2
$\begingroup$

Let $a,b$ be two constant real numbers with $a\neq 0$. Can anyone give a special solution of the functional equation $f(ax+b)=f(x)+1$, where $f:\mathbb{R}\rightarrow \mathbb{R}$?

Note. It is a type of the Abel functional equations, and if $a=1$, then $f(x)=[\frac{x}{b}]$ is an its solution.

$\endgroup$
2
$\begingroup$

For fixed $a\in\mathbb{R}\setminus\{1\}$ and $b\in\mathbb{R}$, let now consider a function $f:\mathbb{R}\to\mathbb{R}$ which satisfies the functional equation $$f(ax+b)=f(x)+1\text{ for all }x\in\mathbb{R}\setminus\left\{\frac{b}{1-a}\right\}\,.\tag{#}$$ Firstly, we assume that $a=0$. Then, we see that $f(x)=f(b)-1$ for any $x\in\mathbb{R}\setminus\{b\}$. Thus, all functions $f:\mathbb{R}\to\mathbb{R}$ with the condition (#) are of the form $$f(x)=\begin{cases}c&\text{if }x=b\,,\\c-1&\text{if }x\neq b\,.\end{cases}$$


Secondly, we assume that $a>0$. Write $I^+:=\left(\dfrac{b}{1-a},+\infty\right)$ and $I^-:=\left(-\infty,\dfrac{b}{1-a}\right)$. For $x\in I^+$, we can see that $$x-\frac{b}{1-a}=a^t\text{ or }t=\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}$$ for some $t\in\mathbb{R}$. Thus, if $g_+(t):=f\left(a^t+\dfrac{b}{1-a}\right)$ for each $t\in\mathbb{R}$, then $$\begin{align}g_+(t+1)&=f\left(a^{t+1}+\frac{b}{1-a}\right)=f\Biggl(a\left(a^t+\frac{b}{1-a}\right)+b\Biggr)\\&=f\left(a^t+\frac{b}{1-a}\right)+1=g_+(t)+1\,.\end{align}$$ Therefore, if $h_+(t):=g_+(t)-t$, then $h_+:\mathbb{R}\to\mathbb{R}$ is periodic with period $1$. That is, $$f(x)=h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\text{ for all }x\in I^+\,.$$ We obtain a similar result for $x\in I^-$. Thus, $$f(x)=\begin{cases} h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ h_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ where $h_+,h_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $c\in\mathbb{R}$ is an arbitrary constant.


Finally, we are dealing with the case $a<0$. We rule out the case $a=-1$, since there does not exist a solution $f$ with the said property. This is because of the contradiction below when $a=-1$: $$f(x)=f\big(-(-x+b)+b\big)=f(-x+b)+1=\big(f(x)+1\big)+1=f(x)+2$$ for all $x\neq \dfrac{b}{2}$. From now on, we assume that $a\neq -1$.

Note that $$\begin{align}f\big(a^2x+(a+1)b\big)&=f\big(a(ax+b)+b\big)=f(ax+b)+1\\&=\big(f(x)+1\big)+1=f(x)+2\end{align}$$ for all $x\neq \dfrac{b}{1-a}$. Let $A:=a^2$, $B:=(a+1)b$, and $\phi(t):=\dfrac{1}{2}\,f(t)$ for all $t\in\mathbb{R}$. Then, $$\phi(Ax+B)=\phi(x)+1$$ for every $x\neq \dfrac{b}{1-a}=\dfrac{B}{1-A}$. Since $A>0$ and $A\neq 1$, we have by the previous section of this answer that $$\phi(x)=\begin{cases} \eta_+\left(\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}\right)+\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}&\text{if }x>\frac{B}{1-A}\,,\\ C&\text{if }x=\frac{B}{1-A}\,,\\ \eta_-\left(\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}\right)+\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}&\text{if }x<\frac{B}{1-A}\,, \end{cases}$$ where $\eta_+,\eta_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $C\in\mathbb{R}$ is an arbitrary constant. Therefore, $$f(x)=2\,\phi(x)=\begin{cases} 2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ 2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ where $c:=2C$.

Recall that $f(ax+b)=f(x)+1$ for $x\neq \dfrac{b}{1-a}$. For $x>\dfrac{b}{1-a}$, we have $$f(ax+b)=f(x)+1=2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{1}$$ Because $ax+b<\dfrac{b}{1-a}$ when $x>\dfrac{b}{1-a}$, we conclude that $$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{\ln|a|}\,.$$ Since $\dfrac{b}{1-a}-(ax+b)=|a|\,\left(x-\dfrac{b}{1-a}\right)$, we obtain $$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}+\frac{1}{2}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{2}$$ Equating (1) and (2), we conclude that $$\eta_+\left(t\right)=\eta_-\left(t+\frac{1}{2}\right)\text{ for each }t\in\mathbb{R}\,.$$ Let $h:\mathbb{R}\to\mathbb{R}$ be given by $$h(t)=2\,\eta_+\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$ Ergo, $h$ is periodic with period $2$ and $$h(t-1)=2\,\eta_-\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$ Hence, $$f(x)=\begin{cases} h\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\ c&\text{if }x=\frac{b}{1-a}\,,\\ h\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}-1\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,, \end{cases}$$ for some real constant $c$ and for some periodic function $h:\mathbb{R}\to\mathbb{R}$ with period $2$.


