Norm not differentiable on a dense set of $\mathbb{R}^n$ Would you have an example of a norm on $\mathbb{R}^n$ ($n \ge 2$) which is not differentiable on a dense set?
 A: I believe the following works: let $f:[-1,1]\to\mathbb R$ be a concave function such that $$f(-x)=f(x)$$ for all $x$, i.e. $f$ is even, and $$f(-1)=f(1)=0.$$ Suppose further that $f$ is not differentiable on a dense set. (This answer should suffice to construct such a function.) Define $$K=\{(x,y)\in\mathbb R^2\mid -f(x) \leq y\leq f(x)\}.$$ There exists a norm $\|\cdot\|$ on $\mathbb R^2$ such that $K$ is precisely the unit ball with respect to $\|\cdot\|$. You may define it by the formula $$\|(x,y)\|=\inf\{r>0\mid (x,y)\in rK\}.$$ (This is the so-called Minkowski functional associated to $K$.)
Now, observe that if $x_0\in[-1,1]$ is a point at which $f$ is non-differentiable, then $\|\cdot\|$ will be non-differentiable along the line $L(x_0)$ through the points $(x_0,f(x_0))$ and $(0,0)$.
Here's a rough sketch of this: suppose $L(x_0)$ is such a line, $a\neq 0$ and $F=\|\cdot\|$ is differentiable at $a(x_0,f(x_0))$. Then, the set of points $(x,y)$ such that $$\langle(\operatorname{grad}F)(a(x_0,f(x_0)),(x,y)\rangle=0$$ provides us with a tangent line to the curve $\|(x,y)\|=a$. (The gradient vector is non-zero, since by homogeneity, $F$ is linear and non-zero along $L(x_0)$ in a small neighborhood of $a(x_0,f(x_0))$.) But the curve $\|x,y\|=a$ is simply a scaled version of the graph of $f$, which is not differentiable at $x$, so it cannot have such a tangent line, contradiction.
Now let $A$ be the set of all points $(a,b)$ lying in some line $L(x)$ such that $f'(x)$ does not exist. Since such $x$ are dense in $[-1,1]$, we can show that $A$ is dense in the plane and we are done.
