TITLE: Is the Euclidean Norm Differentiable at $ 0 $?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.