where $f $ is nice enough) -

Given a function $ f(x,y) : \Omega_1 \times \Omega_2 \to \mathbb{R} $ , and $\alpha,C_1,C_2 $ are some constants, ( $\Omega_i$ is equipped with a probability measure $ \mu_i $ respectively) , then there exists a constant $C_3 $ such that $$ \begin{multline}C_1^2 \int_{\Omega_1 } \left| \int_{\Omega_2} f(x,y) d \mu _2 - \alpha\right| ^2 d\mu_1 + C_2^2 \int_{\Omega_2 } \left| \int_{\Omega_1} f(x,y) d \mu _1 - \alpha\right|^2 d\mu_2 \\ \geq C_3^2 \int_{\Omega_1}\int_{\Omega_2} |f(x,y)-\alpha|^2 d\mu_1 d\mu_2\end{multline}$$ is true.

It seems like it's kind of triangle's inequality. Have you got an idea?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

As written it is not true (if you meant, as I assumed, that $C_1, C_2, C_3$ are positive constants; otherwise just setting $C_3 = 0$ settles all cases with $C_1, C_2 \geq 0$).

Let $\Omega_1 = \Omega_2 = [0,1] \subset \mathbb{R}$ with the Lebesgue measure (making them probability spaces).

Let $f(x,y) = \sin(2\pi x) \sin(2\pi y)$. We have that

$$ \int_0^1 f(x,y) \mathrm{d}y = 0 = \int_0^1 f(x,y)\mathrm{d}x $$

Hence if you take $\alpha = 0$ the left hand side of your desired inequality is identically 0. But the right hand side is given by

$$ \iint_{[0,1]\times[0,1]} \sin(2\pi x)^2 \sin(2\pi y)^2 ~ \mathrm{d}x ~\mathrm{d}y = \frac14 $$

share|improve this answer
    
You're right... I meant indeed positive constants, and I fixed the original formulation of my question (I added a square to the constants) . Actually, in what I'm trying to do, my $\alpha$ should be $\int_{\Omega_1 } \int_{\Omega_2} f(x,y) d\mu_2 d\mu_1 $ ... Does it change the answer? if not, then have you got any idea how to solve the question I posted here: math.stackexchange.com/questions/174950/… ? I saw you've read it... Have you got any idea? Thanks a lot ! –  joshua Jul 25 '12 at 9:35
    
It doesn't change this answer. $\alpha$ is in fact defined to be equal to $\iint f(x,y)$ in my example above. I just saw your other question, and I see where this question is coming from. I don't think it is the best way to approach the problem. I just posted a proof for your other question there. –  Willie Wong Jul 25 '12 at 9:54
    
Thanks a lot @Willie ! –  joshua Jul 25 '12 at 10:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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