Strictly convex renorming of Banach space Banach space $X$ (or its norm) is said to be strictly convex if its unit sphere $S_X$ does not contain any nontrivial line segment. There is also stronger notion of uniform convexivity. We say that space $X$ is uniformly convex if for any $\varepsilon > 0$ there exists $\delta > 0$ such that for any $x,y \in S_X$
$$ \|x - y\| \geq \varepsilon \implies \Bigl \|\frac{x+y}{2} \Bigr \| \leq 1 - \delta. $$
It follows from parallelogram identity that every inner product space is uniformly convex. Moreover, it is known that $L_p$ spaces are uniformly convex for $1 < p < \infty$. On the other side there are classical spaces, such as $C[0,1], L_1, c_0, c$, which are not strictly convex (when equiped with their standard norms). Some of these spaces admit equivalent and strictly convex renorming. For example in $c_0$ space there is norm
$$ \|x\|_{sc} = \sup_{n \in \mathbb{N}}|x_n| + \Bigl( \sum_{n=1}^{\infty} \frac{1}{2^n} |x_n|^2 \Bigr)^{\frac{1}{2}}$$
which is strictly convex and equivalent to the classical norm $\|x\| = \sup_{n \in \mathbb{N}} |x_n|$.
$\textbf{My questions}:$ Does every Banach space admit strictly convex (not necessarily equivalent) renorming? If not, is there some class of B spaces, for which such renorming exists? What about uniformly convex renorming?
 A: 
Any separable Banach space $X$ has an equivalent strictly convex norm.

Let $(x_n)$ be a dense set in $X$. Use H-B 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, 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 $\Gamma = \mathbb N$ and $c_0(\mathbb N) = \{  (x_n) \colon x_n \to 0\}.$
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 .
