In a normed vector space, does $x+r\Gamma = x'+r'\Gamma$ imply $x=x',r=r'$?

Question

Let $X$ denote a real normed vector space and suppose $\Gamma \subseteq X$ is a bounded subset with two or more elements. Consider $r,r' \in \mathbb{R}_{> 0}$ and $x,x' \in X$. Does $x+r\Gamma = x'+r'\Gamma$ imply $x=x',r=r'$?

I am especially interested in the case where $\Gamma$ is the unit ball, open or closed.

Answer 1

If $k$ is the diameter of $\Gamma$, $0 < k < \infty$, then the diameters of the two sets $x+r\Gamma$ and $x'+r'\Gamma$ are $rk$ and $r'k$. So $r=r'$. We used $0<r<\infty, 0<r'<\infty$.

So now we must ask, can a non-trivial translate $y + \Gamma$ of a bounded set $\Gamma \ne \varnothing$ be equal to $\Gamma$? No: let $a \in \Gamma$. Then $\{a+y, a+2y, a+3y,\dots\}$ is unbounded, so not contained in $\Gamma$.

Answer 2

No ! For exemple, take $\Gamma=\{e^{i\theta}\mid\theta\in[0,2\pi[\}$. You have $\Gamma=e^{i\pi}\Gamma$ but $1\neq e^{i\pi}$

Answer 3

If $s>0, a \in X$ then the maps $$ \lambda_s: X\to X, \lambda_s(z)=sz,\quad \tau_a: X\to X,\tau_a(z)=a+z $$ are invertible and satisfy: $$ \tau_a^{-1}=\tau_{-a},\quad \lambda_s^{-1}=\lambda_{s^{-1}}, $$ and $$ \|\tau_a(y)-\tau_a(z)\|=\|y-z\|,\quad \|\lambda_s(z)\|=s\|z\| \quad \forall y,z \in X. $$ We have $$ \tau_x\circ\lambda_r(\Gamma)=\tau_{x'}\circ\lambda_{r'}(\Gamma) \iff f(\Gamma)=\Gamma, $$ where the map $f:=\lambda_{1/r'}\circ\tau_{x-x'}\circ\lambda_r$ is also invertible.

For every $n\in \mathbb{Z}$ we have $$ f^n(\Gamma)=\Gamma. $$ Then for every $x_0,y_0\in \Gamma$ we have $$ \|f(x_0)-f(y_0)\|=\frac{1}{r'}\|\tau_{x-x'}\circ\lambda_r(x_0)-\tau_{x-x'}\circ\lambda_r(y_0)\|=\frac{1}{r'}\|\lambda_r(x_0-y_0)\|=\frac{r}{r'}\|x_0-y_0\|. $$ It follows by induction that $$ \|f^n(x_0)-f^n(y_0)\|=\left(\frac{r}{r'}\right)^n\|x_0-y_0\| \quad \forall n\in \mathbb{Z}. $$ Since $\Gamma$ is bounded, assuming that $x_0\ne y_0$, we have $$ \limsup_{|n|\to \infty} \left(\frac{r}{r'}\right)^n=\limsup_{|n|\to \infty}\frac{\|f^n(x_0)-f^n(y_0)\|}{\|x_0-y_0\|}<\infty \iff \frac{r}{r'}=1\iff r=r'. $$ Now that we have $r=r'$ we deduce that for every $z \in X$: $$ f(z)=\lambda_{1/r}\circ\tau_{x-x'}\circ\lambda_r(z)=r^{-1}(x-x'+rz)=r^{-1}(x-x')+z. $$ Again by induction we have: $$ f^n(z)=nr^{-1}(x-x')+z \quad \forall n \in \mathbb{Z}, z\in X. $$ Since $\Gamma$ is bounded, if $z\in \Gamma$, we have $$ \limsup_{|n|\to\infty}r^{-1}|n|\|x-x'\|=\limsup_{|n|\to\infty}\|f^n(z)-z\|<\infty \iff \|x-x'\|=0 \iff x=x'. $$