Write $z_i(x)=\langle a_i,x\rangle+b_i$. The <log-sum-exp function> is convex and composition with the <affine functions> $z_i$ preserves convexity, so $f_\beta$ is <convex>. Directly, its <Hessian matrix> will also be shown <positive semidefinite matrix>[positive semidefinite] in part d.

Let $M=f(x)=\max_i z_i(x)$. Factoring $e^{\beta M}$ out of the sum gives
$$
f_\beta(x)
=M+\frac1\beta\log\sum_i e^{\beta(z_i(x)-M)}.
$$
At least one term in the sum is $1$, while every term is at most $1$. Therefore
$$
1\leq\sum_i e^{\beta(z_i-M)}\leq m,
$$
and hence
$$
\boxed{f(x)\leq f_\beta(x)\leq f(x)+\frac{\log m}{\beta}.}
$$
Thus $f_\beta$ is a uniform <smooth maximum> of the affine pieces of $f$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 339 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-339/1/c/solution