<p>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 <span class="math-container">$w$</span>.</p>

CC BY-SA 4.0
2021-02-28T18:45:34.140Z
Good point! We can assume that $f$ is strictly quasi-concave instead.
user_lambda
CC BY-SA 4.0
2021-02-28T18:47:57.677Z
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.
Bertrand
CC BY-SA 4.0
2021-02-28T16:49:59.837Z
2021-02-28T16:49:59.837Z
Bertrand
<p>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.</p>
<p>So let <span class="math-container">$f:\mathbb{R}^n_+\to\mathbb{R}_+$</span> be a production function. We let <span class="math-container">$\Delta^n_{++}$</span> be the set of all points in <span class="math-container">$\mathbb{R}^n$</span> with all coordinates being strictly positive and summing to one. It is possible to have a cost-function that is strictly concave on <span class="math-container">$\Delta^n_{++}$</span>.</p>
<p><strong>Proposition:</strong> Suppose <span class="math-container">$f$</span> is differentiable and strictly increasing on <span class="math-container">$\mathbb{R}_{++}$</span>, that <span class="math-container">$f(z)\geq 1$</span> holds for no <span class="math-container">$z$</span> on the boundary of <span class="math-container">$\mathbb{R}_+$</span>, and that <span class="math-container">$f(z)\geq 1$</span> for some <span class="math-container">$z$</span>. Then the function <span class="math-container">$c:\Delta^n_{++}\to\mathbb{R}_{++}$</span> given by <span class="math-container">$$c(w) = \min_{z\geq 0} w\cdot z$$</span>
subject to <span class="math-container">$f(z)\geq 1$</span> is well-defined and strictly concave.</p>
<p><strong>Proof:</strong> That <span class="math-container">$c$</span> is well-defined is straightforward. By assumption, any <span class="math-container">$z$</span> that solves the minimization problem for some <span class="math-container">$w$</span> must be strictly positive. Since <span class="math-container">$f$</span> is differentiable and increasing, the gradient of <span class="math-container">$f$</span> at <span class="math-container">$z$</span> must be collinear with <span class="math-container">$w$</span>, In particular, no point is a minimizer for more than one <span class="math-container">$w$</span>.</p>
<p>Now let <span class="math-container">$w,w'$</span> be two different elements of <span class="math-container">$\Delta^n_{++}$</span> and <span class="math-container">$0&lt;\alpha&lt;1$</span>. Let <span class="math-container">$z$</span> solve the minimization problem at <span class="math-container">$w$</span>, let <span class="math-container">$z'$</span> solve the minimization problem at <span class="math-container">$w'$</span>, and let <span class="math-container">$z''$</span> solve the minimization problem at <span class="math-container">$\alpha w+(1-\alpha)w'$</span>. By what we have shown before, <span class="math-container">$w\cdot z''&gt;w\cdot z$</span> and <span class="math-container">$w'\cdot z''&gt;w'\cdot z'$</span>. Therefore,
<span class="math-container">$$c(w'')=\alpha w+(1-\alpha)w'\cdot z''&gt;\alpha w\cdot z+(1-\alpha)w'\cdot z'=\alpha c(w)+(1-\alpha)c(w').$$</span>
<span class="math-container">$\square$</span></p>
<p>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.</p>

CC BY-SA 4.0
2021-02-28T18:44:14.753Z
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?
user_lambda
CC BY-SA 4.0
2021-02-28T19:05:25.460Z
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.
Michael Greinecker
CC BY-SA 4.0
2021-02-28T19:07:19.530Z
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).
user_lambda
CC BY-SA 4.0
2021-02-28T18:14:42.307Z
2021-02-28T18:14:42.307Z
Michael Greinecker
<p>Define the unit cost function as
<span class="math-container">$$
c(w) = \min_{z\geq 0} w\cdot z
$$</span>
subject to <span class="math-container">$f(z)\geq 1$</span>. Where <span class="math-container">$w$</span> is a vector of input prices, <span class="math-container">$z$</span> is the vector of inputs and <span class="math-container">$f$</span> is a production function. We know from standard results that <span class="math-container">$c$</span> is concave and homogenous of degree one, but I would like to establish conditions on <span class="math-container">$f$</span> such that <span class="math-container">$c$</span> is also strictly concave.</p>

<microeconomics><mathematical-economics><production-function><cost-functions><proof>
Under what condition is a cost function strictly concave in prices?
user_lambda
# Under what condition is a cost function strictly concave in prices?

(Asked by: username_0 on 2021-02-27)

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.

**Tags:** microeconomics, mathematical-economics, production-function, cost-functions, proof

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## 2 answers

### Answer by username_2 on 2021-02-28. Score: 5 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_2**: 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.  
> **username_0**: 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).  

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### Answer by username_1 on 2021-02-28. Score: 7

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 $w$.

> **Comments:**
> **username_0**: Good point! We can assume that $f$ is strictly quasi-concave instead.  
> **username_1**: 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.  

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