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Let $0<a,b<\frac{1}{2}$.

Find all continuous functions $f$ satisfying $f(f(x))=af(x)+bx$.

I proved $f$ is monotonic, but stucked at this point.

How to use the bounds for $a,b$, why is it important to have $a,b\in(0,\frac{1}{2})$?

Ennar
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tong_nor
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  • Well, if $f(x)=\lambda x$ for some constant $\lambda$ then we must have $\lambda = \frac {a\pm \sqrt {a^2+4b}}2$. For $a,b>0$ we always have those solutions anyway. – lulu Sep 24 '17 at 17:37
  • You can show $f$ is injective – Jihoon Kang Sep 24 '17 at 17:40
  • yes, continuous strictly monotonic function is injective, but how to exploit this? – tong_nor Sep 24 '17 at 17:44
  • Indeed $g(y)=\frac{f(y)-ay}{b}$ is a right inverse of $f(x).$ – Thomas Andrews Sep 24 '17 at 17:45
  • @tong_nor you can also show that if $f$ crosses the line $x=y$, it can only do so at $0$. This requires the use of the range of $a,b$ so this might be helpful. – Jihoon Kang Sep 24 '17 at 17:46
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    One thing to notice is that if we define $x_n = f^{\circ n}(x) = f(f(\ldots (f(x))\ldots))$, we have that $x_{n+2} = ax_{n+1}+bx_n$ and $0<a,b<1/2$ makes sure that this sequence converges to $0$. But also by continuity it follows that $0$ is fixed point of $f$, i.e. $f(0) = 0$. Also, obviously $0$ is the only fixed point of $f$. Perhaps Jihoon Kang had something similar in mind. – Ennar Sep 24 '17 at 19:06

2 Answers2

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Assume $f(x_1)=f(x_2)$. Then $$ bx_1=f(f(x_1))-af(x_1)=f(f(x_2))-af(x_2)=bx_2$$ and hence (as $b\ne 0$) $x_1=x_2$. We conclude that $f$ is injective. As $f$ is also continuous, $f$ is strictly monotonous. Thereefore $f\circ f$ is strictly increasing.

Assume $L:=\lim_{x\to +\infty}f(x)$ exists. Then $f(L)=\lim_{x\to+\infty}(af(x)+bx)=\infty$, contradiction. Similarly, $\lim_{x\to -\infty}f(x)$ cannot exist. We conclude that $f$ is a continuous bijection $\Bbb R\to\Bbb R$.

Pick $x_0\in \Bbb R$ and define recursively $x_{n+1}=f(x_n)$. Then $x_{n+2}=ax_{n+1}+bx_n$ for all $n$. It is well-known that the solutiosn of this recursion are of the form $$x_n=\alpha \lambda_1^n+\beta\lambda_2^n, $$ where $\lambda_{1,2}=\frac{a\pm\sqrt{a^2+4b}}{2}$ are the solutions of $X^2-aX-b=0$ and $\alpha,\beta$ are found by solving $x_0=\alpha+\beta$, $x_1=\alpha\lambda_1+\beta\lambda_2$. From the given restrictions for $a,b$, we find $-\lambda_1<\lambda_2<0<\lambda_1<1$. This implies that $x_n\to 0$, independent of $x_0\in\Bbb R$. By continuity, $f(0)=0$.

Assume $x_n>0$ and $x_{n+2}\ge x_n$. As $f\circ f$ is strictly increasing, this implies that the subsequence $x_{n+2k}$ is non-decreasing, contradicting convergence to $0$. Therefore $x_n>0$ implies $x_{n+2}<x_n$. Similarly, $x_n<0$ implies $x_{n+2}>x_n$.

As $f$ is bijective, we can extend the sequence $x_n$ no $n\in\Bbb Z$.

  • If $\alpha\ne 0$, the $\lambda_1^n$ terms is dominant for $n\gg 0$. In particular, all $x_n$ with $n\gg 0$ have the same sign. It follows that the sequences $x_{2n}$ and $x_{2n+1}$ are both decreasing or both increasing.

  • If $\beta\ne 0$, the$\lambda_2^n$ term is dominant for $n\ll 0$. In particular, $x_n$ flips signs as $n\ll 0$. It follows that the sequences $x_{2n}$ and $x_{2n+1}$ have opposite monotonicity.

Both these observations together imply that $\alpha$ and $\beta$ cannot both be non-zero. Therefore, either $f(x)=\lambda_1x$ or $f(x)=\lambda_2 x$. By continuity of $f$, a switch between these two conditions can at most occur at $x=0$, but then $f$ would not be monotonous. We conclude that no switch occurs, so that the only possible solutions are $$ f(x)=\lambda_1x\text{ for all }x\in\Bbb R$$ and $$ f(x)=\lambda_2x\text{ for all }x\in\Bbb R.$$

Hagen von Eitzen
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  • Few points are unclear/ inconsistencies in proof: (1) In your $\beta \neq 0$, you conclude that $x_{2n}$ and $x_{2n+1}$ ahave opposite monotonicity. But as $f \circ f$ is strictly increasing, $x_{2n}$ and $x_{2n+1}$ must be both strictly increasing. So this is a contradiction in itself i.e. $\beta$ must always be $0$. And if so, you miss the negative solution. (2) You have shown that for the set of points $x_n$, you have $f(x) = \lambda_1 x$. But how do you show using continuity that forall x, $f(x) = \lambda x$? – Anon Jun 09 '20 at 06:23
  • (3) In your proof of $f$ being bijective, it is not sufficient to show that the + and - infinity limits do not exists. You also need to show that these limits go to opposite infinities. – Anon Jun 09 '20 at 06:24
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Consider a point $\tilde{x}$ satisfying $f(\tilde{x})=c\tilde{x}$. We note that this can be true of any point $\tilde{x}$ for arbitrary $c$, we find \begin{equation} f(f(\tilde{x}))=f(c\tilde{x})=af(\tilde{x})+b\tilde{x}=(ac+b)\tilde{x}=(a+bc^{-1})(c\tilde{x}), \end{equation} which gives an iterative relation between c and $\tilde{c}$ for the next iterate. Define the iterative function $\tilde{f}(c)=a+bc^{-1}$. Upon iterating this function many times, one approaches one of the fixed points, which are solutions to the quadratic $c^2-ac-b=0$, which are $\frac{a\pm\sqrt{a^2+4b}}{2}$. Therefore, upon iterating arbitrarily many times, one approaches a point for which $f(\tilde{x})\approx c^*\tilde{x}$, where $c^*$ is one of the solutions of the quadratic.

The iterative equation $\tilde{f}$ has an inverse given by $\tilde{f}^{-1}(y)=\frac{b}{y-a}$, with the same fixed points. Therefore, iterating backwards or forwards from any point, one always eventually gets to a point where $f(\tilde{x})\approx c^*\tilde{x}$. Therefore, iterating either forward or backward, one can always get to a contraction mapping that approaches $x=0$. Choosing $x$ infinitesimally small, one can iterate back out, and of course it also must approach a similar $c^*$. Therefore the function should be homogeneous of the form $f(x)=c^*x$ for the solutions of the quadratic.

I don't know if this is completely rigorous but I find it convincing at least.

Teddy Baker
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