Homogeneous Functions

We start with equivalence of two integrals . I assume here that p is a real nonzero number. For p = 1, the second integral reduces to the first and, therefore, is a clear generalization. However, once we know how to compute the more special integral, we may, using a standard substitution, compute the more general one.

Indeed, let u = x/p. Then successively x = pu, dx = pdu, and

Since

this leads to

The situation, where one parameter appears superfluous, is quite common to the class of homogeneous functions.

Definition

A function f(x, y) such that, for t>0, f(tx, ty) = t^af(x, y), is called (positively) homogeneous of order (or degree) a.

Examples

f(x, y) = x² + y² is homogeneous of order 2. Indeed,
f(tx,ty) = (tx)² + (ty)² = t²(x² + y²) = t²f(x,y).
f(x, y) = (x² + y²)/x is homogeneous of order 1
f(x, y) = √x² + y² is also homogeneous of order 1
f(x, y) = (x² - y²)/(x² + y²) has order 0.

Functions in examples 1 and 3 are also symmetric because they do not change when one swaps values of x and y.

The famous Binomial Theorem (discovered and repeatedly used but not proved by I. Newton) may be written in two equivalent forms:

All terms in the latter are homogeneous of order a.

The notion of homogeneity extends to functions of more than 2 variables. For example, all kinds of means are symmetric and naturally homogeneous of order 1. For N variables,

Arithmetic mean, a(x, y, z, ...) = (x + y + z + ...)/N
Geometric mean, g(x, y, z, ...) = (x·y·z· ...)^1/N
Harmonic mean, h(x, y, z, ...) = N/(1/x + 1/y + 1/z + ...)

A famous inequality relates arithmetic and geometric means of nonnegative numbers: a(x, y, z, ...) ≥ g(x, y, z, ...). If we can prove this inequality for powers of 2, N = 2ⁿ, then from a previous result it will follow for all integer N. Let's use induction.

Let n = 1. The inequality is then equivalent to (x + y)² ≥ 4xy which is true because, by rearranging terms, it reduces to (x - y)² ≥ 0 which is obviously true.

For the inductive step, assume the inequality has been proven for N = 2^k, and let there be given N = 2^k+1. Note that N = 2^k+1 = 2·2^k = 2M. Thus we may split the given set of N numbers into two groups of M elements each. Let these be x, y, z, ... and u, v, w, ... We have

a(x, y, z, ..., u, v, w, ...) = (x + y + z + ... + u + v + w + ...)/N

= [(x + y + z + ...)/M + (u + v + w + ...)/M]/2

= (a(x, y, z, ...) + a(u, v, w, ...))/2

= a(a(x, y, z, ...), a(u, v, w, ...)).

Quite similarly g(x, y, z, ..., u, v, w, ...) = g(g(x, y, z, ...),g(u, v, w, ...)). Since, by the inductive assumption,

a(x, y, z, ...) ≥ g(x, y, z, ...) and a(u, v, w, ...) ≥ g(u, v, w, ...)

we finally have

a(x, y, z, ..., u, v, w,...) = a(a(x, y, z, ...), a(u, v, w, ...))

≥ a(g(x, y, z, ...), g(u, v, w, ...))

≥ g(g(x, y, z, ...), g(u, v, w, ...))

= g(x, y, z, ..., u, v, w, ...).

Q.E.D.

References

G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Cambridge University Press (2^nd edition) 1988.
G. Polya, Mathematical Discovery, John Wiley & Sons, 1981

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