Introduction

We say that function \(f\) is exp-concave if \(e^{-f}\) is concave. We are asking if \(f = \sum_i \alpha_i f_i\) is exp-concave if each of the \(f_i\) are.

One way to recognize exp-concave functions is to check the second-order condition: for each \(x\) \[\nabla f(x) \nabla f(x)^\intercal~\preceq~ \nabla^2 f(x) .\] So in our case we would have to check \[\left(\sum_i \alpha_i \nabla f_i(x)\right) \left(\sum_i \alpha_i \nabla f_i(x)\right)^\intercal ~\preceq~ \sum_i \alpha_i \nabla^2 f_i(x) .\] Invoking exp-concavity of each \(f_i\), it suffices to check whether \[\left(\sum_i \alpha_i \nabla f_i(x)\right) \left(\sum_i \alpha_i \nabla f_i(x)\right)^\intercal ~\preceq~ \sum_i \alpha_i \nabla f_i(x) \nabla f_i(x)^\intercal .\] Picking an arbitrary vector \(v\), it is then enough to check \[\left( \sum_i \alpha_i v^\intercal\nabla f_i(x) \right)^2 ~\le~ \sum_i \alpha_i \left(v^\intercal\nabla f_i(x)\right)^2 .\] This is true (by Jensen) if the \(\alpha_i\) are probabilities.