Extension of Euler's Theorem for Homogeneous Functions

Question

We know that Euler's Theorem for Homogeneous Functions states that for homogeneous functions of order k:

$f\left ( \bar{x} \right )=\frac{1}{k}\sum_i x_i\frac{\partial f\left ( \bar{x} \right ) }{\partial x_i} $

I was wondering if the above could be extended for the case where $x_i$ itself is a vector. Could we write the above as:

$f\left ( \bar{x} \right )=\frac{1}{k}\sum_i\sum_j x_{i}^{j}\frac{\partial f\left ( \bar{x} \right ) }{\partial x_{i}^{j}}$
Where
$x_{i}^{j}$ is the jth constituent of $\bar{x_i}$

It seems intuitively correct, and I was able to do it for an elementary homogeneous function, but am unable to prove it mathematically.

Answer 1

If each $x_i$ is a vector (say in $\mathbb R^n$) for $i = 1,\ldots,m$, then $x = (x_1, \ldots, x_m)$ can be thought of as an $m \times n$ matrix, or equivalently an element of $\mathbb R^{m \times n};$ and the functions $x_i^j$ (for $(i,j)\in\{1\,\ldots,m\}\times\{1,\ldots,n\}$) are just the Cartesian coordinate functions of $\mathbb R^{m \times n}$.

Assuming the scalar multiplication you're using in the definition of homogeneity is just componentwise (i.e. $(kx)_i^j = k(x_i^j)$), we see that your second equation is just the standard Euler's theorem for a function defined on $\mathbb R^{m \times n}.$