For convex $f$, the <subgradient inequality> gives
$$
f(x)-f(x_1,\ldots,z,\ldots,x_n)
\leq\partial_if(x)(x_i-z).
$$
Taking the positive supremum over $z\in[0,1]$ and summing squares shows
$$
V^+(x)\leq\sum_i(\partial_if(x))^2\leq1.
$$
The <modified logarithmic Sobolev inequality> from part a with $v=1$ therefore gives
$$
\mathbb P(Z-\mathbb EZ\geq t)\leq e^{-t^2/2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 4 / c / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-208/4/c/i/solution