1
$\begingroup$

enter image description here

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.

|improve this question
$\endgroup$
  • $\begingroup$ 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.. $\endgroup$ – Bertrand Jan 30 at 21:32
  • $\begingroup$ @Bertrand Thanks for a comment. Please leave an answer if you see whats going on. $\endgroup$ – shk910 Jan 30 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.