Let's focus on the following version of Cauchy's Mean Value Theorem:

Cauchy's Mean Value Theorem: Let $f, g$ be functions defined on closed interval $[a, b]$ such that

1) Both $f, g$ are continuous on $[a, b]$

2) Both $f, g$ are differentiable on $(a, b)$

3) $g'$ does not vanish at any point in $(a, b)$

then there is a point $c \in (a, b)$ such that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$

In most good textbooks it is mentioned that this theorem can't be derived from the usual Mean Value Theorem. Using MVT we can get $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{(f(b) - f(a))/(b - a)}{(g(b) - g(a))/(b - a)} = \frac{f'(c_{1})}{g'(c_{2})}$$ and we can't guarantee that $c_{1} = c_{2}$ and hence the above strategy to prove Cauchy's theorem fails.

However I beg to differ and I provide the following argument. The geometrical interpretation of MVT is that if we take any two points on a continuous and smooth curve (with some further restrictions on nature of curve) then between this portion of the curve there is a point at which the tangent to the curve is parallel to the chord joining these two points. Cauchy's theorem is obtained at once if we use the parametric equation of the curve with $x = g(t), y = f(t), dy/dx = f'(t)/g'(t)$ for $t \in (a, b) $ and the endpoints of the curve are $(g(a), f(a)), (g(b), f(b))$. Hence Cauchy's Theorem is derivable from MVT.

The rigorous proof proceeds as follows. Since $g'$ does not vanish it is clear from Rolle's theorem that $g$ is one one function so that $g(a) \neq g(b)$. If $g(a) < g(b)$ then we can show that $g$ is strictly increasing. Let $a \leq c < d \leq b$. Clearly $g(c) \neq g(d) \neq g(a)$. Let us suppose $g(d) < g(a)$ so that $g(d) < g(a) < g(b)$ and hence by IVT there will be a point $x$ in $(d, b)$ with $g(x) = g(a)$ contrary to the fact that $g$ is one one. So we must have $g(a) < g(d)$. Similarly $g(c) < g(b)$. Now suppose that $g(c) > g(d)$ so that $g(a) < g(d) < g(c)$ and by IVT there is an $x \in (a, c)$ with $g(x) = g(d)$ contrary to the fact that $f$ is one one. Hence $g(c) < g(d)$. We have thus shown that $g$ is strictly increasing. If $g(a) > g(b)$ then we can show in same manner that $g$ is strictly decreasing.

Hence by inverse function theorem $h = g^{-1}$ exists and is continuous and differentiable. Suppose $g$ is increasing and let $A = g(a), B = g(b)$ so that $a = h(A), b = h(B)$ and consider the function $F(x) = f(h(x))$ on $[A, B]$. By MVT there is a $C \in (A, B)$ such that $$F(B) - F(A) = (B - A)F'(C)$$ or $$f(h(B)) - f(h(A)) = (g(b) - g(a))f'(h(C))h'(C)$$ If $h(C) = c$ then $c \in (a, b)$ and $h'(C) = 1/g'(c)$ and we get $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$

I would like to get feedback on this proof above. I haven't found any issue with the argument and think it is perfectly valid.

  • $\begingroup$ Looks nice, although if you would spell out the first part of the argument ($g$ is strictly monotonic), the "usual" argument would win in brevity. $\endgroup$ – PhoemueX Aug 1 '15 at 7:06
  • $\begingroup$ There is nothing wrong with the argument math.stackexchange.com/questions/1290172/… $\endgroup$ – A.Γ. Aug 1 '15 at 7:20
  • $\begingroup$ @PhoemueX: Edited my post to include proof of monotone nature of $g$. $\endgroup$ – Paramanand Singh Aug 1 '15 at 7:22
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    $\begingroup$ @A.G.: I checked the linked question and it appears that it repeats the same geometric arguments which I have mentioned in my post. I have tried to make that argument more rigorous by using properties of continuous/differentiable functions. A geometric intuition can't be substitute for a rigorous proof. $\endgroup$ – Paramanand Singh Aug 1 '15 at 7:23
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    $\begingroup$ Your approach is legal: it is found in T. Flett Differential analysis (1980), pp.16-17 . Flett uses this argument to show that Cauchy's theorem is essentially a particular case of Lagrange's ! $\endgroup$ – Tony Piccolo Aug 1 '15 at 14:40

Or we could just consider the function:


Clearly $F(x)$ is defined iff $g(b) $ is different from $g(a)$ , otherwise if $g(a)=g(b)$ then $F(x)$ would satisfy Role's theorem's conditions hence there would exist a point $c$ $ \epsilon$ $(a,b)$ such that $g'(c)=0$ this is a contradiction to our third condition :

3) $g'$ does not vanish at any point in $(a, b)$

Then from the MVT we have :

There exists a point $c$ $\epsilon$ $(a,b)$ such that $F'(c)=0$

thus we get :




  • $\begingroup$ This is the standard proof which most textbooks present. My question was not about it, but rather about the validity of the alternative proof given in my post. $\endgroup$ – Paramanand Singh Mar 26 '18 at 5:29

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