It is not difficult to prove that the three results are indeed solutions to (#). I shall omit the proof of this part as an exercise.

$\endgroup$
  • $\begingroup$ @user7427029 The derivations of all three cases have been supplied. I finished the case $a<0$. Due to slow MathJax, I had to put this result as a separate answer. $\endgroup$ – Batominovski Oct 8 '18 at 18:56
4
$\begingroup$

In this solution, I drop the condition that $a\neq 0$. If $a\neq 1$, then there does not exist such $f$. Taking $x:=\dfrac{b}{1-a}$ in the functional equation yields $$f\left(\dfrac{b}{1-a}\right)=f\left(\dfrac{ab}{1-a}+b\right)=f\left(\dfrac{b}{1-a}\right)+1\,,$$ which is absurd.

If $a=1$, then we have $$f(x+b)=f(x)+1$$ for all $x\in\mathbb{R}$. If $b=0$, then no function works. If $b\neq 0$, then let $g(y):=f(by)-y$ for each $y\in\mathbb{R}$. Thus, $$\begin{align}g(y+1)&=f\big(b(y+1)\big)-(y+1)=f(by+b)-y-1\\&=\big(f(by)+1\big)-y-1=f(by)-y=g(y)\,.\end{align}$$ In other words, $$f(x)=g\left(\frac{x}{b}\right)+\frac{x}{b}\tag{*}$$ for some periodic function $g:\mathbb{R}\to\mathbb{R}$ with period $1$.

In conclusion, for real constants $a$ and $b$, there exists a function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(ax+b)=f(x)+1\text{ for every }x\in\mathbb{R}$$ if and only if $a=1$ and $b\neq 0$. When $a=1$ and $b\neq 0$, all solutions take the form (*). In particular, the solution $f(x)=\left\lfloor\dfrac{x}{b}\right\rfloor$ for all $x\in\mathbb{R}$ obtained by the OP arises from taking $$g(x):=-\left\{x\right\}\text{ for all }x\in\mathbb{R}\,.$$ Here, $\{t\}$ is the fractional part of $t\in\mathbb{R}$.

$\endgroup$
  • $\begingroup$ @ Batominovski . Thanks for your answer. Any solution if we assume that $f:\mathbb{R}\setminus\{\frac{b}{1-a}\}\rightarrow \mathbb{R}$? $\endgroup$ – M.H.Hooshmand Oct 7 '18 at 19:12
  • 2
    $\begingroup$ @M.H.Hooshmand Yes, there are solutions if $a> 0$ is assumed. Let $h_+,h_-:\mathbb{R}\to\mathbb{R}$ be periodic functions with period $1$. Define $$f(x):=\begin{cases}h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\,,&\text{if }x>\frac{b}{1-a}\,,\\ h_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\,,&\text{if }x<\frac{b}{1-a}\,.\end{cases}$$ $\endgroup$ – Batominovski Oct 7 '18 at 19:27
  • 2
    $\begingroup$ @M.H.Hooshmand There are also solutions when $a< 0$. Let $h:\mathbb{R}\to\mathbb{R}$ be a periodic function with period $1$. Define $$f(x):=h\left(\frac{\ln\left|x-\frac{b}{1-a}\right|}{\ln|a|}\right)+\frac{\ln\left|x-\frac{b}{1-a}\right|}{\ln|a|}\text{ for all }x\neq \frac{b}{1-a}\,.$$ $$\phantom{aaa}$$ For $a=0$, if you assume that $f:\mathbb{R}\to\mathbb{R}$ but the functional equation holds for any $x\neq b$, then there are also solutions. Let $c$ be a fixed real number and define $$f(x):=\begin{cases}c\,,&\text{if }x=b\\ c-1\,,&\text{if }x\neq b\,.\end{cases}$$ $\endgroup$ – Batominovski Oct 7 '18 at 19:37
  • 2
    $\begingroup$ I forgot to say: in the first comment ($a>0$), I also assumed that $a\neq 1$. But that should be trivial. I also would like to remark that all solutions $f:\mathbb{R}\setminus\left\{\dfrac{b}{1-a}\right\} \to\mathbb{R}$ for $a\in(0,\infty)\setminus\{1\}$ are in the form I gave. For $a=0$, I also found all solutions. I am, however, not sure yet whether I got every solutions for $a<0$. I have a tingling feeling that there are other solutions. $\endgroup$ – Batominovski Oct 7 '18 at 19:56
  • 1
    $\begingroup$ @Batominovski Just out of interest: How did you derive the rather impressive solutions in your comments? $\endgroup$ – user7427029 Oct 7 '18 at 19:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.