<p>To see why <span class="math-container">$Y$</span> is not convex and <span class="math-container">$V(y)$</span> is convex, remember the definition of <em>convex set</em>, and then consider what are the two sets <span class="math-container">$Y$</span> and <span class="math-container">$V(y)$</span>.</p>
<p>A set (of <span class="math-container">$R^n$</span> or of a more general vector space) is <em>convex</em> if the segment that joins any two points of the set is entirely contained in the set.</p>
<p>The set <span class="math-container">$Y=\{(y, -x):y\leq x^2 \}$</span> is a <strong>portion of plane</strong>: it is the part of plane between the graph of the function <span class="math-container">$y=x^2$</span> and the <span class="math-container">$x$</span> axis, if <span class="math-container">$y\geq 0$</span> (only the negative part, if <span class="math-container">$x\geq 0$</span>).</p>
<p>It is not convex, as you can take two points such that the segment joining  them doesn't  belong entirely to it.</p>
<p><span class="math-container">$V(y)= \{x: x \geq \sqrt y \}$</span> is, instead, an interval,  a <strong>portion of line</strong>, more precisely, it is the half-line <span class="math-container">$[\sqrt y, +\infty)$</span>: the set of all <span class="math-container">$x$</span> equal or greater than <span class="math-container">$\sqrt y$</span>, for a given <span class="math-container">$y$</span>.</p>
<p>A half-line is a convex set: if you take any two points of it, the segment joining them belongs to the half-line.</p>

CC BY-SA 4.0
2023-01-01T12:05:01.323Z
2023-01-02T11:43:24.527Z
2023-01-02T11:43:24.527Z
BakerStreet
<p>The following question is from Microeconomic Analysis by Hal R Varian.</p>
<p><strong>True or false? If V(y) is a convex set, then the associated production set Y must be convex.</strong></p>
<p>The solution available says;<br/>
<em>False. There are many counterexamples. Consider the technology
generated by a production function f(x) = x^2. The production set is
Y = {(y, −x) : y ≤ x^2} which is certainly not convex, but the input requirement set is V (y) = {x : x ≥ √y} which is a convex set.</em></p>
<p>I am unable to understand how f(x) = x^2 is not convex, and V(y) is convex. Could someone please explain? Please give the intuition too :)</p>

CC BY-SA 4.0
2023-01-01T10:41:51.637Z
The production function $f$ here is convex. It is the set $Y$ that is not convex.
Michael Greinecker
<microeconomics><production-function><producer-theory><technology>
Convexity of production sets and input requirement sets
Krishnendu
# Convexity of production sets and input requirement sets

(Asked by: username_0 on 2023-01-01)

The following question is from Microeconomic Analysis by Hal R Varian.

**True or false? If V(y) is a convex set, then the associated production set Y must be convex.**

The solution available says;  
*False. There are many counterexamples. Consider the technology
generated by a production function f(x) = x^2. The production set is
Y = {(y, −x) : y ≤ x^2} which is certainly not convex, but the input requirement set is V (y) = {x : x ≥ √y} which is a convex set.*

I am unable to understand how f(x) = x^2 is not convex, and V(y) is convex. Could someone please explain? Please give the intuition too :)

**Tags:** microeconomics, production-function, producer-theory, technology
> **Comments:**
> **username_1**: The production function $f$ here is convex. It is the set $Y$ that is not convex.  

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## 1 answers

### Answer by username_2 on 2023-01-01. Score: 1 | Accepted answer

To see why $Y$ is not convex and $V(y)$ is convex, remember the definition of *convex set*, and then consider what are the two sets $Y$ and $V(y)$.

A set (of $R^n$ or of a more general vector space) is *convex* if the segment that joins any two points of the set is entirely contained in the set.

The set $Y=\{(y, -x):y\leq x^2 \}$ is a **portion of plane**: it is the part of plane between the graph of the function $y=x^2$ and the $x$ axis, if $y\geq 0$ (only the negative part, if $x\geq 0$).

It is not convex, as you can take two points such that the segment joining them doesn't belong entirely to it.

$V(y)= \{x: x \geq \sqrt y \}$ is, instead, an interval, a **portion of line**, more precisely, it is the half-line $[\sqrt y, +\infty)$: the set of all $x$ equal or greater than $\sqrt y$, for a given $y$.

A half-line is a convex set: if you take any two points of it, the segment joining them belongs to the half-line.


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