$f(x)\geq 0$, $f^{\prime}(x)>0$ and $\frac{f(x)}{f(\frac{x}{2})}=a$, where $a$ is a fixed constant value. Only $f(x)=x^{b}$ be possible? I meet one problem. The function $f(x)$ satisfies $f(x)\geq 0$, $f^{\prime}(x)>0$ and $\frac{f(x)}{f(\frac{x}{2})}=a$, where $a$ is a fixed constant value. It is easy to see $f(x)=x^{b}$ and $a=2^{b}$ meet these conditions. The question is there any other function satisfies these conditions ?
 A: \begin{equation*}
f(x)\geqslant 0,\;\partial _{x}f(x)>0,\;f(x)=af(\frac{x}{2})
\end{equation*}
We show that
\begin{equation*}
f(x)=x^{b},\;a=2^{b}
\end{equation*}
is the only solution.
Note that
\begin{equation*}
\exp [\mu x\partial _{x}]f(x)=f(e^{\mu }x)
\end{equation*}
Thus
\begin{equation*}
f(x)=f(2\frac{x}{2})=f(e^{\ln 2}\frac{x}{2})=\exp [(\ln 2)x\partial _{x}]f(
\frac{x}{2})=af(\frac{x}{2})
\end{equation*}
Consider the eigenvalue equation
\begin{eqnarray*}
x\partial _{x}h_{\lambda }(x) &=&\lambda h_{\lambda }(x)\Rightarrow \partial
_{x}\ln h_{\lambda }(x)=\lambda \Rightarrow \ln h_{\lambda }(x)=c+\lambda
x\Rightarrow h_{\lambda }(x)=dx^{\lambda } \\
\exp [\mu x\partial _{x}]h_{\lambda }(x) &=&h_{\lambda }(\exp [\mu ]x)=\exp
[\lambda \mu ]h_{\lambda }(x)
\end{eqnarray*}
Then, with $\mu =\ln 2$,
\begin{equation*}
\exp [(\ln 2)x\partial _{x}]h_{\lambda }(x)=h_{\lambda }(2x)=\exp [\lambda
\ln 2]h_{\lambda }(x)=2^{\lambda }h_{\lambda }(x)
\end{equation*}
so
\begin{equation*}
h_{\lambda }(x)=f(\frac{x}{2})\Rightarrow f(x)=2^{\lambda }f(\frac{x}{2})=af(%
\frac{x}{2})\Rightarrow a=2^{\lambda }
\end{equation*}
Setting
\begin{equation*}
a=2^{b}\Rightarrow \lambda =b
\end{equation*}
and
\begin{equation*}
f(\frac{x}{2})=d(\frac{x}{2})^{b},\;f(x)=dx^{b}
\end{equation*}
