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Let $I$ be an interval of the real line. Consider two functions $f,g: I \to \mathbb{R}$ such that $f,g$ are strictly increasing and satisfy the following three points property: for all $a,b,c \in I$, we have \begin{equation*} f(a)-f(b) = f(b) - f(c) \Longrightarrow g(a)-g(b) = g(b)-g(c). \end{equation*} If $f$ and $g$ are continuous, I can show that there exists $\mu>0$ and $\lambda \in \mathbb{R}$ such that $f=\mu g + \lambda$. Is the result true in general (without assuming that $f$ and $g$ are continuous)?

My arguments for the continuous case are as follows:

Assume that $I$ has at least two elements $\underline{x}<\overline{x}$. Replacing $f$ by $f-f(\underline{x})$ and $g$ by $g-g(\underline{x})$ if necessary, we can assume that $f(\underline{x})=g(\underline{x})=0$. We shall prove that $f=\mu g$ for some $\mu>0$.

As $f$ is continuous, $f(I)$ is an interval. Therefore there exists a unique $b_{1/2} \in (\underline{x},\overline{x}) \subseteq I$ such that $f(b_{1/2})=(1/2) f(\overline{x})$. Applying the three-point property with $a=\overline{x}$, $b=b_{1/2}$, and $c=\underline{x}$, we deduce that $g(b_{1/2})=(1/2) g(\overline{x})$.

Using the same reasoning, there exists a unique $b_{1/4}\in (\underline{x},\overline{x})$ such that $f(b_{1/4}) = (1/4)f(\overline{x})$. Applying the three-point property with $a=b_{1/2}$, $b=b_{1/4}$, and $c=\underline{x}$, we deduce that $g(b_{1/4})=(1/4) g(\overline{x})$.

Using the same reasoning, there exists a unique $b_{3/4}\in (\underline{x},\overline{x})$ such that $f(b_{3/4}) = (3/4)f(\overline{x})$. Applying the three-point property with $a=\overline{x}$, $b=b_{3/4}$, and $c=b_{1/2}$, we deduce that $g(b_{3/4})=(3/4) g(\overline{x})$.

We omit the simple recurrence argument to prove that for every $n\in \mathbb{N}$ and every $k\in \{1,\ldots,2^n-1\}$, we have $g(b_{k/2^n}) = (k/2^n) f(\overline{x})$ where $b_{k/2^n}$ is the unique point in $(\underline{x},\overline{x})$ such that $f(b_{k/2^n})=(k/2^n) f(\overline{x})$.

Pose $\mu:=f(\overline{x})/g(\overline{x})$. As $f,g$ are strictly increasing and $\overline{x}>\underline{x}$, we must have $\mu>0$. We have proved that $f(y)/g(y) =\mu$ for every $y\in Y:=\{b_{k/2^n} \colon n\in \mathbb{N} \text{ and } k\in \{1,\ldots,2^n-1\}\}$.

Continuity of $f$ implies that the set $Y$ is dense in $[\underline{x},\overline{x}]$. As $f$ and $g$ are continuous, we deduce that $f(x)/g(x)=\mu$ for every $x\in (\underline{x},\overline{x}]$.

Fix $\overline{x}^\prime \in I$ arbitrary satisfying $\overline{x}^\prime \geq \overline{x}$. Applying the previous argument to the interval $[\underline{x},\overline{x}^\prime]$, we deduce the existence of $\mu^\prime>0$ such that $f(x)/g(x)=\mu^\prime$ for every $x\in [\underline{x},\overline{x}^\prime]$. As $\overline{x} \in [\underline{x},\overline{x}^\prime]$, we must have $\mu=f(\overline{x})/g(\overline{x}) = \mu^\prime$.

We have thus proved that $f(x)/g(x)=\mu$ for any $x\in I$ satisfying $x>\underline{x}$. If $I\subseteq [\underline{x},\infty)$, then we are done. Assume that exists $\underline{w}\in I$ satisfying $\underline{w} < \underline{x}$. Following a similar argument as above, we can show that there exists $\tilde{\mu}>0$ such that $f(x)/g(x)=\tilde{\mu}$ for any $x\in I$ satisfying $x<\underline{x}$. The function $f$ is strictly increasing, $f(\underline{x})=0$ and $\underline{x}$ belongs to the interior of $I$. This implies the existence of $a,c\in I$ with $a>\underline{x} > c$ such that $f(c)=-f(a)$. Applying the there-point property with $b=\underline{x}$, we deduce that $g(c)=-g(a)$. We then get that $f(c)/g(c)=f(a)/g(a)$ and $\tilde{\mu}=\mu$.

V. Filipe Martins-da-Rocha
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1 Answers1

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Suppose $f:[0,2]\to\mathbb R$ given by $$f(x):=\begin{cases}x,&\text{if }x\leq1,\\x+2&\text{otherwise}\end{cases}$$ and $g$ satisfy the requirement. Since $f$ is continuous on $[0,1]$ and $(1,2]$, we know by your proof${}^1$ that $g$ has to be affine linear on these segments, too.

But if $a>b>c$ satisfy $f(a)-f(b)=f(b)-f(c)$, then either $a,b,c\leq1$ or $1<a,b,c$: Suppose otherwise, then $2\leq f(a)-f(c)\leq 4$, hence $1\leq f(a)-f(b)\leq 2$, i.e. $f(b)\in(1,3]$, which is not in the range of $f$.

Therefore the two different segments don't interact as far as our property is concerned, and thus we can choose different linear transformations on each interval. For example, $g(x)=x$ or $$g(x)=\begin{cases}5x+9,&\text{if }x\leq1,\\42x-17&\text{otherwise}\end{cases}$$ both work, but are not affine transformations of $f$.

This is essentially the only thing that can go wrong: Since $f$ is monotone, it has at most countably many discontinuities. On every interval where $f$ is continuous, your proof holds, so $g$ has the same discontinuities (or fewer), but we might have different affine transformations after large jumps.

${}^1$You used continuity of $g$ at one point, to deduce $f/g=\mu$ from $f/g|_Y=\mu$. But this isn't actually necessary: By monotonicity and continuity of $f$, \begin{align}g(x)&\geq\sup\{\frac1\mu f(y)\mid y<x\}\\&\geq\sup\{\frac1\mu f(y)\mid Y\ni y<x\}=\frac 1\mu f(x),\end{align} and similarly for the reverse inequality.

anankElpis
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