Solving $\text{D} f \cdot x = f(x)$ Let ${D}f$ denote the Jacobian matrix of $f$. What are the solutions of $Df \cdot x = f(x)$? 
In one-dimension, this is just $f' x = f$ with linear functions as the only solutions. In general, linear maps, i.e., functions of the form $A x$ for some constant matrix $A$, do indeed satisfy the equation, but it is unclear whether these are the only solutions.
 A: Euler's homogeneous function theorem states: if $f: \mathbb{R}^n \setminus \{0\} \to \mathbb{R} $ is continuously differentiable, then $f$ is positive homogeneous of degree $k$ ($f(\alpha x) = \alpha^k f(x)$ for $\alpha>0$) if and only if
$$ x \cdot \nabla f(x) = k f(x). $$
The direction you want is proven by considering $g(\alpha) = f(\alpha x)$, then
$$ \alpha g'(\alpha) = \alpha x \cdot \nabla f(\alpha x) = k f(\alpha x) = kg(\alpha), $$
so $\alpha g'(\alpha) = k g(\alpha)$, so $f(\alpha x) = g(\alpha)=\alpha^k g(1) = \alpha^k f(x)$.
In particular, if $k=1$ then $f(\alpha x) = \alpha f(x)$.
So far, we haven't got to linear functions: $f(x)=\lvert x \rvert$ is a perfectly good solution. So impose that $\nabla f(0)=a$ exists. Clearly $f(0)=0$, so we have
$$ \left\lVert \frac{f(h n)-f(0)-h n \cdot a}{h} \right\rVert  \to 0 $$
as $h \to 0$. But using homogeneity gives that the left-hand side is
$$ \left\lVert f(n)- n \cdot a \right\rVert, $$
which must equal $0$. Hence
$$ f(x) = x \cdot a, $$
and $f$ is linear.
