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What operations is a metric closed under?

Say we have a metric $d: X \rightarrow \mathbb{R}$, and a function $f(d)$ that takes in a metric $d$ and (ideally) spits out another metric.

As an example, let $d$ be a metric and consider the function $f$ where $f(d) = \sqrt{d}$ - then the output $\sqrt{d}$ is also a metric. Similarly, if $f(d) = \frac{d}{1+d}$ then again the output $\frac{d}{1+d}$ is also a metric.

I am wondering if we can say something about what properties of $f$ are necessary / sufficient in order to ensure that the output will be a metric.

Please feel free to make any changes / suggestions the problem since I'm very very new to analysis / functional analysis.

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marked as duplicate by Rudy the Reindeer, Jason DeVito, Pedro Tamaroff, Norbert, Sasha Oct 3 '12 at 3:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The answer to What operations is a metric closed under? describes only sufficient conditions for such functions $f$. But one can see that not all of these conditions are also necessary:

Already be considering $\mathbb R$ with standardmetric and demanding that in this case $f\circ d$ should be a metric, we conclude that $f$ must be a function $[0,\infty)\to [0,\infty)$ with $f(0)=0$ and $f(t)>0$ for $t>0$ and also subadditive: $f(s+t)=(f\circ d)(-s,t)=\le (f\circ d)(-s,0)+(f\circ d)(0,t)=f(s)+f(t)$. The only of the mentoined conditions that is not necessary, is monotonicity.

Indeed, if we merely have $f(0)=0$ and $1<f(t)<2$, then $f\circ d$ is already a metric (inducing discrete topology).

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