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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.

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1 Answer 1

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

See for instance: Lipschitz continuity implies differentiability almost everywhere.

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