<p>The economic intuition here is clear: if the single-input production function is concave, the marginal product is diminishing. So for equal increases of the single input (and given fixed unit cost), say unit per unit, production increases less and less, while cost increases equally. Flip it, and then for sequential unit increases in output, cost increases more and more per incremental output unit. This is the primitive description of a strictly positive second derivative, which is the condition for a strictly convex function.  </p>

CC BY-SA 3.0
2015-10-09T00:35:47.917Z
2015-10-09T00:35:47.917Z
Alecos Papadopoulos
<p>Given the fixed input price $w$, the cost function can be written as
$$
C(q)=f^{-1}(q)\times w
$$
where $f^{-1}$ is the inverse of the production function $f$. From the discussion <a href="https://proofwiki.org/wiki/Inverse_of_Strictly_Increasing_Convex_Real_Function_is_Concave" rel="nofollow noreferrer">here</a>, one can conclude that the inverse of a concave strictly increasing function is convex. Thus, $C(q)$ is convex as well. </p>

<p>Going back to your approach, you might like to have this clearly stated. Let $q''=\alpha q + (1-\alpha)q'$, $f(z)=q$, $f(z')=q'$, and $f(z'')=q''$ Then 
$$
\begin{align}
f(z'')&amp;=&amp;q''\\
&amp;=&amp;(\alpha q + (1-\alpha)q')\\
&amp;=&amp;\alpha f(z)+(1-\alpha) f(z')\\
&amp;\leq&amp;f(\alpha z +(1-\alpha z'))
\end{align}
$$
Then $f(z'')\leq f(\alpha z +(1-\alpha z'))$ implying that $\alpha z +(1-\alpha z')\geq z''$ since $f$ is (strictly) increasing. Hence, 
$$
z''w=C(q'')\leq \alpha C(q)+ (1-\alpha) C(q')
$$</p>

CC BY-SA 3.0
2015-10-08T22:11:11.523Z
Not exactly what I asked for but a very nice proof. Thanks for putting this up.
123
CC BY-SA 3.0
2015-10-08T22:27:09.900Z
@mathtastic sorry, I made some changes.
ramazan
CC BY-SA 3.0
2017-10-14T05:05:32.920Z
But isn't $f$ here only the **left** inverse of $C$? It does work as an inverse on the image, but not for arbitrary production plan that is not in the image? Since it describes the minimal costs for attaining given level of output (but there may be other, worse, plans that attain the same output). And then "the inverse of a concave strictly increasing function is convex" does not apply, does it?
kitsune
CC BY-SA 3.0
2015-10-08T21:59:46.037Z
2017-10-30T07:13:06.700Z
2017-10-30T07:13:06.700Z
ramazan
<p>Let's assume we have an increasing production function $f:\mathbb{R^+} \to \mathbb{R^+}$</p>

<p>Now, assume this production function is concave and that the price of input z is fixed (this is a single-input and single output case). I want to show this implies the corresponding cost function $C^f(w,q)$ is convex. </p>

<p>My thoughts: </p>

<p>let $z,z' \in \mathbb{R^+}$ where W.L.O.G $z'&gt;z$ and let $f(z)=q$ and $f(z')=q$</p>

<p>Since $f$ is concave, we take $\alpha \in [0,1]$  $s.t$:</p>

<p>$$f(\alpha z +(1- \alpha)z') \geq \alpha f(z) +(1- \alpha)f(z')$$ </p>

<p>let $\alpha z +(1- \alpha)z'=z''\in \mathbb{R^+}$ where by necessity $z\leq z'' \leq z'$ </p>

<p>Since cost of $z=w$ is fixed at some $w \in \mathbb{R^+}$</p>

<p>we can rewrite our cost function as $C^f(q)$ which is convex if for $\alpha \in [0,1]$:</p>

<p>$$C(\alpha q +(1- \alpha)q') \leq \alpha c(q) +(1- \alpha)c(q')$$</p>

<p>Now, since $f$ is concave, it cannot be the case that $f$ experiences increasing returns to scale. Then $f$ has constant returns to scale or diminishing returns to scale. </p>

