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Let $\vert x-y\vert$ be the usual distance over $\Bbb{R}$ and $\gamma(x,y)=\Phi(d(x,y))$ where $\Phi(t)=\frac{t}{1+t}$.

I would like to prove that the two distance are not equivalent.

I now the definition to be equivalent is to find two constant $\alpha,\beta$ such that $$ \alpha\cdot\gamma(x,y)\le d(x,y)\le \beta\cdot\gamma(x,y). $$

Effort: Perchance arguing by contradiction. $$ d(x,y)\le \beta\cdot\gamma(x,y) \Longleftrightarrow \vert x-y\vert\le \beta \frac{\vert x-y\vert}{1+\vert x-y\vert} $$ It can be rewritten as $$ 1+\vert x-y\vert\le\beta \quad\text{for}\quad x\ne y $$

Now I am tempted to say it's impossible because $x\mapsto\vert x-y\vert+1$ cannot be bounded by a constant (increasing/decreasing argument). Is it correct?

Additional question if I'm right. What assumptions can be introduced to find a function such that these distances are equivalent?

  • That's perfectly fine. Regarding the other question: If you chose a bounded subset of $\mathbb R$ (say an interval) you may be able to get somewhere ;) – AlexR Oct 05 '14 at 22:16
  • @AlexR Great, thank you very much. I will search. –  Oct 05 '14 at 22:20
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    Happy to help. Note that the metric is translation invariant so you may assume WLOG $y=0$ and look at $|x|$ vs $\frac{|x|}{1+|x|}$. A hint to why they are not equvalent (even on intervals) is that the derivative of the latter blows up near $0$. – AlexR Oct 05 '14 at 22:25

1 Answers1

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According to your definition of equivalence of metrics, the metrics $d$ and $\gamma(x,y)=\Phi(d(x,y))$ are equivalent if and only if there exist $\alpha,\beta$ such that $$\alpha t\le \Phi(t)\le \beta t,\quad t\ge 0 \tag1$$ As you demonstrated, $\Phi(t)/t\to 0$ as $t\to\infty$, violating (1).

A sufficient condition for (1) to hold is that $\Phi'$ is bounded between two positive constants (and $\Phi(0)=0$). But in practice, it is usually easiest to check (1) directly.