Let us take 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 $(l_2, \Vert \cdot \Vert)$ is not strictly convex. By reflexivity, the dual norm is not smooth.
I am trying to find whether $X=(l_2, \Vert \cdot \Vert)$ is smooth Banach space.
My attempt: I assume the space is not smooth. For this, if we take $e_1=(1, 0, 0,…) \in S_{(l_2, \Vert \cdot \Vert)}$, then for $f=(1, 1/4,0,0,…)$ and $g=(1,1/2,0,0,…) \in X^*$, then $f(x)=1=g(x)$. However, I need help to conclude whether $f$ and $g$ belong to the unit sphere of $S_{X^*}$. $\vert f(x) \vert \leq \Vert f \Vert \Vert x \Vert \leq \Vert f \Vert$ and similarly $\vert g(x) \vert \leq \Vert g \Vert$. Then $\Vert f \Vert \geq 1$ and $\Vert g \Vert \geq 1$.
Another way to find the smoothness is to check whether the dual norm is strictly convex. Thank you in advance.
Note: A Banach space is smooth if $J(x)$ is singleton, $J(x)=\{f \in S_{X^*} : f(x)=1\}, x \in S_X$.
Best Answer
Let $z:=\frac12\,(\sqrt3,1,0,0,\dots)$. For $x=(x_1,x_2,\dots)$, let $$f(x):=2x_2,\quad g(x):=\tfrac12\,(\sqrt3\,x_1+x_2).$$ Then $f\ne g$, $\|z\|=1$, $f(z)=1=g(z)$, and $\|f\|=1=\|g\|$.
So, your space is not smooth.
Details: For all $x=(x_1,x_2,\dots)\in B_{\|\cdot\|}$ we have $$f(x)=2x_2\le1,\quad g(x)=\tfrac{\sqrt3}2\,x_1+\tfrac12\,x_2\le \sqrt{x^2+x_2^2}\le1,$$ so that $\|f\|\le1$ and $\|g\|\le1$. Since $\|z\|=1$ and $f(z)=1=g(z)$, we get $\|f\|=1=\|g\|$.