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Is there a name for functions $f : \mathbb{R} \to \mathbb{R}$ such that whenever $|a - b| < |b - c|$ we have $|f(a) - f(b)| < |f(b) - f(c)|$?

It seems like this property should be related to convexity. The above definition was my attempt to make precise the idea of a function that preserves relative closeness of points.

Blue
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Charles Hudgins
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2 Answers2

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Such functions are usually called affine-linear maps.

$f$ must be injective (as for all $b\ne c$ we must have $f(b)\ne f(c)$).

Suppose $f$ is not monotonic, say we have $x<y<z$ and $f(x)<f(y)>f(z)$. If $f(z)<f(x)$ then $|f(z)-f(x)|<|f(z)-f(y)|$ whereas $|z-x|>|z-y|$, contradiction. Similarly, we arrive at a contradiction when $f(z)>f(x)$, or when $f(x)>f(y)<f(z)$. We conclude that $f$ is monotonic.

Consider $a\in \Bbb R$ and let $u:=\inf_{t>a} |f(t)-f(a)|$. For any $c>a$ we can let $b=\frac{c+2a}3$ and, using monotonicity, conclude $|f(c)-f(a)|=|f(c)-f(b)|+|f(b)-f(a)|>2|f(b)-f(a)|\ge 2u$. As this holds for any $c>0$, we conclude that $u\ge2u$, i.e., $u=0$. Similarly, we conclude $\inf_{t<a} |f(t)-f(a)|=0$ and so that $f$ is a continuous monotonic function.

Using continuity, we conclude that $$\tag1|a-b|=|b-c|\implies |f(a)-f(b)|=|f(b)-f(c)|$$ by observing what happens as $c$ varies.

Fix $n\in \Bbb N$ and let $\alpha=n(f(\frac1n)-f(0))$. Then using $(1)$ and injectivity, we find by induction that $$\tag2f(x)=\alpha x+f(0) $$ for all $x=\frac kn$, $k\in \Bbb Z$ As $1\in \frac 1n\Bbb Z$ for all $n$, the $\alpha$ in $(2)$ must be the same for all $n$ and we conclude that $(2)$ holds for all $x\in \Bbb Q$. By continuity, $(2)$ holds for all $x\in \Bbb R$.

Hagen von Eitzen
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  • I wasn't expecting this answer. Very interesting! Thank you for the thorough response. I wonder what happens in general metric spaces, where you no longer have a notion of affine-linear maps. Perhaps all such maps are isometries? – Charles Hudgins Apr 19 '19 at 14:35
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If I am right, no continuous function other than a linear one can satisfy this condition.

Let $c=2b-a+\epsilon$, so that $b-a<c-a$, and assume

$$s_{ab}=\frac{f(b)-f(a)}{b-a}<s_{bc}=\frac{f(c)-f(b)}{c-a}.$$

Then with $c'=2b-a-\epsilon$, so that $b-a>c-a$, $s_{bc'}=s_{bc}-\delta$ but for sufficiently small $\epsilon$ you can ensure

$$s_{bc'}>s_{ab}$$ and $$|b-a|>|c'-b|\land |f(b)-f(a)|<|f(c')-f(a)|.$$

I am not sure that my argument is rigorous. The intuition is that if the slope is not constant, you can find intervals of about the same length such that the images by the function satisfy the $<$ condition, but by a small change you can reverse the comparison on the lengths, while the condition on the function values remains unchanged.