Which functions satisfy $f(cx) = \lambda f(x)$ where $c, \lambda$ are any constants?
The general solution for $c = \lambda$ is given in Steven Stadnicki's answer to this question: Is the function $\frac{x}{2} \cdot (\sin(\ln{x}) - \cos(\ln{x}))$ a fractal?
There are also functions which lead to $\lambda \neq c$.
Take $f(x) = x^b$ for any constant b
Then $c^b x^b = \lambda x^b \implies c^b = \lambda $
What I am particularly looking for is a derivation for the set of functions which solve the equation, which doesn't rely on inspection (if possible).
Extra note:
I have thought of a graphical approach which works for $c,x > 0$ and $\lambda = 1$. Draw any function between $x=1$ and $x=c$ (or specifically, on the interval [1,c) ). Image 1 Now, for every point on our function $(x,f(x)) = (x,f)$, calculate the point $(c x,f)$ and add it to the graph. Now simply repeat over the segment we just added. You can also recursively add the points $(x/c, f)$ to fill in the gap between x=0 and x=1.Image 2
We now have a function $f(x)$ which obeys the equation $f(cx) = f(x), x > 0$.
A plot of f(x) against $\log_c(x)$ will be periodic with period 1 i.e. the following equation will be true:
$f(x) = {\mathcal P}(\log_{c}(x))$ where ${\mathcal P}$ denotes some periodic function, whose periodicity is 1. Image 3
Perhaps then I could guess that the following set of functions work:
$f(x) = x^b {\mathcal P}(\log_c(x))$ where b is any constant
based off the answer we just arrived at above and the answer given by Stadnicki. This would work if $c^b = \lambda$, like before. Is this the general solution for positive c? If so, it would be nice to have an algebraic derivation, in the same vein as Stadnicki's. Also, what about $c < 0$?
Let us suppose $c=e^a>0$ and $x=e^t>0$; the general case will be treated later.
The definition of $a, t$ above immediately leads to
$$ f(e^{t+a}) = \lambda \cdot f(e^t). $$
Defining $\varphi(t)\equiv \log f(e^t)-\frac{\log\lambda}{\log c}t, $ it is easy to see that $\varphi$ is a periodic function with period $a=\log c$. (The definition of $\varphi$ is not ad hoc but can be found in a step-by-step manner, e.g. by testing the functions $f(e^t),\;\log f(e^t),\;\log f(e^t)-bt$ to impose some desirable properties to the function in each step.)
Unwinding the definition of $\varphi$, we know that $f(x)$ is a function of the form
$$ f(x) = x^{\frac{\log\lambda}{\log c}} \cdot e^{\varphi(\log x)},\quad\text{where}\quad \varphi(t+\log c) = \varphi(t). $$
Your solution $f(x)=x^b$ is recovered when $\varphi(t)\equiv 0$ is chosen.
Caveat: One cannot simply take $\log$ of $f$ if $f<0$. Take instead $\log(-f)$ for this case. Finally, if $\lambda<0$, take the logarithm of $e^{i\pi t/a}f(e^t)$ instead of just $f(e^t)$.
Now, one can construct $f(x)$ on $[0,\infty), (-\infty,0]$ individually. Superpose those solutions to get $f$ on the real line. If you want $f$ to be continuous, setting $f(0)=0$ will do.
Lastly, if $c<0$, apply the condition twice to get
$$ f(c^2x)=\lambda^2\cdot f(x). $$
Solve for this equation on $x>0$, following the aforementioned procedure. Extend $f$ onto $x<0$ using the relation $f(cx)=\lambda\cdot f(x)$ and set $f(0)=0$ (if continuity is required). Then $f$ will be consistent all over the real line.