Let $E$ be a normed vector space. One says that the norm $\Vert \Vert$ is strictly convex if $\Vert tx+(1-t)y \Vert < 1, \forall x,y \in E \; \text{with} \; x \neq y, \Vert x\Vert = \Vert y \Vert = 1, \; \forall t \in (0,1)$. One says that a function $\phi : E \to (-\infty , +\infty]$ is strictly convex if $\phi(tx+(1-t)y)<t\phi(x) + (1-t)\phi(y) \; \forall x,y\in E \; \text{with} x\neq y , \; \forall t \in (0,1)$.
How can I show that the norm $\Vert \Vert $ is strictly convex iff the function $\phi(x) = \Vert x \Vert ^p$ is strictly convex for $1<p<\infty$?
(1) If $\phi(x) = \|x\|^p$ is strictly convex, for any $\|x\|=1, \|y\|=1, x \neq y$, we have \begin{equation} \phi(tx + (1-t)y) = \|tx+(1-t)y\|^p < t \phi(x)+ (1-t)\phi(y) = 1. \end{equation}
(2) If norm $\|\cdot\|$ is strictly convex, let $$\bar{x} = \frac{x}{\|x\|}, \quad \bar{y} = \frac{y}{\|y\|},$$ we have \begin{equation} \|t x + (1-t)y\|^p = \|(t\|x\|) \bar{x} + ((1-t)\|y\|) \bar{y}\|^p < (t\|x\| + (1-t)\|y\|)^p. \end{equation} Due to function $f(u)=u^p, 1 < p < \infty$ is convex on domain $\{u \mid u \ge 0\}$. Thus \begin{equation} \|t x + (1-t)y\|^p < t\|x\|^p + (1-t)\|y\|^p. \end{equation}