Convexity of an equivalent norm

Question

Let $X=l_2$ with usual norm $\|\cdot\|_2$. We define a subspace of $X$ as $D=conv (B_{l_2} \cup B),$ where $B = \{ (x_n) \in l_2 : \sum_{n=1}^\infty \frac{n}{2} x_n^2 \leq 1\}$, conv is the convex combination and $B_{l_2}$ is the closed unit ball in $l_2$ with usual norm $\Vert \cdot \Vert_2$. We define Minkowski's functional on set $D$ by $$ \mu_D(x)=\inf \{ t > 0 : x \in tD \}.$$ Then $\Vert \cdot \Vert = \mu_D( \cdot )$ defines an equivalent norm on $(l_2, \Vert \cdot \Vert_2)$, where $\Vert \cdot \Vert_2$ is the usual norm on $l_2$. The norm $\Vert \cdot \Vert$ on $l_2$ is not rotund. Now, I take $$x, x_n \in S_{(l_2, \Vert \cdot \Vert)} \text{ with } \lim_{n \to \infty}\Vert x+x_n \Vert =2. \tag 1$$ I must show that $(x_n)$ has a weakly convergent subsequence converging to an elemnt in $A_0(x)$, $A_0(x)=\{y \in S_X : \Vert \frac{x+y}{2} \Vert =1 \}$. For Minkowski's functional to give a semi-norm, $D$ has to be an absorbing set. However, I do not understand how I can form the seminorm or proceed to prove that $(x_n)$ satisfying equation $(1)$ has a weakly convergent subsequence in $A_0(x)$. Please help me. Thank you.