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Can we impose such condition on function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $(a,b)\mapsto | f(a) - f(b)|$ generates a metric on $\mathbb{R}$?

This question came into my mind when I was working on problem $(a,b)\mapsto | e^{a} - e^{b}|$ is a metric on $\mathbb{R}$. I guess this can be done by taking injective function $f$. But I am not sure whether this will work or not. Certainly, this will help everyone in dealing with such kind of problems. I need help with this.

Thank you very much.

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I think you just need strictly monotonic. –  copper.hat Jun 7 '12 at 7:27
    
@copper.hat Can we take strictly monotonicity as a necessary and sufficient condition? –  srijan Jun 7 '12 at 7:31
    
@copper.hat: Why do you need Strict monotonicity? Isn't just injective $f$ sufficient? –  Ashok Jun 7 '12 at 7:46
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for $\mathbb{R} \to \mathbb{R}$ injective and strict monotonic are equivalent. –  copper.hat Jun 7 '12 at 8:27
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@copper.hat: This is false, e.g. $f(0)=1$, $f(1)=0$, $f(x)=x$ otherwise. –  Chris Eagle Jun 7 '12 at 9:55

2 Answers 2

up vote 9 down vote accepted

Let $f:\Bbb R\to\Bbb R$, and for $x,y\in\Bbb R$ define $d(x,y)=|f(x)-f(y)|$.

First note that for any function $f:\Bbb R\to\Bbb R$ and $x,y,z\in\Bbb R$ we have $$\begin{align*} |f(x)-f(y)|&=\left|\big(f(x)-f(z)\big)+\big(f(z)-f(y)\big)\right|\\ &\le|f(x)-f(z)|+|f(z)-f(y)|\;, \end{align*}$$

so $d$ always satisfies the triangle inequality. It’s also clear that $d(x,x)=0$ for all $x\in\Bbb R$ and that $d$ is symmetric no matter what $f$ we use. Thus, $d$ is always a pseudometric on $\Bbb R$. Finally, in order for $d$ to separate points, so that it’s necessary and sufficient that $f$ be injective: that ensures that if $x\ne y$, then $f(x)\ne f(y)$ and hence $d(x,y)\ne 0$. The function $f$ need not be nice in any other way.

For example, you could use the following function:

$$f(x)=\begin{cases} \tan^{-1}x,&\text{if }x\in\Bbb Q\\ \tan^{-1}(x+1),&\text{if }x\in\Bbb R\setminus\Bbb Q\;. \end{cases}$$

It’s discontinuous at every point, and it’s not surjective, but it is injective, and that’s all that matters.

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Sir, i think by mistake you have written $\mid f(z) - f(z) \mid$. I can't edit. –  srijan Jun 7 '12 at 7:45
    
@srijan: Thanks; you’re right. –  Brian M. Scott Jun 7 '12 at 7:46
    
Sincerely thanks to you sir. I got a very useful result today.:) –  srijan Jun 7 '12 at 7:47
    
So, here $\mathbb R$ is a set, it could be any set. The other interpretation of "a metric on $\mathbb R$" would be where $\mathbb R$ is a topological space. So we want conditions on $f$ that make $|f(x)-f(y)|$ a metric for the usual topology. –  GEdgar Jun 7 '12 at 12:24

It is necessary and sufficient that $f$ be injective. If $f$ is injective, then $|f(a)-f(b)|=0$ iff $a=b$, and otherwise we have some $a\neq b$ such that $|f(a)-f(b)|=0$. Clearly $|f(a)-f(b)|=|f(b)-f(a)|$, so it remains to check the triangle inequality. But this follows from just applying the triangle inequality for $|\cdot |$, so $f$ gives you a metric.

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Thank you very much. –  srijan Jun 7 '12 at 7:48

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