Convexity of $2$-Normed Space

Question

2-Normed Space Uniformly Convex

$2$-Normed Space $(L, \Vert \cdot , \cdot \Vert)$ is said uniformly convex if for every $\epsilon >0$, exist $\delta(\epsilon) >0$ such that $\Vert x,z \Vert = \Vert y,z \Vert = 1$, $\Vert x-y, z \Vert \ge \epsilon, x \not = y$ and $z \not \in V(x,y)$ then $$\Big \Vert \frac{x+y}{2}, z \Big \Vert \le (1-\delta(\epsilon)).$$

2-Normed Space Strictly Convex

Let ($X, \Vert \cdot , \cdot \Vert $) be a $2$-normed space. Then $(X, \Vert \cdot, \cdot \Vert)$ is said to be strictly convex if for every $x,y,z\in X$ satisfying $\Vert x, z \Vert = \Vert y,z \Vert = 1$, $x\not=y$, $z \not \in V(x,y)$, \emph{we have} $\Vert \frac{x+y}{2}, z \Vert < \frac{1}{2}(\Vert x,z \Vert + \Vert y,z \Vert ) <1$.

Theorem: Uniform Convex Implies Strictly Convex

Given $(X, \Vert \cdot, \cdot \Vert)$ uniformly convex. Then for every $\epsilon >0$ exist $\delta(\epsilon) >0$ such that $\Vert x,z \Vert = \Vert y,z \Vert = 1$, $\Vert x-y, z \Vert \ge \epsilon, x \not = y$ dan $z \not \in V(x,y)$. We obtain $$\Big \Vert \frac{x+y}{2}, z \Big \Vert \le (1-\delta(\epsilon)) < 1.$$

Question

What about the converse? Is strictly convex implies uniform convex?