TITLE: Geometric interpretation of Euler's identity for homogeneous functions
QUESTION [14 upvotes]: A function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is called homogeneous of degree $d \geq 0$ if $$f(\lambda x_1, \ldots, \lambda x_n ) = \lambda^d f(x_1, \ldots, x_n)$$  Differentiating both sides with respect to $\lambda$ and then plugging in $\lambda=1$, we obtain the following equality:
$$ \sum_{i=1}^n x_i \frac{ \partial f}{\partial x_i}(x_1, \ldots, x_n) = d \cdot f(x_1, \ldots, x_n) $$ This equation is usually called "Euler's identity." It feels it should have a clean geometrical interpretation, but I'm blanking out on what it might be.

REPLY [4 votes]: Think at it in spherical polar coordinates. For any $0\ne x\in \mathbb{R}^n$, write 
$$
x=r\omega, 
$$
where $r=\lvert x\rvert$ and $\omega=x/\lvert x\rvert$. The function $f\colon \mathbb R^n\to \mathbb R$ is $d$-homogeneous if and only if 
$$\tag{1}
f(r\omega)=r^d f(\omega),$$
and the operator $x\cdot \nabla$ reads
$$
x\cdot \nabla = r\frac{\partial}{\partial r}.$$
So, if $f$ satisfies (1), then 
$$\tag{2}
x\cdot \nabla f (x) = r\partial_r(r^d) f(\omega) = d\cdot f(x).$$ 

EDIT 2018 Actually, the point here is that $x\cdot\nabla$ is the generator of dilations, that is, 
$$
\left.\partial_\epsilon f(e^\epsilon x)\right|_{\epsilon=0} = x\cdot \nabla f(x).$$ 
With an old-fashioned language we can say that $x\cdot \nabla$ is an infinitesimal dilation. Equation (1) expresses a certain behavior with respect to finite dilations, while equation (2) expresses the analogous behavior with respect to infinitesimal dilations. Euler's theorem is the statement that the two points of view are equivalent.
Let me make an example, a kind of a rotation Euler's theorem. Consider in $\mathbb R^3$ the operator $R_z:= x\partial_y - y\partial_x$. This operator is an infinitesimal rotation around the $z$-axis, because 
$$
\partial_\epsilon f( \cos( \epsilon) x - \sin(\epsilon) y, \sin (\epsilon) x +\cos (\epsilon) y, z)|_{\epsilon=0}=R_z f(x, y, z).$$
Now it turns out that 
$$
R_z f(x, y,z ) =0, \quad \iff\quad f=f(x^2+y^2, z).$$ 
Here, again, $f=f(x^2+y^2, z)$ is a global behavior under finite rotations: it expresses the fact that $f$ is invariant under rotations around the $z$ axis. The infinitesimal version of this is $R_z f=0$. 
It is in practice much easier to verify a local statement, such as the latter, instead of a global one.