# Is the maximum of strictly convex functions also strictly convex?

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.

• 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 at 2:16

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.