# Cauchy's MVT-Lagrange's MVT-Rolle's theorem independence

In many textbooks, the former two have been proved with the help of Rolle's theorem. However my teacher(and many sites as well) say that Rolle's theorem is a special case of LMVT and Cauchy's is a generalization.

Can we prove Cauchy's MVT and LMVT without using Rolle's theorem? If not, should we admit that these are just applications of Rolle's theorem and hence yield no extra result?

$$\mathcal{Remark}$$

I found an analogy which could be useful:

Suppose a car is travelling at an average speed of 40 miles/hr. In the course of the travel, it has to, at some point, travel at exactly 40 miles an hour.

This is exactly what LMVT has to say.

• The MVT and Rolles' theorem are equivalent results. One can't get one from the other. Sep 2 '20 at 20:11
• @OliverDiaz en.m.wikipedia.org/wiki/Mean_value_theorem we prove it using Rolle's theorem Sep 3 '20 at 3:16
• what do you mean we? Anyway, as JL showed, you can prove the MVT without resorting to Rolle's theorem. Of course this proof, as well as the modern proof of Rolle's theorem, relay on results that were not available in the 1600's. Sep 3 '20 at 3:29
• @OliverDiaz Yes of course ;) Sep 3 '20 at 3:31

Here is a direct proof of the mean value theorem. The arguments are closed to those given in E. Goulart's Cours d'analyst mathématique, Tome I. The argument same argument can be used to prove Rolle's theorem.

Suppose $$f$$ is continous in $$[a,b]$$ and differentiable in $$(a,b)$$. Define $$\phi(x):=f(x)-\frac{f(b)-f(a)}{b-a}x$$

It is clear that $$\phi$$ is continuous in $$[a,b]$$ and so, it attains its minimum and maximum values in $$[a,b]$$.

If $$\phi$$ attains its maximum and minimum at the endpoints, then $$\phi$$ is constant in $$[a,b]$$ since $$\phi(a)=\phi(b)=\frac{f(a)b-f(b)a}{b-a}$$. Since $$\phi$$ is differentiable in $$(a,b)$$, $$\phi'(x)\equiv0$$.

If $$\phi$$ achieves its wither its maximum or its minimum at some $$c\in (a,b)$$. Since $$\phi$$ is differentiable in $$(a,b)$$, it follows that $$\phi'(c)=0$$.

All this means that

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$ for some $$c\in (a,b)$$.

• Im not satisfied with "$\phi$ attains it max and min values in $[a,b]$". In the proof using Rolle's theorem, you prove that $h(a)=h(b)$ for some definite $a,b$. Sep 2 '20 at 18:22
• @DaBoi: Every continuous function in a compact interval achieves its maximum and minimum value. If both extreme values for $\phi$ are attained at the $\{a,b\}$, say $\phi(a)=\min_{a\leq x\leq b}\phi(x)$ and $\phi(b)=\max_{a\leq x\leq b}\phi(x)$ then $\phi(a)\leq\phi(x)\leq\phi(b)$. But notice that $\phi(a)=\phi(b)$. Sep 2 '20 at 18:27
• @DatBoi: If $\max_{a\leq x\leq b}\phi(x)$ is achieved at some point $c\in (a,b)$, then $\phi'(c)=0$ (this can be easily shown by noticing that or all points $x\in (a,c)$ close enough to $c$, $\frac{\phi(x)-\phi(c)}{x-c}\geq0$ and for points in $(c,b)$ close enough to $x$, $\frac{\phi(x)-\phi(c)}{x-c}\leq0$. Passing to the limit as $x\rightarrow c$ one gets $\phi'(c)=0$. Sep 2 '20 at 18:32
• Yes. But one could also say that a local min/max exists in $(-\infty,\infty)$ Sep 2 '20 at 18:35
• @DatBoi: There could be local maxima and local minima. But there are definitively points at which the "global" minima and "global" maxima are achieved. Again this happens because $\phi$ is continuous in the compact interval $[a,b]$. it is by looking at those points that we can derives the result when the global extrema is attain at the endpoints in which case, $\phi$ is constant (by design actually). When $\phi$ is not constant (and so, both global extrema does not occur at endpoints) you can take local extrema if you want but that is not effect the validity of the argument I presented. Sep 2 '20 at 20:02