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I am looking at for which point(s) in $\mathbb{R}^n$ and $p\geq 1$ such that the map $\mathbb{R}^n\rightarrow\mathbb{R}$ given by

$x\mapsto ||x||_p=(|x_1|^p+\cdots+|x_n|^p)^{\frac{1}{p}}$

is differentiable.

Intuitively, it is not hard to "guess" the answer to be

  1. for all $p$, the map is not differentiable at origin.
  2. for $p=1$, the map is not differentiable everywhere, except for the axis.
  3. for $p>1$, the map is differentiable everywhere except at the origin.

I come up with these answers by checking continuity of partial derivatives, which is pretty tedious. However, I was even asked to verify these by definition.

I am wondering whether there is a very neat and insightful way to see this.

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up vote 5 down vote accepted

In fact you don't need to check all the partial derivatives of all these functions, but is still remains straightforward . . . we can take some shortcuts. Lets see how:

Guess 1.: It is correct. To see this, you can compose all the norms with, for instance, the curve $\gamma(t)=(t,0,\ldots,0)$. It will gives us $\|\gamma(t)\|_p=|t|^{{p}\frac{1}{p}}=|t|$, wich we know is not differentiable at zero.

Guess 2.: It is not precisely correct, but I understand what you want to mean. The function $x\mapsto\|x\|_1$ is differentiable everywhere except on the axis. To see this, take a fixed $x$ outside the axis and for a small neighborhood of it that doesn't intersect the axis,

$$\displaystyle\|x\|_1=\sum_{i=1}^n\pm x_i$$

where the signs $\pm$ are determined by the signs of the coordinates of $x$ and they don't change on a small neighborhood of $x$. Hence, $\|x\|_1$ is differentiable on $x$ outside the axis, because it is just a sum/subtraction of coordinates. And $x\mapsto\|x\|_1$ is not differentiable on the axis. To see this, take a point over the $i$-axis, $(0,\ldots,0,x_i,0,\ldots,0)$ and consider the path


(just put $t$ in another coordinate). Then $\|\gamma(t)\|_1=|t|+|x_1|$ wich is not differentiable at $0$, hence $\|\cdot\|_1$ is not differentiable at $(0,\ldots,0,x_i,0,\ldots,0)$

Edit: as @iloveinna pointed me out, the argument in fact shows that in points wich contains some zero-coordinate, you can perform the same argument, so $\|\cdot\|_1$ is not differentiable not just on axis, but on the hyperplanes generated by the vectors of canonical basis.

Guess 3.: It is correct, because outside the origin, they are all the composition of two functions wich are differentiable out of $0$, say

\begin{eqnarray} x&\longmapsto&|x_1|^p+\cdots+|x_n|^p \phantom{20}\textrm{and}\\ t&\longmapsto&t^\frac{1}{p} \end{eqnarray}

As we can see, we have not calculated the partial derivtives. Instead, we have used paths to show non-differentiability and elementary properties (compositions and signs of coordinates) to see differentiability, but it still remains straightforward.

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signals $\leadsto$ signs :) – Mariano Suárez-Alvarez May 13 '12 at 5:45
@MarianoSuárez-Alvarez Thanks. English is not my native language. – matgaio May 13 '12 at 5:46
Eu sei :) ${}{}$ – Mariano Suárez-Alvarez May 13 '12 at 5:47
Just a point in Guess 2: Your argument actually applies to prove that the map is not differentiable at certain point as long as the point contains a zero-coordinate. So it does not only non-differentiable on axis , but hyperplanes spanned by the axis, right? – iloveinna May 13 '12 at 11:24
It seems it is exactly this, good point. I have not realized this so far. If you are in some of these hyperplanes, say, $x=x_{i_1}e_{i_1}+\cdots+x_{i_k}e_{i_k}$, you can conider the path $$\gamma(t)=x_{i_1}e_{i_1}+\cdots+x_{i_k}e_{i_k}+te_K$$ where $K\neq i_1,\cdots,i_k$ and perform the same analysis. In other words, you can "walk orthogonal to the hyperplane" – matgaio May 13 '12 at 18:01

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