Are 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 $x$.
My question is:

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

This statement is indeed true in the specific case, when $V$ is a Hilbert space.
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 $v \in V \setminus \{0\}$. 

No this is not always true. Take $(V, \Vert \cdot \Vert)=(\mathbb R^2, \sup (\vert x \vert, \vert y \vert))$. The norm is not Fréchet differentiable when $x = \pm y$.
