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?
Thanks in advance ! 
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?
 A: 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


*

*The triangle inequality $|a + b| \leq |a| + |b|$ and the arithmetric-mean-geometric-mean inequality which implies $2|a||b| \leq |a|^2 + |b|^2$. 

*The definition of $g$ and the fact that $\mu_2$ is a probability measure (so has total mass 1). 

*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).
