Lipschitz continuity and differentiability in $\mathbb{R}^n$

Question

Let $\| \cdot\|_1$ and $\\| \cdot\|_2$ be norms on $\mathbb{R}^n$ and let $F:(\mathbb{R}^n,\| \cdot\|_1) \rightarrow (\mathbb{R}^n,\| \cdot\|_2)$ be a homeomorphism which is a Lipschitz map and let $\gamma:[0,a] \rightarrow \mathbb{R}^n$ be a rectifiable curve.

As $F$ and $\gamma$ are Lipschitz hence the curve $F(\gamma) \subset \mathbb{R}^n$ is Lipschitz, hence from Rademacher's theorem $F(\gamma(\cdot))$ is differentiable almost everywhere.

Assume that $0 < b < a$. I have two questions:

1) If $F(\gamma(\cdot))$ is differentiable at $b$ then: Does the limit $$ \lim_{s \rightarrow b } \frac{\gamma(s)-\gamma(b)}{s-b} $$ exist?

2) If $\gamma(\cdot)$ is differentiable at $b$ then: Does the limit $$ \lim_{s \rightarrow b } \frac{F(\gamma(s))-F(\gamma(b))}{s-b} $$ exist?

Answer 1

There is no reason for a Lipschitz homeomorphism to preserve differentiability at a point. Take $n=1$ and $F(x) = 2x+|x|$, which is strictly increasing hence invertible. The inverse is $G(y) = (2x-|x|)/3$.

1. Let $\gamma = G$. Then $\gamma$ is not differentiable at $0$, but $F\circ \gamma$ is the identity map.

2. Let $\gamma $ be the identity map. Then $\gamma$ is differentiable, but $F\circ \gamma = F$ is not differentiable at $0$.

Answer 2

(1) If $f:(\mathbb{R}^n,\|\ \|_1)\rightarrow (\mathbb{R}^n,\|\ \|_2)$ is Lipschitz homeomorphism, then consider $f\circ c$ where $c$ is a rectifiable.

Define $x=f\circ c : [0,1]\rightarrow (\mathbb{R}^n,\|\ \|_2),\ x(t)=(x_1,\cdots,x_n)(t)$.

If $\pi_i : (\mathbb{R}^n,\|\ \|_2)\rightarrow (\mathbb{R},d)$ with $\pi(y)=y_i$, where $d$ is a restriction of $\|\ \|_2$ on coordinate axis, then $\pi $ is $C$-Lipschitz ($C$ may be greater than $1$).

That is by Rademacher, $x_i$ is differentiable a.e.

(2) If $S$ is unit sphere wrt Euclidean distance, then for $p\in S$, a segment $[pq] $ is mapped by $f$ onto a segment $[pr]$ where $q=(-\epsilon,0) $ and $r=(0,0)$.

And $f$ is identity on outside of $S$.

Consider a curve $[(0,1)q]\bigcup [q(0,-1)]$ which is not smooth. But the image of $f$ is smooth.

(3) For $2)$, consider $f^{-1}$.