Is the Euclidean Norm Differentiable at $ 0 $? As the title, is the euclidean norm differentiable at $0$? I tried that prove by contradiction and apply the definition but can't really get a contradiction. Any hints?
 A: It isn't. The definition of differentiable is that the derivative of the norm function (let me call it $N$) at zero would be a vector $v$ such that $$\lim_{x\to 0} \frac{N(x)-x \cdot v}{\left\|x\right\|}=0.$$
(Here, $\left\|x\right\|$ is also the Euclidean norm of $x$, but it plays a different role from $N$, so I used different notation.)
If you let $x\to 0$ along vectors orthogonal to $v$, the limit is $1$, a contradiction.
A: Assume, for a contradiction, that $f(x)=\|x\|$ is differentiable at $0$, ie
$$
f(x)=f(0)+df(0)x+\|x\|\epsilon(x)
$$
for all $x\in\mathbb{R}^n$ with $\lim_{\|x\|\rightarrow 0}\epsilon(x)=0$.
Now fix $x$ such that $\|x\|>0$.
Then for all $t\in\mathbb{R}$
$$
|t|\|x\|=f(tx)=tdf(0)x+|t|\|x\|\epsilon(tx)
$$
so
$$
|t|=t\frac{df(0)x}{\|x\|}+|t|\epsilon(tx)
$$
which proves that 
$$
t\longmapsto |t|
$$
is differentiable at $0$ with derivative $df(0)x/\|x\|$.
As is well-known, the latter is not differentiable at $0$. Contradiction.
Note: this proof works in general for the norm of a normed vector space. No need to restrict to inner product spaces.
A: Another approach, that seems to work:
$g(x)=(\underbrace{1,0,...,0}_\text{n})^T x$ is linear and therefore differentiable (at $0$).
Then $|x_1| = (f \circ g)(x)$
Suppose for a contradiction, that $f$ is differentiable at $0$.
Then by the chain rule $|x_1|$ is differentiable at $0$. However, the first partial derivative $\frac{\partial}{\partial x_1} |x_1|$ does not exists. So $|x_1|$ not differentiable. Contradiction.
