Subtle difference between convex and strictly convex, why?

Question

• A function is convex if

  $f(\theta x + (1-\theta) y) \leq \theta f(x) + (1-\theta) f(y)$, $\theta \in [0,1]$

• A function is strictly convex if

  $f(\theta x + (1-\theta) y) < \theta f(x) + (1-\theta) f(y)$, $\theta \in (0,1)$

What is the big deal with definition of $\theta$ between these two definitions?

Why must $\theta$ not take on the boundary values of $0$ and $1$ in the definition of strict convexity?

Answer 1

If $\theta = 0$ in the second definition, then the condition becomes $f(y) < f(y)$, which is never true. Similarly, if $\theta = 1$, we get $f(x) < f(x)$.

Answer 2

The $[0,1]$ or $(0,1)$ is not that important. You can write $(0,1)$ instead of $[0,1]$ in the first definition.

What is important is that the $\leq$ sign changes to $<$.

Of course, the $<$ inequality cannot hold for $\theta=0$, since the inequality then becomes $f(y)<f(y)$, which is not true for any function, so you cannot write $[0,1]$ instead of $(0,1)$ in the second definition.

Answer 3

If the strict equation was to hold for $\theta = 0$, it would require $f(y) < f(y)$. That is why we want to exclude the boundary values from the definition.

It didn't make much difference with the non-strict version, as $\theta = 0$ gave an identity of $f(y) \leq f(y)$.