# Understanding the proof using the mean value theorem 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.

• 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 Jan 30 '19 at 21:32
• @Bertrand Thanks for a comment. Please leave an answer if you see whats going on. – shk910 Jan 30 '19 at 23:37