Let $X$ be an elliptic curve.

Then a divisor $E$ which is ample in the sense of the first definition is necessarily ample in the usual sense. Conversely, if it is ample in the usual sense and $\deg E \geq 2$ then it is ample in the sense of the first definition.

In fact, assume that $E$ ample in the sense of the first definition. Then $(2)$ implies that $\mathcal{O}_X(E) \neq \mathcal{O}_X$ and $(1)$ implies that $\dim H^0(X, E) \geq 2$, because a non-trivial divisor with $\dim H^0(X, E) =1 $ on an elliptic curve cannot be globally generated. Then by $(2)$ and Riemann-Roch it follows $$\deg E = \dim H^0(X, E) \geq 2,$$ hence $E$ is ample in the usual sense by [Hartshorne, *Algebraic Geometry*, Corollary 3.3 page 308].

Conversely, assume that $E$ is ample in the usual sense and $\deg E \geq 2$. This last condition implies that $|E|$ is base-point free on $X$, hence $E$ is globally generated and $(1)$ holds. Moreover, since $\omega_X = \mathcal{O}_X$, by Kodaira Vanishing Theorem and Serre duality one deduces $(2)$. So $E$ is ample in the sense of the first definition.