0

Let $\{ f_1,f_2,\ldots,f_m \}$ be a set of convex functions, where $f_i : C \subset \mathbb{R}^n \to \mathbb{R}$ and with $C$ a convex set. Then,

$$F(x) := \max \{ f_1(x),f_2(x), \ldots, f_m(x) \}$$ is a convex function. What happens if each $f_i$ is strictly convex? Is $F$ also strictly convex?

I believe that it is not true. If it is the case, a counterexample would be great.

Rodrigo de Azevedo
  • 1
The Student
  • 301
  • 1
    It is true when your family of convex functions is finite, but when taking the (pointwise) supremum of an infinite family, it may not be. For example, if you consider $f_n(x) = \frac{x^2 - 1}{n}$ over the interval $C = [-1, 1]$, the pointwise supremum is constant. – Theo Bendit May 24 '19 at 02:16

1 Answers1

2

Consider $C = \mathbf{R}^n$ for simplicity; let $k = \arg\max_{i \in [m]} f_i(\lambda x + (1 - \lambda) y)$. Then

$$ \begin{align} \max_{i=1, \dots, m} f_i(\lambda x + (1 -\lambda)y) &= f_k(\lambda x + (1 - \lambda)y) \overset{(*)}{<} \lambda f_k(x) + (1- \lambda) f_k(y) \\ &\leq \lambda \max_{i} f_i(x) + (1-\lambda) \max_{i} f_i(y), \end{align} $$ where $(*)$ used the fact that all the $f_i$'s are strictly convex, so $F$ is in fact strictly convex.

Pacciu
  • 6,251
VHarisop
  • 3,840