# 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 $$x = \pm y$$.