Equivalent definition of convexity with $\theta \in (0,1)$ instead of $[0,1]$.

Question

We defined a functional $\Phi: V \to \mathbb{R}$ ($V$ is a Banach space) to be convex if $$ \tag{1} \Phi((1 - \theta) v + \theta w) \le (1 - \theta) \Phi(v) + \theta \Phi(w) \quad \forall v,w \in V \ \forall \theta \in [0,1]. $$ Our lecturer casually added that instead it suffices that $(1)$ holds for all $\theta \in (0,1)$, which I will call $(2)$.

Obviously, $(1)$ implies $(2)$, but I am not sure if they are indeed equivalent. Does equivalence hold if $\Phi$ is continuous?

The setting is the following: $\Phi$ will always be the potential of an operator $A: V \to V^*$, meaning $\Phi$ will be Gateaux-differentiable and $$ \langle A u, v \rangle = \lim_{h \to 0} \frac{\Phi(u + h v) - \Phi(u)}{h} \quad \forall u,v \in V $$ We defined a map $F: X \to Y$, where $X$ and $Y$ are normed spaces to be Gateaux differentiable if $$y \mapsto \lim_{h \to 0} \frac{\Phi(x + h y) - \Phi(x)}{h}$$ is well defined (as in the limit exists) linear and bounded (therefore continuous, right?).

Answer 1

Equivalence is always true because the inequality becomes an equality when $\theta=0$ or $\theta =1$.

Answer 2

When $\theta=1$, (1) gives $$ \Phi(w) \le \Phi(w) $$ which is trivially true, and similarly for $\theta=0$ so the edge cases don't matter really