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, $J(x)=\{f \in S\_{X^\*} : f(x)=1\}, x \in S\_X$.



Smoothness of a Hilbert space under an equivalent norm
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.




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*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 $\|f\|=1=\|g\|$.


