This is a continuation of this other question I asked earlier.
Since @Arnaud D. proved in my previous question that the result can not hold under such weak hypothesis, I reinforced the hypothesis on the function $F$.
A function $F\colon \mathbb R^n\to \mathbb R$ is called a distance function if it verifies the three conditions:
$F(x)\geqslant 0$ and $F(x)=0\iff x=0$,
$F(tx)=\vert t\vert F(x)$,
$F$ is continuous.
I want to prove the following result.
Let $F\colon \mathbb R^n\to \mathbb R$ be a distance function. If $w\in \mathrm{GL}_n(\mathbb R)$, then there exists a constant $C_w$ (which does not depend on $X$), such that $$F(w(X))\leqslant C_wF(X). \quad \forall X \in \mathbb{R}^n$$
I think I should use something like $\Vert w(W)\Vert\leqslant \Vert w\Vert \cdot \Vert X\Vert$, or try to use a matrix $W=(w_{i,j})$ for $w$ and compute the coordinates of $w(x)$ in terms of $w_{i,j}$ and $x_i$.
But I can't figure it out. Any hint or solution would be much appreciated.
Notice that the counter-example provided by @Arnaud D. in my previous question does not work here since it does not verify $F(x)=0\iff x=0$.
I think I finally got it, I just need to act as if $F$ is a norm. I encourage you to criticize me if this proof has flaws.
Let $$C_w:=\sup_{F(Y)=1} F(w(Y)),$$
which does not depend on $X$, and is well defined since $F$ is continuous on the compact unit sphere.
I assume $X\ne 0$. Now, set $Y:=\frac 1{F(X)} X$, which is well-defined since $F(X)\ne 0$ because $X\ne 0$ and $F$ is a distance function.
Then because of the properties of $F$:
$$F(w(X))=F(w(F(X)Y))=F(X)\underbrace{F(w(Y))}_{\leqslant C_W}.$$