# Poincaré Inequality - Product Of Measures

I'm given two euclidean spaces $\mathbb{R}_1 , \mathbb{R}_2$ , with probability measures on them , that satisfy the Poincaré's inequality: $\lambda^2 \int_{\mathbb{R}^k} |f - \int_{\mathbb{R}^k} f d\mu | ^2 d\mu \leq \int_{\mathbb{R}^k} | \nabla f | ^2 d\mu$ for some constants $C_1 , C_2$ respectively.

How can I prove that the space $(\mathbb{R}_1 \times \mathbb{R}_2 , \mu_1 \otimes \mu_2 )$ also satisfies the Poincaré inequality?

My attempt: denote: $g(x) = \int_{\mathbb{R}_2 } f(x,y) d\mu_2$ . We can then apply the inequality in order to get that: $$C_1 ^2 \cdot \int_{\mathbb{R}_1 } \left| \int_{\mathbb{R}_2} f(x,y) d\mu_2 - \int_{\mathbb{R}_1} \int_{\mathbb{R}_2} f(x,y) d\mu_2 d\mu_1 \right| ^2 d\mu_1 \leq \int_{\mathbb{R_1}} \left| \nabla \int_{\mathbb{R}_2} f(x,y) d\mu_2 \right|^2 d\mu_1$$ and we can even say that the RHS is $\leq \int_{\mathbb{R}_1 } \int_{\mathbb{R}_2 } | \nabla f | ^2 d\mu_1 d\mu_2$ where our function is nice enough...But how can I finish the proof?

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Without loss of generality (by replacing $f$ by $f - \alpha$ you can assume that $\iint_{\mathbb{R}_1\times\mathbb{R}_2} f \mathrm{d}(\mu_1 \otimes\mu_2) = 0$.

Let $g(x) = \int_{\mathbb{R}_2} f(x,y)\mathrm{d}\mu_2$ as you did, we note that by the above assumption $\int g(x) \mathrm{d}\mu_1 = 0$. Now we compute

\begin{align} \iint |f(x,y)|^2 \mathrm{d}(\mu_1\otimes\mu_2) & \leq 2\iint |f(x,y) - g(x)|^2 + |g(x)|^2 \mathrm{d}(\mu_1\otimes \mu_2) \tag{1}\\ & = 2 \int_{\mathbb{R}_1} \int_{\mathbb{R}_2} \left| f(x,y) - \left(\int_{\mathbb{R}_2} f(x,y')\mathrm{d}\mu_2\right)\mathrm{d}\mu_2\right|^2 \mathrm{d}\mu_1 + 2\int_{\mathbb{R}_1} \left|g(x) \right|^2 \mathrm{d}\mu_1 \tag{2} \\ & \leq C_2 \int_{\mathbb{R}_1} \int_{\mathbb{R}_2} \left|\nabla_y f(x,y)\right|^2 \mathrm{d}\mu_2\mathrm{d}\mu_1 + C_1\int_{\mathbb{R}_1} \left| \nabla_x g(x)\right|^2 \mathrm{d}\mu_1 \tag{3} \end{align}

where we used

1. The triangle inequality $|a + b| \leq |a| + |b|$ and the arithmetric-mean-geometric-mean inequality which implies $2|a||b| \leq |a|^2 + |b|^2$.
2. The definition of $g$ and the fact that $\mu_2$ is a probability measure (so has total mass 1).
3. In both terms we use Poincare inequality. In the first we use it on the function $f(x,y)$ in the $y$ variable. In the second we use it on the function $g(x)$ in the $x$ variable, noting that $\int g \mathrm{d}\mu_1= 0$.

