Are strictly convex Banach norms Fréchet differentiable? Suppose $(V, \|\cdot\|_V)$ and $(W, \|\cdot\|_W)$ are two Banach spaces and $f: V \to W$ is some function. We call a bounded linear operator $A \in B(V, W)$ Fréchet derivative of $f$ in $x \in V$ iff
$$\lim_{h \to 0} \frac{\|f(x + h) - f(x) - Ah\|_W}{\|h\|_V} = 0$$
We call a $f$ Fréchet differentiable in $x$ iff there exists a Fréchet derivative of $f$ in $x$.
We call a Banach space $(V, \|v\|)$ strictly convex, iff $\forall x \neq y \in V, \lambda \in (0,1)$ if $\|x\|=\|y\|=1$, then $x + \lambda(y-x) < 1$.
Hilbert spaces are a particular case of strictly convex spaces.
Proof:
If $\langle x, x\rangle = 1$ and $\langle y, y \rangle = 1$, then $\langle x + \lambda(y-x), x + \lambda(y-x)  \rangle = (1-\lambda)^2 + \lambda^2 + 2(1-\lambda)\lambda \langle x, y \rangle < (1-\lambda)^2 + \lambda^2 + 2(1-\lambda)\lambda = 1$
My question is:

Suppose $(V, \|\cdot\|_V)$ is a strictly convex Banach space. $f: V \to \mathbb{R}, v \mapsto \|v\|_V$. Is it true, that $f$ is Fréchet differentiable $\forall x \in V \setminus \{0\}$?

If $V$ is a Hilbert space, then it is true.
Proof:
One can manually check, that $h \mapsto \frac{h}{2\sqrt{x_0}}$ is a Fréchet derivative for $x \mapsto \sqrt{|x|}$ in $x_0 \neq 0$. One can also manually check, that $h \mapsto 2\langle v, h \rangle_V$ is a Fréchet derivative for $x \mapsto \langle x, x \rangle_V$ in all $v \in V$. And it is a well known fact, that the composition of Fréchet derivatives of two functions is a Fréchet derivative of their composition. Thus, as $\|v\|_V = \sqrt{\langle v, v \rangle_V}$, we have, that $h \mapsto \ \frac{\langle v, h \rangle_V}{\|v\|_V}$ is a Fréchet derivative of $\|v\|_V$ in all $v \in V \setminus \{0\}$. 
However condition of “strict convexity” can not be omitted here. $(\mathbb{R}^2, l_\infty)$ is a counterexample.

Here is a counterexample on $\mathbb R^2$:
$$
\|(x,y)\| := \sqrt{ \max(x^2 + 2y^2, \ 2x^2 + y^2 )}.
$$
It is the maximum of two striclty convex norms. 
It is strictly convex and not differentiable along for points with $|x|=|y|$.
