# Double Integral Claim

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?

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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

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

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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