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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?

daw
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malenkiy_scot
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1 Answers1

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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 $(0,1]$.

Hagen von Eitzen
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  • Can you provide an example of such smooth and monotonic $f_0$ which is not $x^l$? – malenkiy_scot Aug 11 '14 at 15:19
  • Try $f(x) = {x}^{l} \left( 1+a\sin \left( 2,{\frac {\pi ,\ln \left( x \right) }{\ln \left( m \right) }} \right) \right) $ where $a$ is sufficiently small: I think $|a| < \dfrac{\ell \ln(1/m)}{\ell \ln(1/m) + 2 \pi}$ will do. – Robert Israel Aug 11 '14 at 15:32