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Assume $(X,d)$ is a metric space, and define a new metric $\tilde{d}$ on $X$. Set $\tilde{d} = \frac{d(x,y)}{1+d(x,y)}$. Now with manipulation and since $d$ is a metric, I manage to show that $\tilde{d}$ satiesfies the triangular ineqality. But I read that one can conside the function $f(t) = \frac{t}{1+t}$ and show that this is a strictly increasing function (no problem). But they argue that the triangular inequality follows from $f$ is stricly increasing, why?

I guess I could pick three values $f(t_1), f(t_2), f(t_3)$ which all lies in $\mathbb{R}$ then on $\mathbb{R}$ we have $p(x,y) = |x-y|$ and thus $p(f(t_1),f(t_2)) \leq p(f(t_1), f(t_3)) + p(f(t_3), f(t_2))$ is that it?

Olba12
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2 Answers2

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This is nonsense; you cannot conclude that $\tilde{d}$ is a metric from the fact that $f$ is strictly increasing. For instance, the function $g(t)=t^2$ is strictly increasing on $[0,\infty)$, but if $d$ is a metric then $\tilde{d}(x,y)=g(d(x,y))$ is not necessarily a metric. For instance, if $d$ is the usual metric on $\mathbb{R}$, then $$\tilde{d}(0,2)=4>1+1=\tilde{d}(0,1)+\tilde{d}(1,2),$$ violating the triangle inequality.

Eric Wofsey
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  • But a question was on the concrete function, not a quadratic one. As you can see by my answer, the argument works in this case. – szw1710 Mar 09 '17 at 08:18
  • Even though this does not answer the question, it was still helpfull in a way where it made my confusion go away. – Olba12 Mar 09 '17 at 08:22
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    You can make such an argument in this case, but it is surely deeply misleading to say that "the triangular inequality follows from $f$ is stricly increasing". – Eric Wofsey Mar 09 '17 at 08:23
  • @EricWofsey, subadditivity of $f$ on $[0,\infty)$ played a key-role. There is no such property for a quadratic function. From my answer it is trivially seen that any strictly increasing $f$ which is subadditive on $[0,\infty)$ and s.t. $f(0)=0$ is good. So, this is a sufficient condition for $\tilde{d}$ to be a metric. It is interesting whether we could find any kind of a necessary condition. – szw1710 Mar 09 '17 at 08:58
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Let $u=d(x,z),v=u=d(x,y),w=d(y,z)$. It is enough to check the condition $$u\le v+w\implies f(u)\le f(v)+f(w).$$ By increasingness we have $f(u)\le f(v+w)$. Elementary computation shows that $f(v+w)\le f(v)+f(w)$, which finishes this argument.

szw1710
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  • Your computations was what I actually did, for the specified function. So we have the same argument. I just got confused by the sentence "or on could just notice that the function $f$ is increasing. Great answer. – Olba12 Mar 09 '17 at 08:22