A solution for $f(ax+b)=f(x)+1$

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.

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.

• @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. – Batominovski Oct 8 '18 at 18:56

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 $$t\in\mathbb{R}$$.

• @ Batominovski . Thanks for your answer. Any solution if we assume that $f:\mathbb{R}\setminus\{\frac{b}{1-a}\}\rightarrow \mathbb{R}$? – M.H.Hooshmand Oct 7 '18 at 19:12
• @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 $$f(x):=\begin{cases}h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\,,&\text{if }x>\frac{b}{1-a}\,,\\ h_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\,,&\text{if }x<\frac{b}{1-a}\,.\end{cases}$$ – Batominovski Oct 7 '18 at 19:27
• @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 $$f(x):=\begin{cases}c\,,&\text{if }x=b\\ c-1\,,&\text{if }x\neq b\,.\end{cases}$$ – Batominovski Oct 7 '18 at 19:37
• 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. – Batominovski Oct 7 '18 at 19:56
• @Batominovski Just out of interest: How did you derive the rather impressive solutions in your comments? – user7427029 Oct 7 '18 at 19:59