<p>Then:</p>

<p>Since $z''$ is clearly feasible then we have that:</p>

<p>$$C(w,q'') \leq w*z''$$ 
$$=\alpha w*z + (1- \alpha) w*z'$$
$$= \alpha c(w,q) + (1- \alpha) c(w,q')$$ </p>

<p>Does this look correct? </p>

<production-function><proof-reference>
Concave production function implies convex cost function
123
# Concave production function implies convex cost function

(Asked by: username_0 on 2015-10-08)

Let's assume we have an increasing production function $f:\mathbb{R^+} \to \mathbb{R^+}$

Now, assume this production function is concave and that the price of input z is fixed (this is a single-input and single output case). I want to show this implies the corresponding cost function $C^f(w,q)$ is convex.

My thoughts:

let $z,z' \in \mathbb{R^+}$ where W.L.O.G $z'>z$ and let $f(z)=q$ and $f(z')=q$

Since $f$ is concave, we take $\alpha \in [0,1]$ $s.t$:

$$f(\alpha z +(1- \alpha)z') \geq \alpha f(z) +(1- \alpha)f(z')$$

let $\alpha z +(1- \alpha)z'=z''\in \mathbb{R^+}$ where by necessity $z\leq z'' \leq z'$

Since cost of $z=w$ is fixed at some $w \in \mathbb{R^+}$

we can rewrite our cost function as $C^f(q)$ which is convex if for $\alpha \in [0,1]$:

$$C(\alpha q +(1- \alpha)q') \leq \alpha c(q) +(1- \alpha)c(q')$$

Now, since $f$ is concave, it cannot be the case that $f$ experiences increasing returns to scale. Then $f$ has constant returns to scale or diminishing returns to scale.

Then:

Since $z''$ is clearly feasible then we have that:

$$C(w,q'') \leq w\*z''$$
$$=\alpha w\*z + (1- \alpha) w\*z'$$
$$= \alpha c(w,q) + (1- \alpha) c(w,q')$$

Does this look correct?

**Tags:** production-function, proof-reference

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## 2 answers

### Answer by username_2 on 2015-10-08. Score: 7 | Accepted answer

Given the fixed input price $w$, the cost function can be written as
$$
C(q)=f^{-1}(q)\times w
$$
where $f^{-1}$ is the inverse of the production function $f$. From the discussion [here](https://proofwiki.org/wiki/Inverse_of_Strictly_Increasing_Convex_Real_Function_is_Concave), one can conclude that the inverse of a concave strictly increasing function is convex. Thus, $C(q)$ is convex as well.

Going back to your approach, you might like to have this clearly stated. Let $q''=\alpha q + (1-\alpha)q'$, $f(z)=q$, $f(z')=q'$, and $f(z'')=q''$ Then
$$
\begin{align}
f(z'')&=&q''\\
&=&(\alpha q + (1-\alpha)q')\\
&=&\alpha f(z)+(1-\alpha) f(z')\\
&\leq&f(\alpha z +(1-\alpha z'))
\end{align}
$$
Then $f(z'')\leq f(\alpha z +(1-\alpha z'))$ implying that $\alpha z +(1-\alpha z')\geq z''$ since $f$ is (strictly) increasing. Hence,
$$
z''w=C(q'')\leq \alpha C(q)+ (1-\alpha) C(q')
$$

> **Comments:**
> **username_0**: Not exactly what I asked for but a very nice proof. Thanks for putting this up.  
> **username_2**: @mathtastic sorry, I made some changes.  
> **username_3**: But isn't $f$ here only the **left** inverse of $C$? It does work as an inverse on the image, but not for arbitrary production plan that is not in the image? Since it describes the minimal costs for attaining given level of output (but there may be other, worse, plans that attain the same output). And then "the inverse of a concave strictly increasing function is convex" does not apply, does it?  

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### Answer by username_1 on 2015-10-09. Score: 2

The economic intuition here is clear: if the single-input production function is concave, the marginal product is diminishing. So for equal increases of the single input (and given fixed unit cost), say unit per unit, production increases less and less, while cost increases equally. Flip it, and then for sequential unit increases in output, cost increases more and more per incremental output unit. This is the primitive description of a strictly positive second derivative, which is the condition for a strictly convex function.


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