Define the unit cost function as $$ c(w) = \min_{z\geq 0} w\cdot z $$ subject to $f(z)\geq 1$. Where $w$ is a vector of input prices, $z$ is the vector of inputs and $f$ is a production function. We know from standard results that $c$ is concave and homogenous of degree one, but I would like to establish conditions on $f$ such that $c$ is also strictly concave.


2 Answers 2


As Bertrand pointed out, strict-concavity will necessarily fail along any rays through the origin. But one can have strict concavity for normalized price systems.

So let $f:\mathbb{R}^n_+\to\mathbb{R}_+$ be a production function. We let $\Delta^n_{++}$ be the set of all points in $\mathbb{R}^n$ with all coordinates being strictly positive and summing to one. It is possible to have a cost-function that is strictly concave on $\Delta^n_{++}$.

Proposition: Suppose $f$ is differentiable and strictly increasing on $\mathbb{R}_{++}$, that $f(z)\geq 1$ holds for no $z$ on the boundary of $\mathbb{R}_+$, and that $f(z)\geq 1$ for some $z$. Then the function $c:\Delta^n_{++}\to\mathbb{R}_{++}$ given by $$c(w) = \min_{z\geq 0} w\cdot z$$ subject to $f(z)\geq 1$ is well-defined and strictly concave.

Proof: That $c$ is well-defined is straightforward. By assumption, any $z$ that solves the minimization problem for some $w$ must be strictly positive. Since $f$ is differentiable and increasing, the gradient of $f$ at $z$ must be collinear with $w$, In particular, no point is a minimizer for more than one $w$.

Now let $w,w'$ be two different elements of $\Delta^n_{++}$ and $0<\alpha<1$. Let $z$ solve the minimization problem at $w$, let $z'$ solve the minimization problem at $w'$, and let $z''$ solve the minimization problem at $\alpha w+(1-\alpha)w'$. By what we have shown before, $w\cdot z''>w\cdot z$ and $w'\cdot z''>w'\cdot z'$. Therefore, $$c(w'')=\alpha w+(1-\alpha)w'\cdot z''>\alpha w\cdot z+(1-\alpha)w'\cdot z'=\alpha c(w)+(1-\alpha)c(w').$$ $\square$

It should be noted that one needs an assumption that guarantees an interior solution. If some input is not used at all, raising its price will make the same input combination cost-minimizing, and along such price changes, strict concavity will be necessarily violated. Without the differentiability assumption, it can happen that the same bundle is cost-minimizing for several prices and then for any convex combinations thereof. This, too, would lead to a violation of strict concavity.

  • $\begingroup$ Thank you this is very useful but I'm confused about something. You do not assume that $f$ is linearly homogeneous, which would clash with strict concavity I guess. Maybe strict quasiconcavity + linearly homogenous would do the trick? $\endgroup$ Commented Feb 28, 2021 at 18:44
  • $\begingroup$ I do allow $f$ to be homogenous. The problem is that the cost function itself is homogenous on $\mathbb{R}_+$, and that is not compatible with being strictly concave. If I pick some $w\gg 0$, then the cost function is essentially linear on the set $\{\lambda w\vert \lambda>0\}$ and thus not strictly concave. But this problem does not occur with normalized prices. $\endgroup$ Commented Feb 28, 2021 at 19:05
  • $\begingroup$ Sorry I got confused, I was thinking of a linearly homogenous production function, which is not allowed here (but this is not part of my statement). $\endgroup$ Commented Feb 28, 2021 at 19:07

No function that is homogeneous of degree one, is at the same time strictly concave in its arguments. If the function is differentiable (or non-differentiable at a finite number of points), then the Hessian of a linear homogeneous function is singular. So if you want to end up with a unit cost function that is strictly concave, you have to drop at the same time the linear homogeneity in $w$.

  • $\begingroup$ Good point! We can assume that $f$ is strictly quasi-concave instead. $\endgroup$ Commented Feb 28, 2021 at 18:45
  • $\begingroup$ If $f$ is strictly quasi-concave, this does not change anything in the above statement, the hessian of the cost function wrt $w$ is still singular. $\endgroup$
    – Bertrand
    Commented Feb 28, 2021 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.