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On $\mathbb{R}^{n}$, we have the norm $\| \cdot \|_{\infty}: \mathbb{R}^{n} \rightarrow \mathbb{R}^{>0}$ which sends

$x \mapsto \max{(|x_{i}|)_{1 \leq i \leq n}}$

My calculus is a bit shaky so I apologize if this is a simple question, but I was wondering just in the case for $\mathbb{R}^{2}$ at which points is the function differentiable.

My guess on this issue was to consider the $p$ norm where at $p=1$, the differentiable everywhere except on the axes, and for $p>1$, we have the the form is differentiable everywhere except the origin. The reasoning for this can be found here https://math.stackexchange.com/a/144477/135520

My guess is that this norm is differentiable everywhere except the origin.

I was wondering if this guess was correct, and if someone could shed some light on this.

Thanks!

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user135520
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    no, differentiable except when there are at least two of the $x_i$ with the same absolute value, which is also the maximum. In $R^2,$ differentiable except when $|x|= |y|$ – Will Jagy Oct 09 '15 at 19:41

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As pointed by Will, $\| \cdot \|_{\infty}$ is differentiable at every point $x=(x_1,...,x_n)$ except where there are at least two $x_i$ with the same absolute values $|x_i|$.

Proof:

We first show it's not differentiable if two absolute values conicide and are maximal. Assume w.l.o.g that $|x_1|=|x_2|=\| x\|_{\infty}$. Then if $\| \cdot \|_{\infty}$ was differentiable at $x$, then it would stay differentiable after fixing all coordinates but the first one.

i.e, the function $x \to \|(x,x_2,x_3,...,x_n) \|_{\infty}$ would be differentiable at $x_1$. This follows by the chain rule, since this funciton is a composition of $x \to (x,x_2,x_3,...,x_n)$ and $\| \cdot \|_{\infty}$.

However, $f(x)=\|(x,x_2,x_3,...,x_n) \|_{\infty}=\max\{|x|,|x_2|\}$. This is not differentiable where $|x_1|=|x_2|$. (For instance if $x_1>0$ then $f(x)=\max\{x,|x_2|\}$ in a small enough neighbourhood of $x_1$. But then it equals $x$ for $x > |x_2|$, and $|x_2|$ for $x < x_2$. So the derivatives from the left and right are $0,1$, and are different.

At any other point, suppose $|x_1|=\|x \|_{\infty}$. Then since $|x_1|>|x_i|$ for any $i >1$ than in a small neighbourhood $f((x_1,...x_n))=|x_1|$ which is differntiable when $|x_1| \neq 0$.

Asaf Shachar
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