We consider the normed vector space $(\mathbb{R}^n,\Vert\cdot\Vert)$. Is the map $\Vert\cdot\Vert$ differentiable even if it is not induced by a scalar product?
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Truth to be told, it's not differentiable (in $0$) when it's induced by a scalar product either. – Sep 10 '18 at 11:57
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A norm which is differentiable off the origin is called a "smooth norm". – Theo Bendit Sep 10 '18 at 12:09
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Sometimes yes, sometimes no. For example, consider the $p$-norms: $$ \|(x_1,\dots,x_n)\|_p = \left(\sum_{k=1}^n |x_k|^p\right)^{1/p} . $$ This is differentiable (except at the origin) if $p>1$, but not if $p=1$.
GEdgar
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1It could also be remarked that even the 1-norm is differentiable almost everywhere -- namely at points where all the $x_i$s are nonzero. – hmakholm left over Monica Sep 10 '18 at 12:02
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1And, more generally, any convex function is differentiable almost everywhere? – GEdgar Sep 10 '18 at 12:04
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A norm on $\mathbb{R}^n$ is never differentiable at the origin. Let $x_0\in \mathbb{R}^n\setminus \left\{0\right\}$. Then $$\lim_{t\to 0}\frac{\|(0+tx_0)\|-\|0\|}{t} =\lim_{t\to 0}\frac{\|t\|}{t}\|x_0\|$$ By the norm axiom $x_0\neq 0 \Rightarrow \|x_0\|\neq 0$, and so the limit does not exist, as $$\lim_{t\to 0^+}\frac{\|t\|}{t}=1\neq -1=\lim_{t\to 0^-}\frac{\|t\|}{t} $$
On the other hand, the triangle inequality $$\left|\|x\|-\|y\|\right|\leq \|x-y\| $$ tells us that any norm in $\mathbb{R}^n$ is Lipschitz with Lipschitz constant $L=1$. Thus by Rademacher's theorem, it is differentiable almost everywhere in $\mathbb{R}^n$.
Lorenzo Q
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1I like the reference to Rademacher's theorem but there is a subtlety. The theorem refers to the Euclidean norm on $\mathbb R^n$, but this answer only shows that $\lVert\cdot\rVert$ is 1-Lipschitz with respect to itself. To conclude that $\lVert \cdot \rVert$ is Lipschitz with respect to the Euclidean norm ($\ell^2$), we also need the inequality $$\lVert x\rVert \le C \lVert x\rVert_{\ell^2}.$$This inequality holds because all norms are equivalent on $\mathbb R^n$. – Giuseppe Negro Sep 10 '18 at 12:10
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Absolutely. I noticed that I was working with different norms but since we are working in $\mathbb{R}^n$ I did not worry about it :) but for a reader who is not familiar with this fact this is an important remark. – Lorenzo Q Sep 10 '18 at 12:17
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A simple example of a norm on $\Bbb R^2$ which is non-differentiable at points other than the origin is given by $$||(x,y)||=\max(|x|,|y|).$$
If $h>0$ then $||(1+h,1)||=1+h$, while if $-2<h<0$ then $||(1+h,1)||=1$. Hence the function $h\mapsto||(1+h,1)||$ is not differentiable at $h=0$, so $||.||$ is not differentiable at $(1,1)$.
David C. Ullrich
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