Let $X_1,\ldots,X_N$ be independent with laws $p_1,\ldots,p_N$. We prove the claim by induction. Split the <law of total variance> at the last coordinate:
$$
\operatorname{Var}(f(X))
=\mathbb E\!\left[\operatorname{Var}(f(X)\mid X_1,\ldots,X_{N-1})\right]
+\operatorname{Var}\!\left(\mathbb E[f(X)\mid X_1,\ldots,X_{N-1}]\right).
$$
The first term is at most $c_N^2\mathbb E|\partial_Nf(X)|^2$. Apply the induction hypothesis to $g(x_1,\ldots,x_{N-1})=\mathbb E_{X_N}f(x_1,\ldots,x_{N-1},X_N)$. Differentiation under the expectation and <Jensen inequality> give
$$
|\partial_i g|^2
=|\mathbb E_{X_N}\partial_i f|^2
\leq\mathbb E_{X_N}|\partial_i f|^2.
$$
Writing $c=\max_i c_i$ and combining the terms yields
$$
\operatorname{Var}(f(X))
\leq c^2\mathbb E\sum_{i=1}^N|\partial_i f(X)|^2
=c^2\mathbb E\lVert\nabla f(X)\rVert^2.
$$
This is the <Tensorization of a Poincaré inequality>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 208 / 2 / b / Solution

{c}

If each probability measure $\mu_i$ has Poincaré constant $C_i$, then the product measure $\bigotimes_i\mu_i$ has Poincaré constant at most $\max_iC_i$. This follows by iterating the <law of total variance> and applying <Jensen inequality> to derivatives of conditional expectations.

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 Incoming links: Tensorization of a Poincaré inequality