For the productions $f(k,l) $ that are continuously differentiable, is the proposition that

"$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2>0$"

always true, I have checked for Cobb-Douglas functions and believe it will also hold for all Constant-Elasticity fuction, can someone give me some hints or show some counter-examples?

Here we define $f(k,l)$ to be decreasing return to scale if $ f(tk,tl)<tf(k,l)$ for $t>1$ for all $(k,l)$ in the domain.


3 Answers 3


In general the statement is wrong. Here is a counterexample:

Suppose you have $f(k,l) = -k l^\beta$ with $\beta >0$ and $(k,l)\in\mathbb{R}^2_{++}$ (you can interpret $f$ as a production function for a "bad" commodity). Then you have: $$f(tk,tl) = - t^{1+\beta} k l^\beta = t^{1+\beta} f(k,l) < t f(k,l)$$ so that $f(k,l)$ is decreasing returns to scale.

Now let's evaluate $ f_{ll}f_{kk} - f^2_{kl} $. Since $f$ is linear in $k$ we have that $f_{kk}$ is zero, so it is $ f_{ll}f_{kk}$. On the other hand, we have $f_{kl} = -\beta l^{\beta-1}$ so that overall we have: \begin{equation} f_{ll}f_{kk} - f^2_{kl} = 0 - \left[ -\beta l^{\beta-1} \right]^2 = - \beta^2 l^{2(\beta -1)} <0 \end{equation}

So then, where is the trick?

The fact is that $ f_{ll}f_{kk} - f^2_{kl} > 0$ is not a sufficient condition to conclude that the hessian of $f(k,l)$ is negative semi-definite and so $f$ is concave because you also have to impose restrictions on the sign of the principal minor of order 1 of the hessian matrix.

However, provided the hessian matrix is indeed negative semi-defintie than you conclude $f$ is concave and decreasing returns to scale will follow, the statement then holds.


We examine the function $F(K,L)$ that is homogeneous of degree $\lambda < 1$. Then we have that its partial derivatives are homogeneous of degree $\lambda -1 $.

For a homogeneous function $F(K,L)$ of degree $\lambda$ it holds that

$$K\cdot F_K + L\cdot F_L = \lambda \cdot F(K,L) \tag{1}$$

Analogously for the partial derivatives we have

$$F_L:\;K\cdot F_{LK} + L\cdot F_{LL} = (\lambda -1)F_L <0 \implies K\cdot F_{LK} < -L\cdot F_{LL} \tag{2} $$


$$F_K: L\cdot F_{KL} + K\cdot F_{KK} = (\lambda -1)F_K <0 \implies L\cdot F_{KL} < -K\cdot F_{KK} \tag{3} $$

If the cross-partial is non-negative then necessarily the second partials are negative, and also both sides in the inequalities above are positive. Then multiplying per side we have

$$K\cdot F_{LK} \cdot L\cdot F_{KL} < L\cdot (-F_{LL}) \cdot K\cdot (-F_{KK})$$

$$\implies F_{KK}\cdot F_{LL} > F^2_{KL}$$

So a sufficient condition for (joint) concavity of a production function in two variables that exhibits decreasing returns to scale, is that the cross-partial derivative is non-negative.

Note that this sufficient condition related to the cross-partial guarantees also the "first step" for joint concavity, namely, that the second-order partials are negative (they alone provide for partial concavity of the production function).


For a single output production function, decreasing returns to scale does imply concavity of $f$. I believe you can find this result in Ch5 of MWG, either as a proposition or an exercise. The condition you have comes from considering the Hessian of $f$, and showing that the determinant is positive. For more on the Hessian Conditions, this is a handy reference: https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/t


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