Consider the Hilbert space $l_2$ with an equivalent norm
$$\Vert x \Vert = \max \{2 \Vert x \Vert_1, \Vert x \Vert_2 \},$$
where $\Vert x \Vert_1 =( \sum_{n=2}^\infty x_n^2 )^{\frac{1}{2}}$ and $\Vert x \Vert_2 = ( \sum_{n=1}^\infty x_n^2 )^{\frac{1}{2}}$, for $(x_n)_{n \geq 1} \in l_2$.
The unit ball with respect to norm $\Vert \cdot \Vert$ is $B_{\Vert \cdot \Vert}= Y \cap B_{(l_2, \Vert \cdot \Vert_2)}$, where $Y=\{x \in l_2 : \sum_{n=2}^\infty x_n^2 \leq \frac{1}{4} \}$. Then $(l_2, \Vert \cdot \Vert)$ has property (H) ( here, property (H) means the weak and norm convergence coincides on the unit sphere). Now, I need to find whether the dual of the above norm satisfies property (H).
My approach: First, I took $f_n=e_1+e_n$ and $f=e_1$. Then for $x=e_1 $, $f(x)=f_n(x)=1$. Also, $f \in S_{X^*}$; but $f_n \not \in S_{X^*}$, so the example fails to verify property (H) on the dual space. Next, I tried to prove that the dual has property (H). I took $f_n$ and $f$ on the dual sphere such that $f_n \xrightarrow{w} f$. We need to show: $\Vert f_n - f \Vert \to 0$. But I cannot understand how I can visualise this on the dual space. Kindly guide me.