Two functional equations

Question

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?

Answer 1

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.

Answer 2

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.