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Let $f:S\subseteq \Bbb R^n\to\Bbb R$. One can prove that if $f(\lambda {\bf x})=\lambda^pf({\bf x})$ for each ${\bf x}\in S$ such that $\lambda {\bf x}\in S$, then ${\bf x}\cdot \nabla f({\bf x})=pf({\bf x})$. The proof is not complicated: one defines the function $\varphi(\lambda)=f(\lambda {\bf x})$ for a fixed ${\bf x}$ and evaluates $\varphi'(1)$ in two different ways. I'd like to get a hint to prove the converse: if ${\bf x}\cdot \nabla f({\bf x})=pf({\bf x})$ for each ${\bf x}\in S$, $S$ open, then $f$ is homogeneous of degree $p$ in $S$.

This problem is on Apostol's Mathematical Analysis.

Pedro
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2 Answers2

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Use the same $\varphi(\lambda):=f(\lambda x)$ and the equation to get a differential equation in $\varphi$. You can get that $$p\varphi(\lambda)=\lambda \varphi'(\lambda).$$

Use now that the value of $\varphi(1)=f(x)$ to get the solution for $\lambda^p f(x)=\varphi(\lambda)=f(\lambda x)$.

OR.
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Or consider $\psi(\lambda)=\lambda^{-p}f(\lambda x)$.

Ted Shifrin
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