Poincaré Inequality - Product Of Measures

Question

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?

Answer 1

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

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

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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).