Homogeneous functions

March 21st, 2005 ~ Posted in: General

A function f:\R^m \rightarrow R^k is homogeneous of degree n if f(t\vec{x})=t^nf(\vec{x}). I’ve seen this definition before, in the proof of some inequality about integrals (a really famous one that I can’t recall the name of), but since I didn’t understand the proof, never really got interested in homogeneity until today.

I came across it again in a problem in a book: if z=f(x,y) is homogeneous of degree n, then x\frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x,y) . I still haven’t figured out how to prove this, but it caused me to examine the idea of homogeneity. Like, does this problem assume that f is continuous, or is that a result of the fact it’s homogeneous?

First I tried looking for individual examples of homogeneous functions, to get an overview of what they look like, and I came up with some: (x-y)^n, \frac{(x)^{n+1}}{y}, etc. But then I realized from the first example, that a metric has to be homogeneous of degree 1, by definition— or I thought I did, until I realized this holds true only if t\geq 0.

Then I tried generalizing my thought process :) to higher dim spaces, and realized that because of homogeneity, f has only to be defined on the unit ball, and then for any \vec{x}\in \R^m, f(\vec{x}) = \|x\|^n f(\vec{u}) for some \vec{u} in the unit ball.

This gives us a way to construct arbitrary homogeneous functions, given just their values on the unit ball. Of course, you need to be careful that for each \vec{u}, f(-\vec{u})=(-1)^n f(\vec{u}). So, if f: \R \rightarrow \R and n odd (even), then f has to be odd (even) to be homogenous of degree n, for example.

I realized from this ability to construct arbitrary homogeneous functions that homogeneity does not imply continuity, so the problem that got me started isn’t well-stated.

All of these observations can be made directly from the definition of homogeneity, but I think the thought process that lead to them in this particular instance was pretty neat.

General ~

One Response to “Homogeneous functions”

  • 1. Marc
    March 23rd, 2005 at 8:22 am

    Did you prove the original problem? ie. say f:R^n -> R and f(tx) = t^n f(x). You want to prove in
    general that

    = nf(x)

    Hint: Differentiate the homogentity condition with respect to ‘t’ using the chain rule.
    denotes the standard inner product in R^n.

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