# Are homothetic additively separable preferences always equivalent to CES?

Are homothetic additively separable preferences always a monotonic transformation of CES preferences?

In technical language, the question is the following:

Let $$n>1$$, and let $$f:\mathbb{R}^n_{\ge 0}\rightarrow\mathbb{R}_{\ge 0}$$ be continuously differentiable, concave, and homogeneous of degree one. Here, homogeneity of degree one means that for all $$s\in\mathbb{R}_{\ge 0}$$, and $$x\in\mathbb{R}^n_{\ge 0}$$, $$f(sx)=sf(x)$$.

And suppose that there exist continuously differentiable monotonic increasing functions $$g:\mathbb{R}\rightarrow\mathbb{R}_{\ge 0}$$ and $$h_1,\dots,h_n:\mathbb{R}_{\ge 0}\rightarrow\mathbb{R}$$ such that for all $$x\in\mathbb{R}^n_{\ge 0}$$: $$f(x)=g(h_1(x_1)+\cdots+h_n(x_n)).$$

Must it be the case that there exists $$a_1,\dots,a_n\in\mathbb{R}$$ and $$\rho,b_1,\dots,b_n\in\mathbb{R}_{\ge 0}$$ such that for all $$i\in\{1,\dots,n\}$$, $$h_i(x)=a_i+b_i \frac{x^{1-\rho}-1}{1-\rho}$$? (Where, when $$\rho=1$$, we understand this as stating $$h_i(x)=a_i+b_i\log(x)$$.)

Note: This was first posted to math.se here https://math.stackexchange.com/questions/4648187/homogeneous-of-degree-one-functions-that-are-a-monotonic-transformation-of-an-ad but I think I have a higher chance of getting an answer here.