Homogeneous functions In vertue of the Euler's homogeneous function theorem one can characterize a continuously differentiable positive homogeneous function $f$ of dergree $\gamma$ as follows:
$$\sum _{i=1}^{n} {x_i}\partial_i{f(x)}=\gamma f(x).$$
My question is: what about function satisfaying $$\sum _{i=1}^{n} {\alpha_i x_i}\partial_i{f(x)}=\gamma f(x),$$ is it homogeneous? Or does it have a name?
Thanks in advance.
 A: Provided that the domain of $f$ is the positive orthant, it is a function satisfying $f(\lambda^{\alpha_1}x_1,...,\lambda^{\alpha_n} x_n)=\lambda^\gamma f(x_1,...,x_n)$ for all $\lambda>0$.
The proof is a minor modification of that for the Euler identity: just replace the function $g$ in the proof here with $g(\lambda)=f(\lambda^{\alpha_1}x_1,...,\lambda^{\alpha_n} x_n)-\lambda^\gamma f(x_1,...,x_n)$ for a fixed $(x_1,...,x_n)$.
A: If a function is homogeneous of degree 1, $f(\lambda x) = \lambda f(x)$, and differentiation with respect to $\lambda$ gives $ \lambda x \cdot \nabla f(\lambda x)=f(x)$, and evaluating at $\lambda = 1$ gives the Euler identity, $x \cdot \nabla f(x) = f(x)$.  
To make the math come out the way you suggest, you need a new property, like $f( \lambda^{\alpha} x ) = \lambda^{\phi(\alpha)} f(x)$.  Then differentiating with respect to $\lambda$ gives
$$
\sum_{\ell=1}^N \dfrac{\partial f(\lambda^{\alpha}x)}{\partial x_i} \alpha_i \lambda^{\alpha_i-1}x_i = \phi(\alpha) \lambda^{\phi(\alpha)-1}f(x)
$$
As $1^{\phi(\alpha)-1}=1$ and $1^{\alpha-1}=1$ and $1^{\alpha_i}=1$, you get
$$
\sum_{\ell=1}^N \dfrac{\partial f(x)}{\partial x_i} \alpha_i x_i = \phi(\alpha) f(x)
$$
like you want.
