Can we detect smoothness of a norm by its behavior along paths? We say a norm $\| \cdot \|$ on $\mathbb{R}^n$ is smooth if it is smooth as a function $\mathbb{R}^n\setminus \{0\} \to \mathbb{R}$. (i.e, after restricting the domain).
We say a norm is smooth along paths, if for any smooth path $\alpha:I \to \mathbb{R}^n \setminus \{0\}$, the composition$\|\cdot\| \circ \alpha$ is smooth.
Does smoothness along paths of a norm implies that it's smooth?
(Note that in general differentiability along paths is not enough to detect smoothness, but in this case we are talking about a norm, not a general function).
Remarks:
1) Since any norm is Lipschitz (w.r.t the metric induced by it), it follows that the value of the derivative $\frac{d}{dt}\big|_{t=0} \|\alpha(t)\| $ depends only on $\dot \alpha(0),\alpha(0)$ and not on other properties of the specific path chosen. (See a proof at the end).
Thus, for each $a \in \mathbb{R}^n\setminus\{0\}$, there is a function:
$T_a:\mathbb{R}^n \to \mathbb{R}$ defined by: $T_a(v)=\frac{d}{dt}\big|_{t=0} \|\alpha(t)\|$ where $\alpha(t)$ is any path satisfying $\alpha(0)=a,\dot\alpha(0)=v$. It is proved here that if $T_a$ is linear, then $\|\cdot  \|$ is differentiable at the point $a$.
So, proving linearity of $T_a$ (for every $a$) will show at least that $\|\cdot\|$ is one-time differentiable. (Showing harder degrees of regularity, i.e twice differentiability etc sounds more challenging, but let's start with that).
So far, using homogeneity and triangle inequality, I managed to prove:
$$T_a(\lambda v)=|\lambda|T_{\frac{a}{\lambda}}(v) \, \, \forall \lambda \in \mathbb{R},$$ $$T_a(v+w)\le T_a(v)+\|w\|\, \, , \, \,T_a(v+w) \le T_{\frac{a}{2}}(v)+T_{\frac{a}{2}}(w)$$

Proof of claim (1): (The derivative depends only on the velocity of the path)
$$\big| \|\alpha(t)\|-\|\tilde \alpha(t)\| \big| \le \|\alpha(t)-\tilde \alpha(t) \|,$$ so assuming $\tilde \alpha(0)=\alpha(0),\dot{\tilde \alpha}(0)=\dot \alpha(0)=v \in \mathbb{R}^n$ we get: $$(*)\, \, \bigg| \frac{\|\alpha(t)\| - \|\alpha(0)\|}{t}-\frac{\|\tilde \alpha(t)\| - \|\tilde \alpha(0)\|}{t}\bigg| \le \|\frac{\alpha(t)-\tilde \alpha(t)}{t} \| $$
Now since $\lim_{t \to 0}\frac{\alpha(t)-\tilde \alpha(t)}{t} =\lim_{t \to 0}\big( \frac{\alpha(t)- \alpha(0)}{t}-\frac{\tilde \alpha(t)-\tilde \alpha(0)}{t}\big)=v-v=0$
we get by the continuity of the norm, together with $(*)$ we obtain the required equality
$$ \frac{d}{dt}\big|_{t=0} \|\alpha(t)\|  = \lim_{t \to 0} \frac{\|\alpha(t)\| - \|\alpha(0)\|}{t}=\lim_{t \to 0} \frac{\|\tilde \alpha(t)\| - \|\tilde \alpha(0)\|}{t}=\frac{d}{dt}\big|_{t=0} \|\tilde \alpha(t)\|$$
 A: It'll be $C^1$ but not necessarily smooth.
First differentiability. A supporting hyperplane of $B_R=\{x\mid \|x\|\leq R\}$ at a given point is in fact a tangent plane because directional derivatives along this plane where it touches the unit ball must be zero: a differentiable function $\gamma:(-1,1)\to\mathbb R$ with $\gamma(0)=1$ and $\gamma(t)\geq 1$ must have $\gamma'(0)=0.$
To get continuity of the derivative, consider a simpler case of a tangent plane $y=c$ in two dimensions. We can pivot this slightly around the point $(-W,c)$ where $W$ is the Euclidean radius of $B_R$; the line through $(-W,c)$ and $(-W,c+\epsilon)$ does not intersect $B_R.$ This shows that varying the angle of the tangent plane slightly doesn't increase the norm by more than a small amount depending on the Euclidean radius of $B_R$.
Now a non-$C^\infty$ example in $\mathbb R^3.$ It's enough to modify the function $(x,y)\mapsto \|(x,y,1)\|$ near the origin in $\mathbb R^2$ to be non-smooth while still being convex and smooth along paths. Add a smooth non-negative bump at $(3^{-n},4^{-n})$ of height $c16^{-n}$ and radius $4^{-n-10}.$ For small enough $c$ this will still be convex on any half-plane not including zero by a second derivative test, and it has a supporting hyperplane at zero because it's non-negative. It is smooth along paths that don't go through zero, and a smooth path through zero will only hit finitely many of the bumps so will also have a smooth composition with this function. But the function itself is not $C^3$ because it differs from a $C^3$ function by bumps of scale $\Theta(\|x\|_2^{\log 16/\log 3})$ in the positive orthant only. Here $x\in\mathbb R^2$ and $\|\cdot\|_2$ is Euclidean norm.
This doesn't rule out the possibility that norms that are smooth along paths are $C^2.$ Note we do have a.e. twice differentiability (Alexandrov theorem), and the distributional Hessian is a Radon measure - see for example Evans and Gariepy, Measure theory and fine properties of functions.
