<blockquote>
  <p><a href="https://i.sstatic.net/TRGaD.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TRGaD.png" alt="enter image description here"></a></p>
</blockquote>

<p>This is from Hopkins and Kornienko (2004) : Running to keep in the same place. The above proof is for the proposition 1 in this paper. I don't understand how the mean value theorem is applied in this case. </p>

<p>As far as I am concerned, the mean value theorem is <span class="math-container">$\frac{f(b) - f(a)}{b-a} = f'(c)$</span> for <span class="math-container">$c \in (a,b)$</span>. Rearranging this, it should be <span class="math-container">$f(b)-f(a) - f'(c)(b-a) = 0$</span>.</p>

<p>However,  this definition is not exactly used in this case. If we see the first application of the mean value theorem,  I think that <span class="math-container">$f'(c)$</span> is equivalent to <span class="math-container">$(V_1(x_1, z-px_1) - pV_2(x_1, z-px_1))(\alpha + G(\check{z}))$</span>. I don't understand why it is <span class="math-container">$G(\check{z})$</span> instead of <span class="math-container">$G(z)$</span>. </p>

<p>Also, I think <span class="math-container">$f(b)- f(a)$</span> is equivalent to <span class="math-container">$    V(x(z), z-px(z)) (G(z) - G(\check{z})) $</span> in this case. But, I don't understand how do this. </p>

<p>I appreciate if you give some help or hint. </p>

CC BY-SA 4.0
2019-01-30T21:32:55.033Z
Well, I confess that it is not fully obvious to me either. In the case where they set 
$a=x(z)$ and $b=x(\check z)$ it is not possible to keep the other argument 
$z$ in $V$ constant. May be should I read the paper, to see what is going on..
Bertrand
CC BY-SA 4.0
2019-01-30T23:37:19.063Z
@Bertrand Thanks for a comment. Please leave an answer if you see whats going on.
shk910
<microeconomics>
Understanding the proof using the mean value theorem
shk910
# Understanding the proof using the mean value theorem

(Asked by: username_0 on 2019-01-30)

> [![enter image description here](https://i.sstatic.net/TRGaD.png)](https://i.sstatic.net/TRGaD.png)

This is from Hopkins and Kornienko (2004) : Running to keep in the same place. The above proof is for the proposition 1 in this paper. I don't understand how the mean value theorem is applied in this case.

As far as I am concerned, the mean value theorem is $\frac{f(b) - f(a)}{b-a} = f'(c)$ for $c \in (a,b)$. Rearranging this, it should be $f(b)-f(a) - f'(c)(b-a) = 0$.

However, this definition is not exactly used in this case. If we see the first application of the mean value theorem, I think that $f'(c)$ is equivalent to $(V\_1(x\_1, z-px\_1) - pV\_2(x\_1, z-px\_1))(\alpha + G(\check{z}))$. I don't understand why it is $G(\check{z})$ instead of $G(z)$.

Also, I think $f(b)- f(a)$ is equivalent to $ V(x(z), z-px(z)) (G(z) - G(\check{z})) $ in this case. But, I don't understand how do this.

I appreciate if you give some help or hint.

**Tags:** microeconomics
> **Comments:**
> **username_1**: Well, I confess that it is not fully obvious to me either. In the case where they set  $a=x(z)$ and $b=x(\check z)$ it is not possible to keep the other argument  $z$ in $V$ constant. May be should I read the paper, to see what is going on..  
> **username_0**: @Bertrand Thanks for a comment. Please leave an answer if you see whats going on.  

---

## 0 answers

