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Is there a systematic approach that can be used to solve these two functional equations?

$$af(x) = f(bx), \quad\text{where }\ f\colon \mathbb{R}\to\mathbb{R} \tag{1}$$

$$ag(y) + ay = g(ay),\quad\text{where }\ g\colon\mathbb{R}\to\mathbb{R}\tag{2}$$

I pluged them into wolfram and got the result, but I am not sure how to do it systematically. Moreover, how to prove that the solutions are unique?

MarianD
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pizet
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  • Are there any other restrictions on these functions? In particular, that $f$ and $g$ are continuous? If not, then there are uncountably many functions that satisfy either of these equations. – Paul Sinclair Mar 18 '19 at 23:35
  • Sorry, yes, I would like the functions to be continuous. I would like to get the solutions that wolfram gives. :) – pizet Mar 19 '19 at 08:45
  • Well as I point out in my post, there are uncountably many solutions. I do not know what additional assumptions Wolfram is making. Perhaps that $f$ is analytic? – Paul Sinclair Mar 19 '19 at 15:46
  • By the way, it is related to the following question https://math.stackexchange.com/questions/3151352/invariant-sets-of-2-by-2-linear-mapping?noredirect=1#comment6494265_3151352. These functional equations arise when you try to determine the invariant curves of a linear mapping given by a 2 by 2 matrix. – pizet Mar 20 '19 at 12:05
  • Okay, but as I've never studied dynamical systems, I have no idea what invariance you are talking about. – Paul Sinclair Mar 20 '19 at 17:11

2 Answers2

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What I said in the comment is a common condition that must be satisfied to get a unique solution in a functional equation (that is, unique up to a finite number of parameters). However, in this case, it is not sufficient. Your condition in (1) is just not restrictive enough.

It is very easy to see by induction that if $a\ne 0, b \ne 0$, then for any $x$ and any $n \in \Bbb Z$, $$f(b^nx) = a^nf(x)$$ Suppose $b > 1$, and let $\varphi : (-b, -1] \cup [1,b) \to \Bbb R$ be an arbitary function. Define $$\mu: (0, \infty) \to \Bbb Z : x \mapsto \min\{n \in \Bbb Z \mid b^nx \ge 1\}$$ Note that for all $x > 0, 1 \le b^{\mu(x)}x < b$.

Then we can define a function $f : \Bbb R \to \Bbb R$ by $$f(x) = a^{-\mu(|x|)}\varphi(xb^{\mu(|x|)})$$ and $f(0) = 0$.

No matter what $\varphi$ we pick, we get a function that satisfies your equation. If we want $f$ to be continuous, then all we need to do is require that

  1. $\varphi$ is continuous.
  2. $\lim_\limits{x \to b} \varphi(x) = a\varphi(1)$ and $\lim_\limits{x \to -b}\varphi(x) = a\varphi(-1)$

There are still plenty of choices for $\varphi$ that satisfy these conditions. If $b < 1$, the only difference is that $\varphi$ is defined on $[-1, -b) \cup (b, 1]$ and $\mu(x) = \min\{n \in \Bbb Z \mid b^nx \le 1\}$.

I suspect the same is true for the second equation, but I haven't bothered to examine it closer.

Paul Sinclair
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There are some standard elementary tricks for such functional equations. Mostly just experience and intution and trying obvious things. For example, in equation (2) if $\,y=0\,$ then $\,a g(0) = g(0)\,$ which forces $\,a=1\,$ if $\,g(0)\ne 0\,$ and $\,g(y)+y = g(y)\,$ which can't be true for $\,y\ne 0\,$, and so define $\,f(x) := g(x)/x\,$ if $\,x\ne0.\,$ But now assuming $\,ay\ne 0\,$ we get $\,1+f(y) = f(ay)\,$ which reminds us of logarithms. So $\,f(y) = \log_a(y) + b\,$ is one possible solution depending on $\,b.\,$ Uniqueness usually comes only if the functions are "nice" enough, for example, continuous, otherwise there are "bad" counterexamples. In our particular case of $\,f(y)\,$ the $\,b\,$ is arbitrary and hence not unique, but since $\,f(1)=b\,$ even if the value of $\,f(1)\,$ were given, then there would still be "bad" counterexamples.

Somos
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