HOMOGENEITY AND INEQUALITIES

KENNETH MAPLES

1. Homogeneity of functions

Let’s consider a function f : Rn → R from n-dimensional Euclideanspace to the real numbers.

Definition. A function f : Rn → R is (positively) homogeneous ofdegree k if

f(αv) = αkf(v)

for all scalars α > 0 and nonzero vectors v ∈ Rn.

Homogeneous functions are severely constrained. Suppose that weknow the values of f for any vector on the unit sphere. Let v ∈ Rn beany nonzero vector (not necessarily on the unit sphere). We can findthe value of f at v with homogeneity, as

f(v) = f

(‖v‖ v

‖v‖

)= ‖v‖kf

(v

‖v‖

)Because v/‖v‖ has length 1 the value of f(v/‖v‖) is known, so we havederived the value of f(v). In fact, the function is uniquely (and freely)determined by its values on the unit sphere.

Note that although the values of the function are constrained on eachray starting at the origin, this does not imply any relationship betweenfunction values on different rays. In particular, the function might stillbe highly discontinuous.

Here are some examples of homogeneous functions:

• Let ‖·‖2 : Rn → [0,∞) be the usual Euclidean norm. Thenthe homogeneity property of the norm precisely says that thefunction is homogeneous of degree 1; in fact,

‖αv‖2 =√α2v2

1 + · · ·+ α2v2n = α‖v‖2

for any α > 0.

Date: January 31, 2008.1

2 KENNETH MAPLES

• We have a similar result for other norms on Euclidean space:for example, if we consider the 1 norm,

‖αv‖1 = |αv1|+ · · ·+ |αvn| = α‖v‖1• Any polynomial that is homogeneous of degree k is also homo-

geneous by our definition. In fact, we can define the polynomialp : Rn → R by

p(x1, ..., xn) =∑|α|=k

cαxα11 · · ·xαn

n

where the sum is taken over all n-tuples (α1, ..., αn) such thateach αj is a nonnegative integer and |α| = α1 + · · · + αn = k(such tuples are called multi-indices) For example,

p(x1, x2) = x21 + 2x1x2 + x2

2

is a homogeneous function of degree 2.• Not all polynomials are homogeneous. In fact, q(x) = x+ x2 is

a polynomial of one variable which is not homogeneous.

2. Homogeneity of inequalities

Homogeneity solves problems in the theory of inequalities by reduc-ing complicated expressions to a form we can manipulate.

To illustrate the technique, we will prove the Cauchy-Schwarz in-equality; for numbers x, y ∈ Rn,

n∑k=1

xkyk ≤

√√√√ n∑k=1

x2k

√√√√ n∑j=1

y2j

Putting on our homogeneity thinking caps, we notice that each sideof the equation is a homogeneous function of degree 1 in both of thevectors x and y. We therefore suspect that leveraging this homogeneitywill make the inequality easier to prove.

To this end, we will prove the inequality in two steps:

(1) We will establish the inequality given∑x2k =

∑y2j = 1.

(2) We will extend it to general x and y.

Let’s get started! We first need to show thatn∑k=1

xkyk ≤ 1

for all x, y ∈ Rn such that ‖x‖2 = ‖y‖2 = 1; i.e., such that∑x2k =∑

y2j = 1. To prove this we recall that the square of any real number

HOMOGENEITY AND INEQUALITIES 3

cannot be a negative number. Therefore,

(xk − yk)2 ≥ 0

for each k. Expanding the square and rearranging,

2xkyk ≤ x2k + y2

k

After adding all of the inequalities together we apply our hypothesis,

2n∑k=1

xkyk ≤n∑k=1

x2k +

n∑k=1

y2k ≤ 2

Dividing by 2, we deduce thatn∑k=

xkyk ≤ 1

as required.Now we will show that this is sufficient to prove the inequality for all

possible x, y ∈ Rn. The inequality is obvious for zero x or y, so we canassume that both of the vectors are nonzero. If we write x = x/‖x‖2and y = y/‖y‖2 we can rearrange the left hand side,

n∑k=1

xkyk = ‖x‖2‖y‖2n∑k=1

xkyk

The theorem is true for unit-length x and y, so

≤ ‖x‖2‖y‖2

=

√√√√ n∑k=1

x2k

√√√√ n∑k=1

y2k

as promised.This proof can be imitated to prove analogous results for infinite

series and integrals; for example, given square-integrable f and g,∫f(x)g(x) dx ≤

√∫f(x)2 dx

√∫g(x)2 dx

Although we decided to look at the Euclidean norm in this example,we could have just as easily chosen to require that |x1|+ · · ·+ |xn| = 1or any other convenient scaling factor. The form we chose was tailoredto the Cauchy-Schwarz inequality.