# Is any Banach norm in $\mathbb{R}^n$ almost everywhere 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 $$x$$.

If $$(V, \|\cdot\|_V)$$ is a Banach space, then define $$nnd(V, \|\cdot\|_V)$$ as the set of all points of $$V$$, where $$f: v \mapsto \|v\|_V$$ is not Fréchet differentiable.

Suppose $$\|\cdot\|$$ is a Banach norm on $$\mathbb{R}^n$$. Is it true, that $$\mu(nnd(\mathbb{R}^n, \|\cdot\|)) = 0$$?

Here $$\mu$$ stands for Lebesgue measure.

I know, that if the norm in question is a Hilbert norm, then our statement will be true (and moreover any Hilbert norm is Fréchet differentiable everywhere except $$0$$).

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, this stronger property is not always true in the general statement of my problem. For example:

$$nnd(\mathbb{R}^2, \|(x,y)\| := \sqrt{ \max(x^2 + 2y^2, \ 2x^2 + y^2 )}) = \{(x,y)| |x| = |y|\}$$

It still however satisfies the condition I am asking about.

Yes. Indeed, a norm on $$\mathbb{R}^n$$ is a locally Lipschitz function (wrt eg the Euclidean norm), so is differentiable almost everywhere.