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

Write $X=\mu +\sigma Z$ with $Z$ standard normal and set $g(z)=f(\mu +\sigma z)$. The Gaussian Poincaré inequality and the chain rule giveHence $N(\mu ,{\sigma }^2)$ satisfies a $\sigma$-Poincaré inequality in the question's convention, where the constant is squared on the right-hand side.

Let $X_1,\dots ,X_N$ be independent with laws $p_1,\dots ,p_N$. We prove the claim by induction. Split the law of total variance at the last coordinate:The first term is at most $c_N^2\mathbb{E}|{\partial }_{N}f(X){|}^2$. Apply the induction hypothesis to $g(x_1,\dots ,x_{N-1})={\mathbb{E}}_{X_N}f(x_1,\dots ,x_{N-1},X_N)$. Differentiation under the expectation and Jensen inequality giveWriting $c={\max }_i{c}_i$ and combining the terms yieldsThis is the Tensorization of a Poincaré inequality.

Let $g:\mathbb{R}\to \mathbb{R}$ be continuously differentiable. Applying the Poincaré inequality for $p$ to $g\circ \varphi$ and using the chain rule givesThus the Pushforward of a Poincaré inequality by a Lipschitz function gives a $cL$-Poincaré inequality for $q$.

The sharp Poincaré inequality for the uniform distribution on an interval isTensorizing these $n$ identical one-dimensional inequalities givesfor the uniform distribution on $[0,1{]}^n$. Therefore one may take $c=1/\pi$; this value is sharp, as functions depending only on one coordinate attain the one-dimensional constant.