This is the multivariate follow-up question to Examples of (univariate) locally homogeneous functions.
Positive homogeneous functions are functions $f:\mathbb{R}^d\to\mathbb{R}$ that satisfy $$ f(c\mathbf{x}) = cf(\mathbf{x}) $$ for all $\mathbf{x}\in \mathbb{R}^d$ for $c>0$. Now I was reading the paper http://arxiv.org/abs/1611.02862, and there the definition of locally positive homogeneous functions was given as $$ f(c\mathbf{x}) = cf(\mathbf{x}) $$ for all $x\in\mathbb{R}^d$ and $|c-1|\leq \epsilon\ll 1$.
In Examples of (univariate) locally homogeneous functions we got that the univariate functions are continuous, piecewice linear functions, and that global and local homogeneous functions are the same. Do we get the same result for multivariate functions, i.e. are the sets of globally and locally positive homogeneous functions the same?
What do I know of these functions:
- $\nabla_\mathbf{x}f(\mathbf{x})\cdot \mathbf{x}=f(\mathbf{x})$
- $f(\mathbf{0})=0$
We can show that again global = local.
Let $z>0$. There exists a $k\in\mathbb{N}$ and an $\epsilon\in\mathbb{R}$ with $|\epsilon|\ll1$ such that $z=(1+\epsilon)^k$. Set $c=1+\epsilon$, and observe that for all locally positive homogeneous functions we have that $$ f(c\mathbf{x})=cf(\mathbf{x}) $$ for all $\mathbf{x}\in\mathbb{R}^d$. Observe that $$ zf(\mathbf{x}) = c^kf(\mathbf{x}) = c^{k-1}f(c\mathbf{x}) = c^{k-2}f(c^2\mathbf{x}) = \dots = f(c^k\mathbf{x}) = f(z\mathbf{x}) $$ holds for all $\mathbf{x}\in\mathbb{R}^d$. Since $z$ was arbitrary, we must have that $$ zf(\mathbf{x}) = f(z\mathbf{x}) $$ for all $\mathbf{x}\in\mathbb{R}^d$ and $z>0$. But this is exactly the definition of a positive homogeneous function. Therefore, all locally positive homogeneous functions are globally homogeneous functions, and vice versa.