0
$\begingroup$

According to page 7 of this slide, "A convex production set Y implies that the associated input requirement set V(y) is convex".

How can one go about proving it?

$\endgroup$
1
  • 2
    $\begingroup$ Have you tried applying the definitions (convexity, input requirement set) directly? Could you edit the problems you encounter with that approach into the question? $\endgroup$
    – Giskard
    Oct 23, 2019 at 19:39

1 Answer 1

1
$\begingroup$

Start with definitions:

  1. Production (possibilities) set: $Y$ which you know is convex
  2. Input requirement set: $V(y)=\{\mathbf{x}:(y,−\mathbf{x})∈Y\}$

On page 7. you can see: $\mathbf{y}\in Y$ and $\mathbf{y'} \in Y$ which then implies $t\mathbf{y}+(1-t)\mathbf{y'} \in Y$.

Hint 1:

Remember that: $\mathbf{y}=(y,−\mathbf{x})$ !

What does it mean that $Y$ is a convex set?

$Y$ being a convex set means that if $\mathbf{y}=(y,−\mathbf{x}) \in Y$ and $\mathbf{y'}=(y,−\mathbf{x'}) \in Y$ then $t(y,−\mathbf{x})+(1-t)(y,−\mathbf{x'}) \in Y$ as page 7. would suggest.

Okay, but what does that actually mean?

$t(y,−\mathbf{x})+(1-t)(y,−\mathbf{x'}) \in Y \implies (ty+(1-t)y, -t\mathbf{x}-(1-t)\mathbf{x'}) \in Y \implies (y,-t\mathbf{x}-(1-t)\mathbf{x'}) \in Y $

Which is just equivalent of saying that:

$t\mathbf{x}+(1-t)\mathbf{x'} \in V(y)$ because both $\mathbf{x}$ and $\mathbf{x'}$ are in $V(y)$. (Look: second definition)

$\endgroup$

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.