Homogeneous Functions
A function f : Rn → R is said to be homogeneous of degree k if
f (t~x) = tk f (~x) for any scalar t. The following result is one of many due to
Euler.
Theorem 1. Suppose f : Rn → R is continuously differentiable on Rn . Then
f is homogeneous of degree k if and only if
kf (~x) =
n
X
∂f (~x)
i=1
∂xi
xi
(1)
for all ~x ∈ Rn .
Proof. For any ~x ∈ Rn , let g~x (t) = f (t~x) − tk f (~x). Note that g~x = 0 for all
~x exactly when f is homogeneous of degree k. Suppose that equation (1) is
true. By the chain rule,
n
dg~x (t) X ∂f (t~x)
=
xi − ktk−1 f (~x)
dt
∂xi
i=1
1
=
t
n
X
∂f (t~x)
i=1
∂xi
!
k
txi − kt f (~x)
1
kf (t~x) − ktk f (~x)
t
k
= g~x (t).
t
=
So g~x0 (t) = kt g~x (t), which means that y(t) = g~x (t) satisfies the first-order
differential equation dy
= kyt . This equation has general solution y = Ctk , so
dt
g~x (t) = Ctk . But g~x (1) = f (~x)−f (~x) = 0, so C = 0 and thus g~x is identically
zero; hence f is homogeneous of degree k.
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Now suppose that f is homogeneous of degree k. Then g~x (t) is identically
zero, so g~x0 (t) = 0. But as above we have
g~x0 (t)
=
n
X
∂f (t~x)
∂xi
i=1
so
n
X
∂f (t~x)
i=1
∂xi
xi − ktk−1 f (~x),
xi = ktk−1 f (~x).
Set t = 1 to get equation (1).
Corollary 1. If f is homogeneous of degree k, then each partial derivative
∂f
is homogeneous of degree k − 1.
∂xj
Proof. Suppose f is homogeneous of degree k. Then equation (1) is true,
and we can differentiate both sides with respect to xj to get
n
X ∂ 2f
∂f
∂f
=
xj +
,
k
∂xj
∂xj ∂xi
∂xj
i=1
where we have used the fact that
( ∂2f
xi ,
∂
∂f
j ∂xi
xi = ∂x
∂2f
∂xj ∂xi
x +
∂2x j
j
(2)
if i 6= j,
∂f
,
∂xj
if i = j.
Now equation (2) is equivalent to
n
X
∂ 2f
∂f
xi = (k − 1)
,
∂xi ∂xj
∂xj
i=1
which by Theorem 1 implies that
∂f
∂xj
(3)
is homogeneous of degree k − 1.
A homogeneous polynomial of degree k is a polynomial in which
each term has degree k, as in
 
x
f y  = 2x2 y + 3xyz + z 3 .
z
2
A homogeneous polynomial of degree k is a homogeneous function of degree
k, but there are many homogenous functions that are not polynomials. For
example,
x3 + x2 y + xy 2 + y 3
x2 + y 2
p
is homogeneous of degree 1, as is x2 + y 2 . Also, to say that g is homogeneous of degree 0 means g(t~x) = g(~x), but this doesn’t necessarily mean g is
constant: for example, consider
x2 − y 2
x
.
g
= 2
y
x + y2
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Lagrange Multipliers
Now let f : Rn → R be homogeneous of degree k. Suppose we want to find
the maximum or minimum of f subject to a linear constraint c1 x1 + c2 x2 +
· · · + cn xn = M . Lagrange’s equations are
∂f
= λci ,
∂xi
i = 1, 2, . . . , n.
Multiply by xi and sum on i to get
n
X
i=1
n
X
∂f
xi
=λ
ci x i
∂xi
i=1
Now by Euler’s theorem the left-hand side is kf , and the constraint equation
says the right-hand side is λM . So λ = kf /M , and we have
∂f
kf (~x)
ci ,
=
∂xi
M
i = 1, 2, . . . , n.
(4)
Example 1. For example, suppose we want to minimize
f (x1 , x2 , x3 , x4 ) = x21 + x22 + x23 + x24
subject to the constraint 2x1 − x2 + 4x3 + 5x4 = 20. Then f is homogeneous
of degree 2, and the equations (4) are
2x1 =
2f (~x)
2,
20
2x2 =
2f (~x)
(−1),
20
3
2x3 =
2f (~x)
4,
20
2x4 =
2f (~x)
5.
20
Then the constraint gives
f (~x)
(4 + 1 + 16 + 25) = 20,
20
= 200
, and this implies x1 = 20
,
so the minimum value is f (~x) = 20·20
46
23
23
10
40
50
x2 = − 23 , x3 = 23 , and x4 = 23 .
Example 2. Suppose we want to maximize the Cobb-Douglas production
function
Y (K, L) = AK 1−α Lα
subject to the constraint wL + pK = M . Since Y is homogeneous of degree
1, the equations (4) are
(1 − α)K −α Lα =
or
AK 1−α Lα
AK 1−α Lα
p and αK 1−α Lα−1 =
w,
M
M
1−α
p
=
K
M
and
α
w
=
,
L
M
that is,
K=
2
(1 − α)M
p
and L =
αM
.
w
Homogeneous Functions and Scaling
The degree of a homogenous function can be thought of as describing how the
function behaves under change of scale. In thermodynamics all important
quantities are either homogeneous of degree 1 (called extensive, like mass, energy and entropy), or homogeneous of degree 0 (called intensive, like density,
temperature and specific heat).
In economics, the Cobb-Douglas production function
Y (K, L) = AK 1−α Lα
gives output Y as a function of capital investment K and labor investment L.
(The exponent α is empirically determined: it’s usually taken to be between
2
and 34 .) As noted above, Y is homogeneous of degree 1:
3
Y (tK, tL) = At1−α K 1−α tα Lα = tY (K, L).
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Another way of thinking about this fact is the following: if we increase both
capital and labor investments by some proportion (say 10%), the output Y
increases by the same proportion. Economists call this “constant returns to
scale”.
We note an immediate implication of constant returns to scale. If we look
at the per capita output Y /L, we have
LY K
,
1
K
Y (K, L)
L
=
=Y
,1 ,
L
L
L
that is, per capita output depends only on the quantity of capital employed
by each worker.
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Exercises
1. Give an example of a homogeneous function of degree −1. (Suggestion:
take a partial derivative of a nonconstant function of degree 0).
2. Show that, if f is homogeneous of degree 1, then the Hessian Hf (~x) is
degenerate at every ~x 6= ~0.
3. Maximize x3 y 2 z on the plane x + 2y + 3z = 6.
4. The constant elasticity of substitution (CES) production function is
1
Q(K, L) = c[(1 − α)K r + αLr ] r ,
where c, α, and r are constants. Show that the CES function is homogeneous
of degree 1.
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