# Does Euler's Theorem for homogeneous functions require continuous differentiability?

Euler's Theorem for homogeneous functions states that $$f: \mathbb{R}^n \to \mathbb{R}$$ is homogeneous of degree $$k$$ ($$f(cx) = c^k f(x)$$ for all $$c > 0$$, $$x \in \mathbb{R}^n$$), if and only if the partial derivatives of $$f$$ satisfy $$k f(x) = \sum_{i=1}^n x_i \frac{\partial f(x)}{\partial x_i}$$ Clearly this theorem requires differentiability of $$f$$. However, it seems to often be stated with the requirement of continuous differentiability, e.g., here. On the other hand, it is also sometimes stated without this requirement, and I don't see how any of the proofs actually use continuity of partial derivatives (e.g., the proof here).

I would greatly appreciate if anyone could clear this up!

Fix a point $$x$$ where $$f$$ is differentiable. Consider the function $$g(c)=f(cx)$$.
Homogeneity implies that if $$f$$ is differentiable at $$x$$, then it is differentiable at $$tx$$ for every $$t\neq 0$$, because $$\begin{eqnarray*} f(tx+h)-f(tx)&=&t^k(f(x+h/t)-f(x))\\ &=&t^{k}(f'(x)(h/t)+o(\|h\|_2/t))\\ &=&t^{k-1}f'(x)(h)+t^ko(\|h\|_2/t)\\ \end{eqnarray*}$$ so that $$f'(tx)=t^{k-1}f'(x)$$ for every $$t\neq 0$$. The notation $$f'(x)(w)$$ in this context means the dot product between the gradient of $$f$$ at $$x$$ and the vector $$w$$.
It follows that $$g$$ is differentiable at every $$c\neq 0$$, and by the chain rule, $$g'(c)=f'(cx)(x)=c^{k-1}f'(x)(x)$$ On the other hand, since $$g(c)=f(cx)=c^{k}f(x)$$, direct differentiation gives $$g'(c)=kc^{k-1}f(x)$$, and we deduce the announced formula, $$kf(x)=f'(x)(x)$$.