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I am presently working on an exercise from Kaplansky, I. Set Theory and Metric Spaces (ex. 7, pg 70).

The question is as follows:

Suppose that: $f$ is concave, $f(0) = 0$, $f(x) > 0$ for $x > 0$, and $f$ is montone in the weak sense. Let $M$ be a metric space with distance function $D$. Then $f(D)$ is also a distance function on $M$.

The basic properties are easy to prove, where I am having difficulty is the triangle inequality, which asserts:

$$f( D(a,c) ) \le f ( D(a,b) ) + f ( D(b,c) ) \tag{1}$$

Since $f$ is weakly monotonic, I know that following must hold:

$$f( D(a,c) ) \le f [ D(a,b) + D(b,c) ] \tag{2}$$

I have been attempting to use the definition of concavity to demonstrate that $(2) \implies (1)$, but have been unable to figure anything out.

I would appreciate some assistance on what to do.

GovEcon
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1 Answers1

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If $f$ is concave and $f(0)=0$, then $f$ is sub-additive, i.e. $$ f(x+y) \leq f(x)+f(y). $$ (see the wikipedia article http://en.wikipedia.org/wiki/Concave_function for a proof.)

Applying this to your line (2) finishes the argument.

GCD
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