Homogeneity and Euler's Theorem

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Let us start with a definition:

Homogeneity: Let ¦ :Rⁿ ® R be a real-valued function. Then ¦ (x₁, x₂ ...., x_n) is homogeneous of degree k if l^k¦(x) = ¦(l x) where l ³ 0 (x is the vector [x₁...x_n]).

In other words, a function is called homogeneous of degree k if by multiplying all arguments by a constant scalar l , we increase the value of the function by l^k, i.e.

l^k¦(x₁, x₂,..., x_n) = ¦(lx₁, lx₂,...., lx_n)

If k = 1, we call this a linearly homogenous function.  If we interpret ¦(x) as a production function, then k = 1 implies constant returns to scale (as l^k = l), k > 1 implies increasing returns to scale (as l^k > l) and if 0 < k < 1, then we have decreasing returns to scale (as l^k < l).

Phillip Wicksteed (1894) stated the "product exhaustion" thesis implied by the marginal productivity theory of distribution - namely, that if all agents were paid their marginal product, then total costs would exhaust the entire product. Wicksteed assumed constant returns to scale - and thus employed a linear homogeneous production function, a function which was homogeneous of degree one.  It was A.W. Flux (1894) who pointed out that Wicksteed's "product exhaustion" thesis was merely a restatement of Euler's Theorem. Euler’s Theorem states that under homogeneity of degree 1, a function ¦ (x) can be reduced to the sum of its arguments multiplied by their first partial derivatives, in short:

Theorem: (Euler's Theorem) Given the function ¦ :Rⁿ ® R, then if ¦ is positively homogeneous of degree 1 then:

¦ (x₁, x₂, ...., x_n) = x₁ [¶ ¦ /¶ x₁] + x₂ [¶ ¦ /¶ x₂] + ...... + x_n [¶ ¦ /d¶x_n]

or simply:

¦ (x) = å ⁿ_i=1 [d¦ (x)/dx_i]·x_i

Proof: By definition of homogeneity of degree k, letting k = 1, then l ¦ (x) = ¦ (l x) where x is a n-dimensional vector and l is a scalar. Differentiating both sides of this expression with respect to x_i and using the chain rule, we see that:

[¶ l ¦ (x)/¶ ¦ (x)]·[¶ ¦ (x)/¶ x_i] = [¶ ¦ (l x)/¶ (l x_i)]·[¶ (l x_i)/¶ x_i]

as [¶ l ¦ (x)/¶ ¦ (x)] = l and ¶ (l x_i)/¶ x_i = l then l [¶ ¦ (x)/¶ x_i] = [¶ ¦ (l x)/¶ (l x_i)]l then:

¶ ¦ (x)/¶ x_i = ¶ ¦ (l x)/¶ (l x_i)                                                         (E.1)

Now, differentiating both sides of the original expression l ¦ (x) = ¦ (l x) with respect to l , we get:

¶ l ¦ (x)/¶ l = å ⁿ_i=1[¶ ¦ (l x)/¶ (l x_i)]·[¶ (l x_i)/¶ l ]

As ¶ l ¦ (x_i)/¶ l = ¦ (x_i) and ¶ (l x_i)/¶ l = x_i for all i = 1,..., n, then this expression reduces to:

¦ (x) = å ⁿ_i=1[¶ ¦ (l x)/¶ (l x_i)]·x_i

Now using the equality in (E.1), we can substitute ¶ ¦ (x)/¶ x_i for ¶ ¦ (l x)/¶ (l x_i). Thus, this becomes:

¦ (x) = å ⁿ_i=1[¶ ¦ (x)/¶ x_i]·x_i

which is Euler’s Theorem.§

One of the interesting results is that if ¦(x) is a homogeneous function of degree k, then the first derivatives, ¦_i(x), are themselves homogeneous functions of degree k-1.  So, for the homogeneous of degree 1 case, ¦ _i(x) is homogeneous of degree zero.  Consequently, there is a corollary to Euler's Theorem:

Corollary: if ¦ :Rⁿ ® R is homogenous of degree 1, then å ⁿ_i=1[¶²¦(x)/¶ x_i¶x_j]x_i = 0 for any j.

Proof: By Euler’s Theorem, ¦ (x) = å ⁿ_i=1[¶ ¦ (x)/¶ x_i]·x_i . Differentiating with respect to x_j yields:

¶ ¦ (x)/¶ x_j = [¶ ²¦ (x)/¶ x₁¶x_j]x₁ + ..... + [¶ ²¦ (x)/¶ x_j¶x_j]x_j + ¶ ¦ (x)/¶ x_j + ..... + [¶ ²¦ (x)/¶ x_n¶x_j]x_n

or rewriting:

¶ ¦ (x)/¶ x_j = å ⁿ_i=1[¶ ²¦ (x)/¶ x_i ¶x_j]x_i + ¶ ¦ (x)/¶ x_j

where, note, the summation expression sums from all i from 1 to n (including i = j). Nonetheless, note that the expression on the extreme right, ¶ ¦ (x)/¶ x_j appears on both sides of the equation. Thus:

å ⁿ_i=1[¶ ²¦ (x)/¶ x_i¶x_j]x_i = 0

which is what we sought.§

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