Yes, my question is really a dumb one. I forgot the condition in the definition of strict convexity.
Recall the definition of strict convexity:
If $f,g\in E^{*}$ such that $f\neq g$ and $\|f\|=\|g\|=1$, then for $0<\lambda<1$, we have $\|\lambda f+(1-\lambda)g\|< 1.$
Thus, this proof actually begins with choosing $f,g\in E^{*}$ such that $f\neq g$ and $\|f\|=\|g\|=1$, not assuming $\|f\|=\|g\|=1$ in the middle. Then, everything follows smoothly:
Suppose $F(x)$ contains at least two points. Let $f,g\in F(x)$ be such that $f\neq g$ and $\|f\|=\|g\|=1$, and let $0<\lambda<1$ be an arbitrary real number.
Then, since $F(x)$ is convex, we have $$\lambda f+(1-\lambda)g\in F(x).$$
Thus, we have $$\|f\|=\|g\|=\|x\|=\|\lambda f+(1-\lambda)g\|=1,$$ which contradicts the hypothesis that $E^{*}$ is strictly convex.
Thus, $F(x)$ contains at most $1$ point, but $F(x)$ is not empty, so $F(x)$ contains only one point.
Any separable Banach space $X$ has an equivalent strictly convex norm.
Let $(x_n)$ be a dense set in $X$. Use the Hahn-Banach theorem to find $f_n \in X^*$ with norm one such that $f_n(x) =\|x_n\|$. Let $T \colon X \to \ell_2$ be given by
$$X \ni x \mapsto ( \tfrac {1}{2^n} f_n(x))_{n=1}^\infty \in \ell_2.$$
Then $T$ is bounded and injective. Show that $|x|:= \|x\|+\|Tx\|$ is an equivalent strictly convex norm. Hint: Use the fact that $(X,|\cdot|)$ is strict convexity iff $|x+y|=|x|+|y|$ implies that $x=ay$ for some $a>0$.
For $\Gamma$ uncountable, $\ell_{\infty}(\Gamma)$ admits no strictly convex norm.
See Chapter II,7 in [1]
There exist reflexive, non separable Banach spaces that admit no uniformly convex norm
See [2] and [3]
$\textbf{Edit:}$
As noted by OP, although $\ell_\infty(\mathbb N)$ is not separable, it admits a strictly convex renorm. Day [4] showed that
If there exists an injective bounded linear operator
$$T \colon X \to c_0(\Gamma)$$
for some set $\Gamma$, then $X$ admits a strictly convex renorm. Here, $c_0(\Gamma)$ is the set of all bounded real functions $f$ such that for every $ε>0$ the set $(|f|>ε)$ is finite.
$\ell_\infty(\mathbb N)$ falls into that category with $\Gamma = \mathbb N,$ $c_0(\mathbb N) = \{ (x_n) \colon x_n \to 0\}$ and
$$T((x_n)_{n=1}^\infty)=(\tfrac 1n x_n)_{n=1}^\infty$$
If you are willing to drop uniform convexity for locally uniform convexity, i.e.,
$$ \| x_n\| = \|x\|=1 \text { and } \|x_n +x\| \to 2 ~ \text{ imply } ~ \|x_n-x\| \to 0$$
then things are much nicer:
Every separable Banach space admits an equivalent norm that is locally uniformly convex.
See Theorem 2.6 (i) in [1].
This does not extend to the non separable case, as shown by the example of $\ell_\infty(\Gamma)$ with $\Gamma$ uncountable.
In fact, Troyonski [5] showed that
If $X$ is reflexive, then both $X$ and $X^*$ admit an equivalent locally uniformly convex (and Frechet differentiable) norm.
[1] Smoothness and Renormings in Banach Spaces
[2] Reflexive Banach spaces not isomorphic to uniformly convex spaces
[3] Banach spaces which can be given an equivalent uniformly convex norm
[4] Day, M. M. (1955). Strict Convexity and Smoothness of Normed Spaces. Transactions of the American Mathematical Society, 78(2), 516–528.
[5] S . L . T ROYANSKI , ‘On locally uniformly convex and dif ferentiable norms in certain non-separable Banach spaces’, Studia Math . 37 (1971) 173 – 180 .
Best Answer
The duality map of a Banach space $E$ is the map that maps each $x \in E$ to the set $J_x=\{f \in E' ; f(x)=||x||^2, ||f||=||x||\}$. As you pointed out, $E$ is smooth iff for each $x\in X, J_x$ is a singleton, say $\{J(x)\}$. In that case, we identify $J_x$ with $J(x)$. Notice that if $E$ is a Hilbert space then $J$ is in fact the identify map (under that identification), and is thus linear. In fact, Hilbert spaces are the only ones with that property, that is, if the duality map of a Banach space is linear, then it is a Hilbert space.