Consider the following functional equation:
\begin{equation}f(x)=kf(mx)\end{equation}
where $x \in [0,1]$; $k>0$; $0<m<1$; and $f(1)=1$.
It has at least one solution: $f(x)=x^l$, where $l=log_m({1/k})$.
Assume further that $f$ is smooth on $(0,1)$. Is $f$ unique? If not, does adding monotonicity restriction on $f$ guarantee uniqueness?
Let $f_0\colon [m,1]$ be any continuous (and maybe monotonic) function with $f(m)=\frac1k$, $f(1)=1$. Then we can define $f(x)=k^{-r}f_0(m^{-r}x)$ where $r=\lfloor\log_mx\rfloor $ and obtain a continuous solution on $(0,1]$ and every continuos solution on $(0,1]$ can be obtained this way. If $k>1$, this solution extends to $[0,1]$. If $k<1$, there cannot be a continuous solution as necessarily $f({m^r})=k^{-r}\to\infty$. If $k=1$, the solution is continuous at $0$ only if it is constant.
If we require not only that $f_0$ is continuos, but is smooth and the derivatives at $m$ and $1$ "fit", we obtain a smooth solution on $(0,1]$.