To finish the proof, we note that by Minkowski's inequality

$$\left(\int_{\mathbb{R}_1} |\nabla g|^2\mathrm{d}\mu_1\right)^\frac12 \leq \int_{\mathbb{R}_2} \sqrt{ \int_{\mathbb{R}_1} |\nabla_x f(x,y)|^2 \mathrm{d}\mu_1}\mathrm{d}\mu_2$$

and by Holder's inequality on a probability space

$$\int_{\mathbb{R}_2} \sqrt{ \int_{\mathbb{R}_1} |\nabla_x f(x,y)|^2 \mathrm{d}\mu_1}\mathrm{d}\mu_2 \leq \left(\int_{\mathbb{R}_2} \int_{\mathbb{R}_1} |\nabla_x f(x,y)|^2 \mathrm{d}\mu_1 \mathrm{d}\mu_2 \right)^\frac12$$

which leads to our final conclusion that

$$\iint |f(x,y)|^2 \mathrm{d}(\mu_1\otimes\mu_2) \leq \max(C_1,C_2) \iint |\nabla f(x,y)|^2 \mathrm{d}(\mu_1\otimes \mu_2)$$

In the $L^2$ case using the properties of the inner product we can get rid of a factor of 2 if we replace step (1) by

\begin{align} \iint |f(x,y)|^2 \mathrm{d}(\mu_1\otimes \mu_2) &= \iint \left( f(x,y) - g(x) + g(x) \right)^2 \mathrm{d}(\mu_1 \otimes \mu_2) \\ &= \iint |f - g|^2 + |g|^2 + 2 g(f-g) \mathrm{d}(\mu_1\otimes \mu_2) \end{align}

Now, observe that $\int_{\mathbb{R}_2} f(x,y)-g(x) \mathrm{d}\mu_2 = 0$ by definition, and hence also $\int_{\mathbb{R}_2} g(x)\cdot (f(x,y) - g(x)) \mathrm{d}\mu_2$. This implies that $$\iint |f(x,y)|^2 \mathrm{d}(\mu_1\otimes \mu_2) = \iint |f - g|^2 + |g|^2 \mathrm{d}(\mu_1\otimes \mu_2)$$ which is one factor of 2 better than (1).

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Thanks a lot @WIllie! ( at lease I was close ...I think that with the WLOG assumption, it's pretty easy, but thanks a lot ! it's a really nice property I wanted to know how to prove) –  joshua Jul 25 '12 at 10:23
Final question: @Willie: When you did (3) - why did you ommit the multiplication by 2? it seems like it disappeard... (And if it should indeed stay there also in (3), then we get that the constant of the tensor will be $2 max(C_1 , C_2 )$ which is bigger than the constant $max(C_1 , C_2)$ we should get ). Can you see any way to fix this? Thanks ! –  joshua Jul 26 '12 at 15:12
Ah, my $C_1$ is (probably) not your $C_1$ (I hadn't noticed that you defined a $C_1$: I just saw the $\lambda^2$ in your inequality and stopped reading). So yes, $C_1/2$ would be the appropriate constant for Poincare on $\mathbb{R}_1$. –  Willie Wong Jul 26 '12 at 15:16
The factor of two is removable: observe that if $h(x)$ is a function and $h_0 = \int_{\Omega} h(x) \mathrm{d}\mu$, where $\mu$ is a probability measure, we have that $$\int (h(x) - h_0) h_0 \mathrm{d}\mu = 0$$ Apply this to the $y$ integral of the cross term $f(x,y)g(x)$ when you FOIL. (This works in the $L^2$ case because the square can be interpreted as an inner product. For $L^p$ when $p \neq 2$ it doesn't work as well.) –  Willie Wong Jul 26 '12 at 15:19
Hi again @Willie-I still can't figure it out- (how did you remove the 2 constant) - When you're saying I should apply this to the $y$ integral of the cross term, you mean the integral after the first inequality in your first answer? I still can't see it... Can you please explain it again (the equality you wrote for the integral if completely understandble...But I can't see where I should apply it and what does it give me ) Thanks (a lot! ) –  joshua Jul 26 '12 at 15:32
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