Solutions to $f(x) = \nabla f(x)\cdot x$

Question

On $\mathbb{R}^n$, consider the differential equation $f(x) = \nabla f(x)\cdot x$ for some $f\in C^1(\mathbb{R}^n)$. This is obviously solved by $f(x)= C\cdot x$ for any $C\in\mathbb{R}^n$. I would like to know if other solutions exist. If I assume that $f\in C^2$, then I get \begin{align*} f(x) = \nabla f(x)\cdot x = \nabla (\nabla f(x)\cdot x)\cdot x = x\cdot \text{Hess}(f)(x)x + \nabla f(x)\cdot x \end{align*} so that $\text{Hess}(f) = 0$ and $f$ must be a linear function with $f(0) = 0$, i.e. $f(x) = C\cdot x$ as desired. However, maybe there exist solutions which are not twice differentiable. I would bet that this is not the case, as say $f(x) = \frac{1}{2}\lvert x\rvert^2$ does not solve the equation, but I do not know for sure. Thanks for any help!

Answer 1

Look here: https://en.wikipedia.org/wiki/Homogeneous_function#Positive_homogeneity

In the article you'll find that:

Positively homogeneous functions are characterized by Euler's homogeneous function theorem. Suppose that the function $f : \mathbb R^n \mapsto \mathbb R$ is continuously differentiable. Then $f$ is positively homogeneous of degree $k$ if and only if

$$kf(x) = \nabla f(x) \cdot x$$