Solution (source code)

= Solution

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 $\lambda$.