Functions for which $\lambda f(x)=f(\alpha_\lambda x + \beta_\lambda)$

Question

How many functions $f:\mathbb{R}\to R\subseteq\mathbb{R}$ are there such that there is an $S\subseteq\mathbb{R}$ containing an interval containing $1$ such that for any $\lambda\in S$ there exist $\alpha_\lambda,\beta_\lambda\in\mathbb{R}$ with $\lambda f(x)=f(\alpha_\lambda x+\beta_\lambda)$ for all $x\in\mathbb{R}$?

Obviously, $f(x)=x$ satisfies this with $R=S=\mathbb{R}$, $\alpha_\lambda=\lambda$, $\beta_\lambda=0$, and more generally $f(x)=|x|^s$ satisfies it with $R=S=(0;\infty)$, $\alpha_\lambda=\lambda^{1/s}$, $\beta_\lambda=0$.

Similarly, $f(x)=\exp(x)$ satisfies this with $R=S=(0;\infty)$, $\alpha_\lambda=1$, $\beta_\lambda=\log\lambda$, and more generally $f(x)=\exp(rx)$ satisfies it with $R=S=(0;\infty)$, $\alpha_\lambda=1$, $\beta_\lambda=\frac{1}{r}\log\lambda$.

But are there any other families of such functions? I see no obvious way to find them.

Answer 1

I will construct $2^\mathfrak{c}$ examples with $S=\mathbb{R}$ and $\beta_\lambda = 0$ for all $\lambda$, clearly this is the biggest we can hope for since this is how many functions $\mathbb{R}\to \mathbb{R}$ are there. We will demand our functions being odd, for $\lambda = 0$ we can take $a_\lambda = 0$, for $\lambda < 0$ we will take $a_\lambda = - a_{-\lambda}$. So, it remains to deal with $\lambda > 0$. Additionally, our functions will be non-zero for $x \neq 0$.

Put $g(x) = \log f(e^x)$. Function $g$ uniquely determines function $f$ (since we assumed that $f$ is odd). Then the equation reads $g(x)+\log \lambda = g(x+\log a_\lambda), x\in \mathbb{R}$ (we will also assume that $a_\lambda > 0$ for $\lambda > 0$).

Pick a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, which has cardinality $\mathfrak{c}$. Since $\mathfrak{c} = \mathfrak{c}^2$, let us index it by $\mathfrak{c}^2$. By grouping them by the first coordinate we get an isomorphism of Abelian groups $\mathbb{R} = \bigoplus_{t\in \mathbb{R}} \mathbb{R}_t$ (subscript $t$ only to avoid ambiguity). On each of these $\mathbb{R}_t$ we can put $g(x) = c_t x$ for any $c_t>0$ and then extend by linearity to the whole $\mathbb{R}$ (so, our function $g$ would be $\mathbb{Q}$-linear! So the function $f$ is multiplicative $f(ab)=f(a)f(b)$). The number of choices we have is $\mathbb{R}^\mathbb{R} = 2^{\mathfrak{c}}$.

For given $\lambda$ let us pick any $t$ and consider the element of the corresponding $\mathbb{R}_t$ which in it is equal to $\frac{\log \lambda}{c_t}$, its preimage under the isomorphism above will be $\log a_\lambda$ that we want (so, $a_\lambda$ is its exponent). Since we can do this for any $t$, in particular for any $\lambda \neq 0$ there are continuum-many choices of $a_\lambda$.

Bottom line: without any additional conditions like continuity or measurability or something, nothing useful can be said likely.