1

In the following post how come scaling to $x_0=0$ and $r=1$ can help to prove for any $r,x_0$?

Is there a general rule for when is it OK to scale?

moshe
  • 183

1 Answers1

1

You can get away with scaling and translating whenever your problem is invariant under transformations of the form $x\mapsto \alpha x + c$ where $\alpha$ is a real number (or, more generally, an element of the underlying field) and $c$ is a vector. Given how many problems respect such transformations, and how averse some mathematicians are to constants other than $0$ and $1$, this is a pretty common trick. More generally, this is what is happening whenever someone writes "without loss of generality" - they are using a symmetry in the problem to extrapolate a proof from fewer cases.

With your question in particular, consider the map $$f(x)=\frac{1}{r}\cdot (x-x_0)$$ You can check that $f$ has an inverse of a similar form. Then, you have that $f(x_0)$ is the origin and $f[C]$ is a convex set in the same space. Moreover, as the above scales distances by $\frac{1}r$, you get that $d(f(x_0),f[C])=1$. Finally, if a ball of radius $r$ around $x_0$ intersects $C$ in $n$ places, then the ball of radius $1$ around $f(x_0)$ intersects $f[C]$ in $n$ places as well.

At this point, you can just apply the argument that you know: If $x_0$ is the origin and $d(x_0,C)=1$, then $\{y\in X:\|y\|\leq r\}\cap C$ has at most one point. However, since every other situation can be reduced to one like this without changing the size of this intersection, the statement is true in general.

Milo Brandt
  • 60,888
  • Why do i need to check that f has an inverse? I didn't get the meaning of the existence of an inverse to f. – moshe May 24 '16 at 15:07
  • 1
    @moshe I suppose one doesn't directly need to in this case (since scaling distances is sufficient), but in general, it's helpful - basically, having an inverse means that it's a symmetry, so nothing is one can recover the original case by undoing the map. This is to contrast it with, say, having $f$ be a projection or general linear map, where the truth of the statement is not preserved by $f$. – Milo Brandt May 24 '16 at 15:11