Apply the <Poincaré inequality in probability theory> to $g=e^{\lambda f/2}$. Since $\lVert\nabla f\rVert\leq1$,
$$
F(\lambda)-F(\lambda/2)^2
\leq C_P(X)\frac{\lambda^2}{4}F(\lambda).
$$
Hence
$$
F(\lambda)\leq
\left(1-\frac{\lambda^2C_P(X)}4\right)^{-1}
F(\lambda/2)^2.
$$
Iterating this estimate $m$ times yields
$$
F(\lambda)\leq
\prod_{k=0}^{m-1}
\left(1-\frac{\lambda^2C_P(X)}{4^{k+1}}\right)^{-2^k}
F(\lambda/2^m)^{2^m},
$$
valid under the stated bound on $\lambda$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 208 / 2 / a / Solution

A probability distribution $\mu$ has Poincaré constant $C_P$ when every sufficiently regular $f$ satisfies
$$
\operatorname{Var}_\mu f\leq C_P\int\lVert\nabla f\rVert^2d\mu.
$$


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 Incoming links: Poincaré inequality in probability theory