Let $X=l_2$ with usual norm $||.||$. We define a subspace of X as $D= (B_{l_2} \cup B),$ where $B = \{ (x_n) \in l_2 : \sum_{n=1}^\infty \frac{n}{2}x_n^2 \leq 1\}$ 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 $$x, x_n \in S_{(l_2, \Vert \cdot \Vert)} \text{with} \displaystyle \lim_{n \to \infty}\Vert x+x_n \Vert =2.------(1)$$ I must show that $(x_n)$ has a weakly convergent subsequence. 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. Please help me. In addition, it has given that this norm is not LUR at $e_1$ (i.e., there exists sequence $(x_n)$ with $\Vert e_1 +x_n \Vert \to 2$; but $(x_n)$ does not converge to $e_1$). I assume the sequence $(x_n)$ to be $e_1 + e_n$. But I do not understand how I can find out the norm here. Thank you.