The mean value theorem, found in every calculus textbook since the time of Cauchy (or before), says the following:

(MVT) Suppose $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}$.

An immediate corollary is the mean value inequality:

(MVI) Suppose $f : [a,b] \to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose further than $|f'| \le M$ on $(a,b)$. Then $|f(b) - f(a)| \le M |b-a|$.

MVI is in some sense a weaker statement than MVT, since for instance, analogous versions of MVI hold in higher dimensions, while analogous versions of MVT fail there. So I would not really expect that we can use MVI to easily prove MVT.

Now, every mathematically interesting consequence of the MVT I've ever seen can actually be proved using MVI. (By "mathematically interesting" I mean to exclude exercises designed specifically to illustrate the theorem, as well as statements like "there was some time at which your speedometer read exactly 100 km/hr" that really just restate the result). In particular, I think MVI can prove all the "interesting" statements in the question Applications of the Mean Value Theorem.

Are there any mathematically interesting consequences of MVT which cannot be proved from MVI?

(You can interpret "cannot" in as formal or informal a sense as you want. Anywhere from precise statements in model theory or reverse mathematics, to "I don't see any easy way to prove this from MVI".)


3 Answers 3


Consider the following proposition discussed by Hardy in A Course of Pure Mathematics.

Suppose $\lim_{x \to \infty} f(x) = L$ and $\lim_{x \to \infty} f'(x)$ exists. Then $\lim_{x \to \infty} f'(x)= 0.$

A classic proof uses the clever trick of applying L'Hospital's rule to $e^x f(x) / e^x.$

My proof would be, by the MVT there exists $\xi_x \in (x,x+1)$ such that

$$f(x+1) - f(x) = f'(\xi_x).$$


$$0 = \lim_{x \to \infty} [f(x+1) - f(x)] = \lim_{x \to \infty}f'(\xi_x) = \lim_{x \to \infty}f'(x). $$

Knowing only that $|f'(x)| \leqslant M$, since the (finite) limit exist, and $|f(x+1) - f(x)| \leqslant M$ does not appear to help.

  • 1
    $\begingroup$ Nice example. Nice proof too. $\endgroup$ Mar 26, 2016 at 6:22
  • $\begingroup$ I think I can prove this with MVI. Suppose to the contrary that $\lim_{x \to \infty} f'(x) = c \ne 0$. Consider the function $g(x) = f(x) - cx$; then $\lim_{x \to \infty} g'(x) = 0$. Thus there is some sufficiently large $x_0$ with $|g'(x)| \le \frac{|c|}{2}$ for all $x \ge x_0$. Hence by MVI, $|g(x) - g(x_0)| \le \frac{|c|}{2}|x-x_0|$ for all $x \ge x_0$. That is, $$|(f(x) - cx) - (f(x_0) - cx_0)| \le \frac{|c|}{2}|x-x_0|.$$ Dividing both sides by $x$ and letting $x \to \infty$, this yields $|c| \le \frac{|c|}{2}$ which is absurd. $\endgroup$ May 3, 2020 at 20:15

Suppose $f:[a,b]\to R$ is continuous, with $a<b,$ and $f''(x)\geq 0$ for all $x\in (a,b)$. Then whenever $a\leq x<y\leq b$ and $s\in [0,1]$ we have $$s f(x )+(1-s) f(y))\geq f(x s +y(1-s)).$$

This does not require $f'(x)$ to be bounded above or below for $x\in (a,b).$ For example, let $[a,b]=[-1,1]$ and $f(x)=-\sqrt {1-x^2}.$

To prove the assertion,let $z=x s+(1-s)y$ and suppose the assertion is false.

Apply the MVT to $f$ for $x, z$ and for $z,y,$ obtaining $z_1\in(x,z)$ and $z_2\in (z,y)$ with $(f(z)-f(x))/(z-x)=f'(z_1)$ and $(f(y)-f(z))/(y-z)=f(z_2).$

Apply the MVT to $f'$ for $z_1,z_2$ obtaining $z_3\in (z_1,z_2)$ with $(f'(z_2)-f'(z_1))/(z_2-z_1)=f''(z_3).$ The details show that $f'(z_2)<f'(z_1),$ but obviously $z_2>z_1$ so $f''(z_3)<0,$ a contradiction.

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    $\begingroup$ This is good because it is a case where $f'$ is unbounded on the interval. $\endgroup$
    – RRL
    Mar 26, 2016 at 6:24

A consequence of the MVT that cannot be proven from the MVI is Hôpital's rule. In the proof of this rule one makes explicit use of the point $c$ guaranteed by the MVT. Going back one sees that this point comes from Rolle's theorem, which is strictly for real-valued functions. As a consequence Hôpital's rule is not valid for complex or vector-valued functions.

On the other hand the MVI can be proven from scratch (see Rudin's PMA, Theorem 5.20), and is therefore valid as well for complex or vector-valued functions.


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