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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 ?

congmingniao
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1 Answers1

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\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*}

Urgje
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