Confusion over the definition of strongly convex

Question

Recall that

Let $D$ be a convex set, then a function $f: D \subseteq \mathbb{R}^n \to \mathbb{R}$ is convex if $f(\sigma x + (1-\sigma)y) \leq \sigma f(x) + (1-\sigma)f(y)$ for all $x,y \in D$, $\sigma \in [0,1]$ and strictly convex if holds strictly whenever $x \neq y$ and $\sigma \in (0,1)$

I am having trouble with the definition of strongly convex. Some people online defines strongly convex as (Fundamentals of Resource Allocation in Wireless Networks: Theory and Algorithms By Slawomir Stanczak, Marcin Wiczanowski, Holger Boche)

Notice that $\mu \in (0,1)$

Others provides the same definition but includes $0$ and $1$ i.e. $\mu \in [0,1]$ for example https://arxiv.org/pdf/1507.07144.pdf

Which definition is correct?

Answer 1

The definitions are equivalent. Using the notation of the second definition, note that $$f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)-\frac{\sigma}{2}\lambda(1-\lambda)\|x-y\|^2$$ is true no matter what $f$ is for $\lambda=0$ or $\lambda=1$: for $\lambda=0$ both sides just become $f(y)$, and for $\lambda=1$ both sides just become $f(x)$. So the statement is true for all $\lambda\in[0,1]$ iff it is true for all $\lambda\in(0,1)$.