# Is “$f(k,l)$ is decreasing return to scale $\Leftrightarrow f_{ll}f_{kk}-f_{kl}^2>0$” always true?

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 $$f(tk,tl) for $$t>1$$ for all $$(k,l)$$ in the domain.

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

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

and

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

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